### Saturday, November 29, 2008

## Congruence based factoring algorithm

Contrary to conventional wisdom I've found a congruence based way to factor a target composite T, which thankfully is also very simple as in challenging orthodox views I've found that simple is best.

With all positive integers, given a target composite T to be factored, first find a prime number p near to, or less than sqrt(T). Then pick k such that k = p-1. Make sure k is coprime to T.

Next you calculate positive integer 'a', with n=1, using

a^2 = (nT)(k^2)^{-1} - 1 mod p

where if it does not exist, you can change n and try again, or decrement k, and try again, where there should be a 50% probability for each try that it will exist as there are (p-1)/2 quadratic residues mod p. The ambiguity there in process has to do with the apparent equivalence of either approach.

That is the one key equation which you'll notice relies on quadratic residues, but then also you are considering nT mod p, so the issue of focusing on factoring nT has to be addressed.

Note that (k^2){-1} mod p is the modular inverse.

If you find 'a', you check ak to see if it is greater than p and if not, you decrement k, and try again.

If it is greater than p you check for a factorization into f_1 and f_2,where

f_1*f_2 = nT

with

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p.

If you've succeeded when you take ak modulo p, you will have a factor of T.

Note there is an existence condition, where you can find a value I call k_0, which is determined by finding k such that

abs(nT - (1 + a^2)k^2)

is a minimum, as a k that will factor nT non-trivially must be within floor(k_0/2p) steps from k_0 where each step is in increments of 2ap, so if you find that you do not have a solution you can check to see how close you were, which is how the focus is on nT. The k_0 result is a remarkable and surprising one where I will try to show some of the theory and the underlying equations at the end.

The algorithm will factor nT mod p, so the goal is to have it factor nT, where n is a minimum, as the algebra is using n mod p as well, and key to that I'd hypothesize is finding a small 'a' with a large k, where of course =E1k is greater than p.

So no matter what, if you get to f_1 you are factoring nT mod p, but your hope is to factor nT and it's not completely clear to me how hard it is to get that happening.

Example: Let T=119. Then p=11 is greater than sqrt(119), and trying k=10 gives

a^2 = (119)(10^2)^{-1} - 1 mod 11 = 8 mod 11,

but 8 is not a quadratic residue modulo 11, so no 'a' exists for this case. Trying now k=9, gives

a^2 = (119)(9^2)^{-1} - 1 mod 11 = 4 mod 11

so a=2 is a solution. And ak = 18, which is greater than 11, so

f_1 = ak = 18 mod 11 = 7 mod 11.

And you have a non-trivial factorization, as 7 is a factor of 119.

The equations result trivially from considering a simple system of equations:

z^2 = y^2 + nT

where n is some non-zero integer, you can introduce z = x+ak, and substitute out:

(x+ak)^2 = y^2 + nT

multiplying out and simplifying I then have

x^2 = y^2 + nT - 2axk - a^2*k^2

and you could could factor nT - 2axk - a^2*k^2 to get y, but the problem with that is that you don't know x, so I use a trick:

2ax = k + p*r.

Then the above becomes

x^2 = y^2 + nT - (1 + a^2)k^2 - k*p*r

which removes x, and that's the point, as now I can move modulo a prime p of my choice. If k = 2ax, then

x^2 = y^2 + nT - (1 + a^2)k^2.

And finally

nT - (1 + a^2)k^2 = 0 mod p.

Existence of k_0 is found by considering

x^2 = y^2 + nT - (1 + a^2)k^2 - k*p*r

as r = 0 at the correct k, as let k = k_0 + p*j, where j is an integer, then

x^2 = y^2 + nT - (1 + a^2)(k_0^2 + 2p*j*k_0 + p^2*j^2) - (k_0 + p*j)p*r

and as k_0 is constant as is x, y, a and nT, as you move k either forward or back from k_0 the p^2*j^2 term will tend to dominate so that r will tend to be negative, assuming k is positive, to counterbalance it.

It can be shown that the k_0 is the minimum value for k.

It is my hope that I've generated some interest in evaluating this congruence algorithm for factoring. According to the theory it scales well with the only additional time with increasing size of T coming from finding p and calculating 'a' which requires finding a quadratic residue modulo p.

I would appreciate comments or criticisms on this approach and apologize if it's not close to some standard form. I am cutting and pasting from several sources.

With all positive integers, given a target composite T to be factored, first find a prime number p near to, or less than sqrt(T). Then pick k such that k = p-1. Make sure k is coprime to T.

Next you calculate positive integer 'a', with n=1, using

a^2 = (nT)(k^2)^{-1} - 1 mod p

where if it does not exist, you can change n and try again, or decrement k, and try again, where there should be a 50% probability for each try that it will exist as there are (p-1)/2 quadratic residues mod p. The ambiguity there in process has to do with the apparent equivalence of either approach.

That is the one key equation which you'll notice relies on quadratic residues, but then also you are considering nT mod p, so the issue of focusing on factoring nT has to be addressed.

Note that (k^2){-1} mod p is the modular inverse.

If you find 'a', you check ak to see if it is greater than p and if not, you decrement k, and try again.

If it is greater than p you check for a factorization into f_1 and f_2,where

f_1*f_2 = nT

with

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p.

If you've succeeded when you take ak modulo p, you will have a factor of T.

Note there is an existence condition, where you can find a value I call k_0, which is determined by finding k such that

abs(nT - (1 + a^2)k^2)

is a minimum, as a k that will factor nT non-trivially must be within floor(k_0/2p) steps from k_0 where each step is in increments of 2ap, so if you find that you do not have a solution you can check to see how close you were, which is how the focus is on nT. The k_0 result is a remarkable and surprising one where I will try to show some of the theory and the underlying equations at the end.

The algorithm will factor nT mod p, so the goal is to have it factor nT, where n is a minimum, as the algebra is using n mod p as well, and key to that I'd hypothesize is finding a small 'a' with a large k, where of course =E1k is greater than p.

So no matter what, if you get to f_1 you are factoring nT mod p, but your hope is to factor nT and it's not completely clear to me how hard it is to get that happening.

Example: Let T=119. Then p=11 is greater than sqrt(119), and trying k=10 gives

a^2 = (119)(10^2)^{-1} - 1 mod 11 = 8 mod 11,

but 8 is not a quadratic residue modulo 11, so no 'a' exists for this case. Trying now k=9, gives

a^2 = (119)(9^2)^{-1} - 1 mod 11 = 4 mod 11

so a=2 is a solution. And ak = 18, which is greater than 11, so

f_1 = ak = 18 mod 11 = 7 mod 11.

And you have a non-trivial factorization, as 7 is a factor of 119.

The equations result trivially from considering a simple system of equations:

z^2 = y^2 + nT

where n is some non-zero integer, you can introduce z = x+ak, and substitute out:

(x+ak)^2 = y^2 + nT

multiplying out and simplifying I then have

x^2 = y^2 + nT - 2axk - a^2*k^2

and you could could factor nT - 2axk - a^2*k^2 to get y, but the problem with that is that you don't know x, so I use a trick:

2ax = k + p*r.

Then the above becomes

x^2 = y^2 + nT - (1 + a^2)k^2 - k*p*r

which removes x, and that's the point, as now I can move modulo a prime p of my choice. If k = 2ax, then

x^2 = y^2 + nT - (1 + a^2)k^2.

And finally

nT - (1 + a^2)k^2 = 0 mod p.

Existence of k_0 is found by considering

x^2 = y^2 + nT - (1 + a^2)k^2 - k*p*r

as r = 0 at the correct k, as let k = k_0 + p*j, where j is an integer, then

x^2 = y^2 + nT - (1 + a^2)(k_0^2 + 2p*j*k_0 + p^2*j^2) - (k_0 + p*j)p*r

and as k_0 is constant as is x, y, a and nT, as you move k either forward or back from k_0 the p^2*j^2 term will tend to dominate so that r will tend to be negative, assuming k is positive, to counterbalance it.

It can be shown that the k_0 is the minimum value for k.

It is my hope that I've generated some interest in evaluating this congruence algorithm for factoring. According to the theory it scales well with the only additional time with increasing size of T coming from finding p and calculating 'a' which requires finding a quadratic residue modulo p.

I would appreciate comments or criticisms on this approach and apologize if it's not close to some standard form. I am cutting and pasting from several sources.

## JSH: Wow, maybe factoring algorithm DOES work!

Those who know how I use the newsgroups know that I like to put out early research results to help in critiquing them. It speeds things up. For years I've worked at finding a new way to factor with so many failed approaches that I have often been nearly at my own limit, so more so than with other research I just toss things out there.

But not only hasn't anyone come back to show this approach really works badly I replied a few minutes ago to a poster who may have worked at finding a picked example with a rare case where things don't work with the basic algorithm!!! Cool. Maybe the damn thing DOES work then as it looks like he worked for a while to try and find an example where it didn't!

Here's the algorithm, as a re-cap, and then I'll talk some more about what these last few days of posting may mean:

Simply enough I found that you can solve for f_1 and f_2 when

f_1*f_2 = nT,

where T is some target composite to be factored and n is a non-zero integer control variable, modulo some prime number p, where

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p

and 'a' is related to k mod p by

a^2 = (nT)(k^2)^{-1} - 1 mod p.

You pick a prime p greater than sqrt(T), and pick a positive integer k, as close to p as you can get, like k=p-1, and then try to solve for 'a', and if you can, and also ak is greater than p, then you will have factored nT mod p, with high likelihood you factored nT itself.

Here's my pet example again to demonstrate:

Example: Let T=119. Then p=11 is greater than sqrt(119), and trying k=10, with n=1 gives

a^2 = (119)(10^2)^{-1} - 1 mod 11 = 8 mod 11,

but 8 is not a quadratic residue modulo 11, so no 'a' exists for this case. Trying now k=9, gives

a^2 = (119)(9^2)^{-1} - 1 mod 11 = 4 mod 11

so a=2 is a solution. And ak = 18, which is greater than 11, so

f_1 = ak = 18 mod 11 = 7 mod 11.

And you have a non-trivial factorization, as 7 is a factor of 119.

If that works for small numbers it should work for big numbers as there's a scaling built-in which I talk about which has to do with a number called k_0 and floor(k_0/2p) so it could be a REALLY big deal if all that is actually working because it's also polynomial time. The gist of it is that the bigger the number the bigger the prime p that you choose, and that's it. Automatically scales everything for you.

Folks, if the last few days of posting are an indicator then I may have found the world's first polynomial time factoring algorithm!

It could be a remarkable discovery where there will be no debate, no arguing, and no questions about whether or not mathematicians can decide if it's of interest or not.

My hope is that math people who have argued with me for years will show maturity in this situation, and simply congratulate me on a hard victory, which took years to accomplish, and enjoy the wonder of a simple answer where no one thought one could be.

Indications are growing that the world has its first polynomial time factoring algorithm. Wow.

But not only hasn't anyone come back to show this approach really works badly I replied a few minutes ago to a poster who may have worked at finding a picked example with a rare case where things don't work with the basic algorithm!!! Cool. Maybe the damn thing DOES work then as it looks like he worked for a while to try and find an example where it didn't!

Here's the algorithm, as a re-cap, and then I'll talk some more about what these last few days of posting may mean:

Simply enough I found that you can solve for f_1 and f_2 when

f_1*f_2 = nT,

where T is some target composite to be factored and n is a non-zero integer control variable, modulo some prime number p, where

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p

and 'a' is related to k mod p by

a^2 = (nT)(k^2)^{-1} - 1 mod p.

You pick a prime p greater than sqrt(T), and pick a positive integer k, as close to p as you can get, like k=p-1, and then try to solve for 'a', and if you can, and also ak is greater than p, then you will have factored nT mod p, with high likelihood you factored nT itself.

Here's my pet example again to demonstrate:

Example: Let T=119. Then p=11 is greater than sqrt(119), and trying k=10, with n=1 gives

a^2 = (119)(10^2)^{-1} - 1 mod 11 = 8 mod 11,

but 8 is not a quadratic residue modulo 11, so no 'a' exists for this case. Trying now k=9, gives

a^2 = (119)(9^2)^{-1} - 1 mod 11 = 4 mod 11

so a=2 is a solution. And ak = 18, which is greater than 11, so

f_1 = ak = 18 mod 11 = 7 mod 11.

And you have a non-trivial factorization, as 7 is a factor of 119.

If that works for small numbers it should work for big numbers as there's a scaling built-in which I talk about which has to do with a number called k_0 and floor(k_0/2p) so it could be a REALLY big deal if all that is actually working because it's also polynomial time. The gist of it is that the bigger the number the bigger the prime p that you choose, and that's it. Automatically scales everything for you.

Folks, if the last few days of posting are an indicator then I may have found the world's first polynomial time factoring algorithm!

It could be a remarkable discovery where there will be no debate, no arguing, and no questions about whether or not mathematicians can decide if it's of interest or not.

My hope is that math people who have argued with me for years will show maturity in this situation, and simply congratulate me on a hard victory, which took years to accomplish, and enjoy the wonder of a simple answer where no one thought one could be.

Indications are growing that the world has its first polynomial time factoring algorithm. Wow.

### Friday, November 28, 2008

## Simple integer factorization algorithm

I've found that you can solve directly for f_1 and f_2 when f_1*f_2 = T, where T is a composite to be factored.

With all positive integers:

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p

and 'a' is related to k mod p by

a^2 = (T)(k^2)^{-1} - 1 mod p

where p is what I call a helper prime.

Notice that because f_1 = ak mod p, ak must be greater than p, if f_1 is less than p, if k is coprime to T.

Note that a^{-1} mod p is the modular inverse mod p, as is (k^2)^{-1} mod p.

To get the direct solution you pick p a prime where p>sqrt(T) and pick for k a positive integer near p, coprime to p, if you accidentally have a factor in common with T then you have factored it, which I mention only because in programming the algorithm you need to check for that case.

You then find if 'a' exists for your k and if it does you check if ak is greater than p, if it is, then you solve for f_1, if not you decrement k, and try again. There is a 50% chance of success for each k that 'a' exists.

Example: Let T=119. Then p=11 is greater than sqrt(119), and trying k=10 gives

a^2 = (119)(10^2)^{-1} - 1 mod 11 = 8 mod 11,

but 8 is not a quadratic residue modulo 11, so no 'a' exists for this case. Trying now k=9, gives

a^2 = (119)(9^2)^{-1} - 1 mod 11 = 4 mod 11

so a=2 is a solution. And ak = 18, which is greater than 11, so

f_1 = ak = 18 mod 11 = 7 mod 11.

And you have a non-trivial factorization, as 7 is a factor of 119.

The algorithm you'll note is polynomial time and uses residues when conventional wisdom was that you couldn't solve for integer factors with residues, but the key here appears to be quadratic residues, where you get an interlocking mechanism.

With all positive integers:

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p

and 'a' is related to k mod p by

a^2 = (T)(k^2)^{-1} - 1 mod p

where p is what I call a helper prime.

Notice that because f_1 = ak mod p, ak must be greater than p, if f_1 is less than p, if k is coprime to T.

Note that a^{-1} mod p is the modular inverse mod p, as is (k^2)^{-1} mod p.

To get the direct solution you pick p a prime where p>sqrt(T) and pick for k a positive integer near p, coprime to p, if you accidentally have a factor in common with T then you have factored it, which I mention only because in programming the algorithm you need to check for that case.

You then find if 'a' exists for your k and if it does you check if ak is greater than p, if it is, then you solve for f_1, if not you decrement k, and try again. There is a 50% chance of success for each k that 'a' exists.

Example: Let T=119. Then p=11 is greater than sqrt(119), and trying k=10 gives

a^2 = (119)(10^2)^{-1} - 1 mod 11 = 8 mod 11,

but 8 is not a quadratic residue modulo 11, so no 'a' exists for this case. Trying now k=9, gives

a^2 = (119)(9^2)^{-1} - 1 mod 11 = 4 mod 11

so a=2 is a solution. And ak = 18, which is greater than 11, so

f_1 = ak = 18 mod 11 = 7 mod 11.

And you have a non-trivial factorization, as 7 is a factor of 119.

The algorithm you'll note is polynomial time and uses residues when conventional wisdom was that you couldn't solve for integer factors with residues, but the key here appears to be quadratic residues, where you get an interlocking mechanism.

## JSH: Stories are true and then some

The weirdness of being able to easily prove major mathematical finds and find that even that ease is used against you—I've heard it often that my proofs don't "look" like mathematical proofs which is partly why I defined mathematical proof!

(Google: "definition of mathematical proof")

As a major discoverer I'm confronted by the reality of the stories that I gave in a previous post where I have to take as normal bizarre behavior where mathematicians will literally leave the country for periods of time when confronted by my research as if in physically going SOMEWHERE they can escape the truth, and then they come back and go on with their lives as if nothing happened.

The integer factorization research I've used to show "hidden variables" is remarkable for many reasons not only because it hints at physical models far beyond anything used at this time, but because it is necessary for the next level of science and technology, so as this impasse continues the future still remains in doubt.

Imagine Sir Isaac Newton contemplating a situation like this one in his time, and what world we would have today if he had failed.

The people who are resisting this research are fighting to hold on to ideas that have failed. Kind of like Ptolemaic systems resisting Copernican.

If they can succeed they can have ingrained processes and ideas indefinitely that cannot work, no matter how "pure" they claim them to be, as they are purely wrong.

I think ultimately it is a middle class problem: like the American middle class could not fathom how badly things could get if they just decided that facts didn't matter, so who cared who was president?

The middle class view is the desired one, but it can fail in the face of challenges that require you consider the worst case.

Here the worst case to consider is the end of mathematical and scientific advancement as we know it.

The field has already been corrupted by "theory" that has predictions impossible to test by experiment, as mathematicians infect over into other fields.

The problem is that while some may feel anxious about the future of the physics field it is impossible for them to imagine its actual destruction by attitudes more akin to religion than science, so they do not do enough to stop it.

As the United States did so little to stop the end of the positive view it long held in the world because its people as a group could not see defense of that goodwill as a major value—because of their middle class complacency—until it was gone, and I don't think my country is still really aware yet of how great has been its loss.

If the physics field is lost the humanity that finally becomes aware of it will be a species in pain like nothing we have today, when it will not be a debate but a reality, when people who can solve problems do not exist, as no one has learned the true art, but instead live in the fantasy of answers where none exist, learned from the mathematicians who currently practice it.

To understand the future of your students in physics then if the worst case plays out, consider those mathematicians now who when confronted with proofs leave the country, as if that mattered, or simply just go on about their business, continuing research in areas where that research is no longer needed because they are lost as researchers.

They invest time and energy doing nothing of value but pretending that it is valuable. Imagine if that were the future for your physics students.

So what I'm trying to do for you now is give you fear. Because without the proper fear to value what you currently have, you cannot work to prevent its loss. And the real researcher doing research of value may be as lost to our future as the dodo, killed off because of too much trust and an inability to comprehend a world where a "top" mathematician confronted with one of the greatest mathematical proofs in human history, would try his best to ignore it, fight it, run away from it, rather than celebrate the truth.

You must learn fear, or lose so much that you truly value. It's like learning why you keep looking both ways before you cross the road, no matter how middle class your society is.

You must fear losing what you value most, or you will not defend it.

(Google: "definition of mathematical proof")

As a major discoverer I'm confronted by the reality of the stories that I gave in a previous post where I have to take as normal bizarre behavior where mathematicians will literally leave the country for periods of time when confronted by my research as if in physically going SOMEWHERE they can escape the truth, and then they come back and go on with their lives as if nothing happened.

The integer factorization research I've used to show "hidden variables" is remarkable for many reasons not only because it hints at physical models far beyond anything used at this time, but because it is necessary for the next level of science and technology, so as this impasse continues the future still remains in doubt.

Imagine Sir Isaac Newton contemplating a situation like this one in his time, and what world we would have today if he had failed.

The people who are resisting this research are fighting to hold on to ideas that have failed. Kind of like Ptolemaic systems resisting Copernican.

If they can succeed they can have ingrained processes and ideas indefinitely that cannot work, no matter how "pure" they claim them to be, as they are purely wrong.

I think ultimately it is a middle class problem: like the American middle class could not fathom how badly things could get if they just decided that facts didn't matter, so who cared who was president?

The middle class view is the desired one, but it can fail in the face of challenges that require you consider the worst case.

Here the worst case to consider is the end of mathematical and scientific advancement as we know it.

The field has already been corrupted by "theory" that has predictions impossible to test by experiment, as mathematicians infect over into other fields.

The problem is that while some may feel anxious about the future of the physics field it is impossible for them to imagine its actual destruction by attitudes more akin to religion than science, so they do not do enough to stop it.

As the United States did so little to stop the end of the positive view it long held in the world because its people as a group could not see defense of that goodwill as a major value—because of their middle class complacency—until it was gone, and I don't think my country is still really aware yet of how great has been its loss.

If the physics field is lost the humanity that finally becomes aware of it will be a species in pain like nothing we have today, when it will not be a debate but a reality, when people who can solve problems do not exist, as no one has learned the true art, but instead live in the fantasy of answers where none exist, learned from the mathematicians who currently practice it.

To understand the future of your students in physics then if the worst case plays out, consider those mathematicians now who when confronted with proofs leave the country, as if that mattered, or simply just go on about their business, continuing research in areas where that research is no longer needed because they are lost as researchers.

They invest time and energy doing nothing of value but pretending that it is valuable. Imagine if that were the future for your physics students.

So what I'm trying to do for you now is give you fear. Because without the proper fear to value what you currently have, you cannot work to prevent its loss. And the real researcher doing research of value may be as lost to our future as the dodo, killed off because of too much trust and an inability to comprehend a world where a "top" mathematician confronted with one of the greatest mathematical proofs in human history, would try his best to ignore it, fight it, run away from it, rather than celebrate the truth.

You must learn fear, or lose so much that you truly value. It's like learning why you keep looking both ways before you cross the road, no matter how middle class your society is.

You must fear losing what you value most, or you will not defend it.

## JSH: Let's look at the behavior

I am serious in that the question driving things now is, how could professional mathematicians ignore major mathematical proofs for years, and continue doing work in areas where those proofs removed the need for further research or invalidated their current research paths?

The naive or cynical seem to believe that my posting is in order to gain acceptance for my work, but supposedly a more direct route would be fighting for more publication or working to convince individual mathematicians, for instance, at the University of California at Berkeley which is only a few minutes away from me. I have actually visited that campus but when classes were not in session as I was with a woman taking tennis lessons from the Berkeley coach.

I am less interested in talking to individual mathematicians directly than solving this intriguing problem.

I have tried several theories out in various experiments. I want to talk about the results.

In one case a math professor at Morehouse College informed me quite confidently that no prime counting function of that type that I had emailed him existed. I noted that what I sent him clearly DID exist. He did not reply further.

A professor at a northern California university who was asked to look into my prime counting function by a colleague who was himself a sociology professor, said he'd check it, then abruptly went out of the country on a six month sabbatical, and on his return claimed to not remember being asked, but then would not consider it now.

The sabbatical seems to be a consistent theme. The chief editor Ioannis Argyros of the now defunct math journal SWJPAM, left the country after he pulled my paper on non-polynomial factorization after emails from some sci.math posters claimed it was in error.

Less dramatic opting out has been seen from a math grad student from Cornell University who emailed me out of the blue offering help in resolving arguments with my math research which he'd noticed on sci.math, and when I challenged him, what was in it for me, he offered that having a Cornell math grad and possibly that math department in my corner would help, so I sent him the non-polynomial factorization argument, and he proceeded to work through it, translating it into his own words, where he slowed down precipitously as he got closer to the end. After several months of this, he was sending me emails talking about long walks staring up at the sky in the early morning hours. I kid you not. A few more months later he opined that he didn't have the expertise to complete the argument claiming he needed to do further research on algebraic integers.

Consistently what I see when I present my research to mathematicians is this remarkable behavior. So maybe it wouldn't matter anyway if I, say, went to Berkeley and talked to the math people there?

Email allows me to contact mathematicians in every corner of the globe. I've seen this behavior on just about every continent.

I used to liken it to a snapping, as in, I'd say their minds snap. But is that accurate.

Cognitive dissonance is one clear possibility but why does it last so long? Shouldn't any of these people find the results nagging at them until they have to resolve the issue?

Why haven't any of the traditionally available things happened, like grad students anxious to further their careers jumping on results? Or even undergrads at the cutting edge out of curiosity pushing the older establishment to embrace new things that work better?

What is the mechanism for the dysfunction?

And it is global. I am a global player in terms of putting information out there. This month alone my math blog according to Google Analytics shows visits from 66 countries, though only 792 unique visitors.

But assuming that number is correct those were 792 chances in countries around the globe to have a person take up this research as an important cause.

Oh, of course, there are also those who dismiss any statistics and simply state that I am wrong or must be wrong, but my favorite thing to show is the non-rational rejection of Google search rankings.

Posters have proclaimed consistently and with confidence that Google search rankings do not matter, despite my noting that if they did not matter then Google wouldn't be known as a major search engine!!!

Yet they remain adamant and in considering this behavior maybe it's best to end with some enlightening searches where my research should show up in the top 10.

You need to use Google as the search engine for these:

deep variable

define mathematical proof

integer factorization algorithm

spherical packing problem

wrapper theorem

millenium problems

millenium prize mystery

solving quadratic Diophantine equations

I could go on, but those should give some sense of what I mean, if, in your region or country, they show up in the top 10 as they do here in California.

The mystery is a fascinating one: professionals in a major field aren't behaving as expected. Given proofs they "snap", run away, even take sabbaticals rather than face the truth. In the face of what would seemingly be overwhelming evidence to most they maintain denial. And even challenge conventional wisdom with the major search engine Google, maintaining that search rankings are meaningless.

Ok, problem space should be more defined.

Here we have a real world problem. Not some textbook nicety.

Even for the physicists it should be a massively curious problem, as it goes to the question of the value of their own research: is it actually valuable if no one believes it?

The naive or cynical seem to believe that my posting is in order to gain acceptance for my work, but supposedly a more direct route would be fighting for more publication or working to convince individual mathematicians, for instance, at the University of California at Berkeley which is only a few minutes away from me. I have actually visited that campus but when classes were not in session as I was with a woman taking tennis lessons from the Berkeley coach.

I am less interested in talking to individual mathematicians directly than solving this intriguing problem.

I have tried several theories out in various experiments. I want to talk about the results.

In one case a math professor at Morehouse College informed me quite confidently that no prime counting function of that type that I had emailed him existed. I noted that what I sent him clearly DID exist. He did not reply further.

A professor at a northern California university who was asked to look into my prime counting function by a colleague who was himself a sociology professor, said he'd check it, then abruptly went out of the country on a six month sabbatical, and on his return claimed to not remember being asked, but then would not consider it now.

The sabbatical seems to be a consistent theme. The chief editor Ioannis Argyros of the now defunct math journal SWJPAM, left the country after he pulled my paper on non-polynomial factorization after emails from some sci.math posters claimed it was in error.

Less dramatic opting out has been seen from a math grad student from Cornell University who emailed me out of the blue offering help in resolving arguments with my math research which he'd noticed on sci.math, and when I challenged him, what was in it for me, he offered that having a Cornell math grad and possibly that math department in my corner would help, so I sent him the non-polynomial factorization argument, and he proceeded to work through it, translating it into his own words, where he slowed down precipitously as he got closer to the end. After several months of this, he was sending me emails talking about long walks staring up at the sky in the early morning hours. I kid you not. A few more months later he opined that he didn't have the expertise to complete the argument claiming he needed to do further research on algebraic integers.

Consistently what I see when I present my research to mathematicians is this remarkable behavior. So maybe it wouldn't matter anyway if I, say, went to Berkeley and talked to the math people there?

Email allows me to contact mathematicians in every corner of the globe. I've seen this behavior on just about every continent.

I used to liken it to a snapping, as in, I'd say their minds snap. But is that accurate.

Cognitive dissonance is one clear possibility but why does it last so long? Shouldn't any of these people find the results nagging at them until they have to resolve the issue?

Why haven't any of the traditionally available things happened, like grad students anxious to further their careers jumping on results? Or even undergrads at the cutting edge out of curiosity pushing the older establishment to embrace new things that work better?

What is the mechanism for the dysfunction?

And it is global. I am a global player in terms of putting information out there. This month alone my math blog according to Google Analytics shows visits from 66 countries, though only 792 unique visitors.

But assuming that number is correct those were 792 chances in countries around the globe to have a person take up this research as an important cause.

Oh, of course, there are also those who dismiss any statistics and simply state that I am wrong or must be wrong, but my favorite thing to show is the non-rational rejection of Google search rankings.

Posters have proclaimed consistently and with confidence that Google search rankings do not matter, despite my noting that if they did not matter then Google wouldn't be known as a major search engine!!!

Yet they remain adamant and in considering this behavior maybe it's best to end with some enlightening searches where my research should show up in the top 10.

You need to use Google as the search engine for these:

deep variable

define mathematical proof

integer factorization algorithm

spherical packing problem

wrapper theorem

millenium problems

millenium prize mystery

solving quadratic Diophantine equations

I could go on, but those should give some sense of what I mean, if, in your region or country, they show up in the top 10 as they do here in California.

The mystery is a fascinating one: professionals in a major field aren't behaving as expected. Given proofs they "snap", run away, even take sabbaticals rather than face the truth. In the face of what would seemingly be overwhelming evidence to most they maintain denial. And even challenge conventional wisdom with the major search engine Google, maintaining that search rankings are meaningless.

Ok, problem space should be more defined.

Here we have a real world problem. Not some textbook nicety.

Even for the physicists it should be a massively curious problem, as it goes to the question of the value of their own research: is it actually valuable if no one believes it?

### Thursday, November 27, 2008

## JSH: Problem Solving 101

All human beings do things from emotion. Emotion is what drives you, without it, you do nothing. Emotion demands a goal to satisfy the emotion so you can have a new one, to be driven to do something else. Understanding what drives you is the first step in figuring out how you get what you want, and why you do what you do.

If social approbation is a goal then there are many ways to gain it. One may seem to be problem solving, except that social approbation is about belief, of others. So as a goal of an emotion, it is satisfiable by the perception that you solved the problem, whether you really did or not.

Then how does the problem solver find motivation? What is the emotion that drives?

The answers there will determine how successful you can be and may decide what other answers you can or cannot find, in the lifelong quest where you never arrive.

For every problem you solve, another will present. For every answer you find, another will be waiting out of your reach for the moment, until it's found and yet another awaits.

So then is it futile? Like the mythical Sisyphus pushing the boulder up the hill only to have it fall down the other side?

Why, ask why?

For those who pursue mathematical problems one of the wonders is the ability to get absolute answers. To pursue perfect answers. To be absolutely right.

To know that the problem was solved, and in being solved, is forever solved.

So mathematical problem solvers are a special breed because they pursue absolute perfection.

We are driven by emotion. But mathematics is not about human needs or wants. The mathematical problem has no concerns with satisfying an emotion. The problems do not accept you, like you, or care about you, and ultimately they do not in any way depend on you. They just are.

So the first problem to solve for the wouldbe discoverer is, why bother?

If social approbation is a goal then there are many ways to gain it. One may seem to be problem solving, except that social approbation is about belief, of others. So as a goal of an emotion, it is satisfiable by the perception that you solved the problem, whether you really did or not.

Then how does the problem solver find motivation? What is the emotion that drives?

The answers there will determine how successful you can be and may decide what other answers you can or cannot find, in the lifelong quest where you never arrive.

For every problem you solve, another will present. For every answer you find, another will be waiting out of your reach for the moment, until it's found and yet another awaits.

So then is it futile? Like the mythical Sisyphus pushing the boulder up the hill only to have it fall down the other side?

Why, ask why?

For those who pursue mathematical problems one of the wonders is the ability to get absolute answers. To pursue perfect answers. To be absolutely right.

To know that the problem was solved, and in being solved, is forever solved.

So mathematical problem solvers are a special breed because they pursue absolute perfection.

We are driven by emotion. But mathematics is not about human needs or wants. The mathematical problem has no concerns with satisfying an emotion. The problems do not accept you, like you, or care about you, and ultimately they do not in any way depend on you. They just are.

So the first problem to solve for the wouldbe discoverer is, why bother?

## Solving directly for factors of a composite

I've found that you can solve directly for f_1 and f_2 when f_1*f_2 = T, where T is a composite to be factored:

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p

and 'a' is related to k mod p by

a^2 = (T)(k^2)^{-1} - 1 mod p

where p is what I call a helper prime.

To get the direct solution you pick p a prime where p>sqrt(T) and pick for k a positive integer near p.

Example: Let T=119. Then p=11 is greater than sqrt(119), and trying k=10 gives

a^2 = (119)(10^2)^{-1} - 1 mod 11 = 8 mod 11,

but 8 is not a quadratic residue modulo 11, so no 'a' exists for this case. Trying now k=9, gives

a^2 = (119)(9^2)^{-1} - 1 mod 11 = 4 mod 11

so a=2 is a solution. And ak = 18, which is greater than 11, so

f_1 = ak = 18 mod 11 = 7 mod 11.

And you have a non-trivial factorization, as 7 is a factor of 119.

It's an oddly simple method given the history in this area. I'm popularizing it in the hopes that eventually I can get the mathematical community to acknowledge this surprising result.

They missed it, and with something simple unfortunately modern mathematicians are the hardest to convince as they believe everything simple has already been discovered, and they don't believe you can solve directly for factors using congruence relationships.

But the simple mathematics above proves that conventional wisdom in this area has been wrong.

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p

and 'a' is related to k mod p by

a^2 = (T)(k^2)^{-1} - 1 mod p

where p is what I call a helper prime.

To get the direct solution you pick p a prime where p>sqrt(T) and pick for k a positive integer near p.

Example: Let T=119. Then p=11 is greater than sqrt(119), and trying k=10 gives

a^2 = (119)(10^2)^{-1} - 1 mod 11 = 8 mod 11,

but 8 is not a quadratic residue modulo 11, so no 'a' exists for this case. Trying now k=9, gives

a^2 = (119)(9^2)^{-1} - 1 mod 11 = 4 mod 11

so a=2 is a solution. And ak = 18, which is greater than 11, so

f_1 = ak = 18 mod 11 = 7 mod 11.

And you have a non-trivial factorization, as 7 is a factor of 119.

It's an oddly simple method given the history in this area. I'm popularizing it in the hopes that eventually I can get the mathematical community to acknowledge this surprising result.

They missed it, and with something simple unfortunately modern mathematicians are the hardest to convince as they believe everything simple has already been discovered, and they don't believe you can solve directly for factors using congruence relationships.

But the simple mathematics above proves that conventional wisdom in this area has been wrong.

### Wednesday, November 26, 2008

## JSH: Solving for a factor

The great thing about my research into factoring is that it lets you solve for f_1 when f_1*f_2 = nT, where T is the target to be factored, and n is an integer control variable.

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p

and 'a' is related to k mod p by

a^2 = (nT)(k^2)^{-1} - 1 mod p

where p is what I call a helper prime. The equations are perfect and all issues with them come down to the issue of how fast you can use them, and with what probability you factor nT, as, of course, with mod you are looking at nT mod p.

Conventional wisdom among mathematicians has been that because given xy = r mod p, there are p-1 solutions for x and y, modular arithmetic doesn't offer the solution to solving for factorizations, but the mathematics appears to have been just slightly more subtle than they realized as of course, 'a' and k above are related by quadratic residues!!!

And in fact, it's not always possible to solve for r^2 = x mod p for any residue x, as for an odd prime there are in fact only (p-1)/2 quadratic residues, so you have an interlocking mechanism in the equation above, where k locks to particular values of 'a' through quadratic residues dependent on nT mod p, so for some values you will not have solutions, which is where the control variable n can come in, or you can pick another k.

It is remarkable that one simple additional idea could be so critical in just SOLVING for the factors of a composite: quadratic residues as key to the solution.

And just solving for one of the factors of a composite nT, has never before been done.

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p

and 'a' is related to k mod p by

a^2 = (nT)(k^2)^{-1} - 1 mod p

where p is what I call a helper prime. The equations are perfect and all issues with them come down to the issue of how fast you can use them, and with what probability you factor nT, as, of course, with mod you are looking at nT mod p.

Conventional wisdom among mathematicians has been that because given xy = r mod p, there are p-1 solutions for x and y, modular arithmetic doesn't offer the solution to solving for factorizations, but the mathematics appears to have been just slightly more subtle than they realized as of course, 'a' and k above are related by quadratic residues!!!

And in fact, it's not always possible to solve for r^2 = x mod p for any residue x, as for an odd prime there are in fact only (p-1)/2 quadratic residues, so you have an interlocking mechanism in the equation above, where k locks to particular values of 'a' through quadratic residues dependent on nT mod p, so for some values you will not have solutions, which is where the control variable n can come in, or you can pick another k.

It is remarkable that one simple additional idea could be so critical in just SOLVING for the factors of a composite: quadratic residues as key to the solution.

And just solving for one of the factors of a composite nT, has never before been done.

## Hidden variables, factoring and new perspective

As a physics undergrad I remember being taught that the "hidden variables" idea for physics was not considered to be valuable though it had its proponents, and now many years later I'm fascinated by what might otherwise be considered "pure math", where it has me re-thinking my physics on that question as well.

Simply enough I found that you can solve for f_1 and f_2 when f_1*f_2 = nT, where T is some target composite to be factored and n is a non-zero integer control variable, modulo some prime number p, where

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p

and 'a' is related to k mod p by

a^2 = (nT)(k^2)^{-1} - 1 mod p.

And it is a remarkable case where what could be considered previously hidden variables are brought to light.

For those wondering if those equations are valid or not, or simply the musings some deluded crackpot, I can assure you they do work, and getting them to factor is relatively simple, as you pick a prime p greater than sqrt(T), and pick a positive integer k, as close to p as you can get, like k=p-1, and then try to solve for 'a', and if you can, and also ak is greater than p, then you will have factored nT mod p, with high likelihood you factored nT itself.

Here's an example to demonstrate and then I'd like to talk about how the solution is derived:

Example: Let T=119. Then p=11 is greater than sqrt(119), and trying k=10, with n=1 gives

a^2 = (119)(10^2)^{-1} - 1 mod 11 = 8 mod 11,

but 8 is not a quadratic residue modulo 11, so no 'a' exists for this case. Trying now k=9, gives

a^2 = (119)(9^2)^{-1} - 1 mod 11 = 4 mod 11

so a=2 is a solution. And ak = 18, which is greater than 11, so

f_1 = ak = 18 mod 11 = 7 mod 11.

And you have a non-trivial factorization, as 7 is a factor of 119. Which can almost seem magical until you know the underlying mathematics, and also understand just how powerful the result is in terms of a high probability of success, as in fact, a k that will non-trivially factor can be shown to be in the neighborhood of k such that

abs(nT - (1 + a^2)k^2)

is a minimum, and I call that value k_0. Those needing further convincing before continuing can simply try more examples of their own, as now I'd like to talk about how these equations result and then return again to the ramifications for physics theory.

The equations come from an analytical study of the relatively simple quadratic system:

x^2 = y^2 + S

and

z^2 = y^2 + nT

where traditionally mathematicians have approached factoring with methods based on the even simpler

x^2 = y^2 mod N

where N is the composite to be factored.

I went to a more complex system because I was questioning whether or not you could factor one number through factoring another which I designated surrogate factoring, so 'S' is the surrogate to be factored, where you find z from finding x and y with the prior equation, in my initial probes of the efficacy of this approach.

However with time—the final result is the distillation of over 4 years of basic research—I found that instead of factoring S itself, I could force a prime p of my choice into S—forcing a factor—which gave me the system:

x^2 = y^2 mod p

and

z^2 = y^2 + nT

and then simply by adding a few constraints I was able to derive the equations above. Those constraints are:

But its ramifications are huge, as clearly there are variables that are hidden unless you look for them by asking the right questions! Where here they powerfully guide integer factorization itself where factorizations are connected to each other in this remarkable way not known until you derive a solution from these otherwise hidden variables.

If the same can be true for physics systems then what questions are asked can decide what solutions can be found, as in this case with a purely mathematical system, I simply asked the question: can you factor one number through factoring another?

And found a system where previously hidden variables emerged where p is such a remarkable one that I call it a helper prime.

Imagine if, for instance, quantum chromodynamics could be simplified to the level of quantum electrodynamics simply by asking the right questions and finding systems dependent on previously hidden variables including helper ones like p.

The physical reality then might be that disparate groups of atomic nuclei could be interacting with each other on levels not previously imagined, which of course Einstein called "spooky action at a distance", like composites factor dependent on each other in a way not observable without the proper analytical approach!

It is my hope that I have made this research result relevant to the physics community as it is the community from which I was trained, and it is I think the best hope for a continuation of this line of thought and a proper evaluation of the possibilities as my efforts to garner appropriate interest from the mathematical and cryptological communities have been disappointing.

For the physics community the question of what hidden variables can look like in a simple system can be seen now, as well as a relevance between seemingly disparate elements.

Mathematicians for centuries have looked at factoring composites as individual accomplishments—number by number. But considering factorizations in pairs revealed a surprisingly simple solution for f_1 and f_2, when f_1*f_2 = nT.

What might we learn from our physical world if we simply ask questions in a similar way?

It is my hope that there will be people around the world driven by the need to know the answer.

Thank you for your time and attention.

Simply enough I found that you can solve for f_1 and f_2 when f_1*f_2 = nT, where T is some target composite to be factored and n is a non-zero integer control variable, modulo some prime number p, where

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p

and 'a' is related to k mod p by

a^2 = (nT)(k^2)^{-1} - 1 mod p.

And it is a remarkable case where what could be considered previously hidden variables are brought to light.

For those wondering if those equations are valid or not, or simply the musings some deluded crackpot, I can assure you they do work, and getting them to factor is relatively simple, as you pick a prime p greater than sqrt(T), and pick a positive integer k, as close to p as you can get, like k=p-1, and then try to solve for 'a', and if you can, and also ak is greater than p, then you will have factored nT mod p, with high likelihood you factored nT itself.

Here's an example to demonstrate and then I'd like to talk about how the solution is derived:

Example: Let T=119. Then p=11 is greater than sqrt(119), and trying k=10, with n=1 gives

a^2 = (119)(10^2)^{-1} - 1 mod 11 = 8 mod 11,

but 8 is not a quadratic residue modulo 11, so no 'a' exists for this case. Trying now k=9, gives

a^2 = (119)(9^2)^{-1} - 1 mod 11 = 4 mod 11

so a=2 is a solution. And ak = 18, which is greater than 11, so

f_1 = ak = 18 mod 11 = 7 mod 11.

And you have a non-trivial factorization, as 7 is a factor of 119. Which can almost seem magical until you know the underlying mathematics, and also understand just how powerful the result is in terms of a high probability of success, as in fact, a k that will non-trivially factor can be shown to be in the neighborhood of k such that

abs(nT - (1 + a^2)k^2)

is a minimum, and I call that value k_0. Those needing further convincing before continuing can simply try more examples of their own, as now I'd like to talk about how these equations result and then return again to the ramifications for physics theory.

The equations come from an analytical study of the relatively simple quadratic system:

x^2 = y^2 + S

and

z^2 = y^2 + nT

where traditionally mathematicians have approached factoring with methods based on the even simpler

x^2 = y^2 mod N

where N is the composite to be factored.

I went to a more complex system because I was questioning whether or not you could factor one number through factoring another which I designated surrogate factoring, so 'S' is the surrogate to be factored, where you find z from finding x and y with the prior equation, in my initial probes of the efficacy of this approach.

However with time—the final result is the distillation of over 4 years of basic research—I found that instead of factoring S itself, I could force a prime p of my choice into S—forcing a factor—which gave me the system:

x^2 = y^2 mod p

and

z^2 = y^2 + nT

and then simply by adding a few constraints I was able to derive the equations above. Those constraints are:

- z = x + ak mod p
- 2az = k mod p.

But its ramifications are huge, as clearly there are variables that are hidden unless you look for them by asking the right questions! Where here they powerfully guide integer factorization itself where factorizations are connected to each other in this remarkable way not known until you derive a solution from these otherwise hidden variables.

If the same can be true for physics systems then what questions are asked can decide what solutions can be found, as in this case with a purely mathematical system, I simply asked the question: can you factor one number through factoring another?

And found a system where previously hidden variables emerged where p is such a remarkable one that I call it a helper prime.

Imagine if, for instance, quantum chromodynamics could be simplified to the level of quantum electrodynamics simply by asking the right questions and finding systems dependent on previously hidden variables including helper ones like p.

The physical reality then might be that disparate groups of atomic nuclei could be interacting with each other on levels not previously imagined, which of course Einstein called "spooky action at a distance", like composites factor dependent on each other in a way not observable without the proper analytical approach!

It is my hope that I have made this research result relevant to the physics community as it is the community from which I was trained, and it is I think the best hope for a continuation of this line of thought and a proper evaluation of the possibilities as my efforts to garner appropriate interest from the mathematical and cryptological communities have been disappointing.

For the physics community the question of what hidden variables can look like in a simple system can be seen now, as well as a relevance between seemingly disparate elements.

Mathematicians for centuries have looked at factoring composites as individual accomplishments—number by number. But considering factorizations in pairs revealed a surprisingly simple solution for f_1 and f_2, when f_1*f_2 = nT.

What might we learn from our physical world if we simply ask questions in a similar way?

It is my hope that there will be people around the world driven by the need to know the answer.

Thank you for your time and attention.

### Sunday, November 23, 2008

## JSH: Integer factorization algorithm

The usefulness of the factoring equations in terms of how effective they are against large numbers still remains to be determined, however, a fairly simple algorithm using them is readily apparent.

Given a target composite T to be factored, first find a prime number p greater than sqrt(T). Then pick a positive integer k = p-1. Make sure k is coprime to T.

Next you calculate positive integer a, with n=1, using

a^2 = (nT)(k^2)^{-1} - 1 mod p

where if it does not exist, you can change n and try again, or decrement k, and try again, where there should be a 50% probability for each try that it will exist.

If you find 'a', you check ak to see if it is greater than p and if not, you decrement k, and try again.

If it is greater than p you check for a factorization into f_1 and f_2,where

f_1*f_2 = nT

with

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p.

If you've succeeded when you take ak modulo p, you will have a factor of T.

If you wish more criteria before you check you can find k such that

abs(nT - (1 + a^2)k^2)

is a minimum, which is what I call k_0. A k that will factor non-trivially must be within (k_0/2p) steps from k_0 where each step is in increments of 2ap, so if you find that you do not have a solution you can check to see how close you were.

The algorithm will factor nT mod p, so the goal is to have it factor nT, where n is a minimum, as the algebra is using n mod p as well, and key to that I'd hypothesize is finding a small 'a' with a large k, where of course ak is greater than p.

So no matter what, if you get to f_1 you are factoring nT mod p, but your hope is to factor nT and it's not completely clear to me how hard it is to get that happening.

Example: Let T=119. Then p=11 is greater than sqrt(119), and trying k=10, with n=1 gives

a^2 = (119)(10^2)^{-1} - 1 mod 11 = 8 mod 11,

so it does not exist. Trying now k=9, gives

a^2 = (119)(9^2)^{-1} - 1 mod 11 = 4 mod 11

so a=2.

And ak = 18, which is greater than 11, so

f_1 = ak = 18 mod 11 = 7 mod 11.

And you have a non-trivial factorization, as 7 is a factor of 119.

Note that k_0, from using

abs(119 - 5k^2)

is k_0 = 5, as that is the value of k that gives the minimum value, and 5 is within 0 steps in increments of 44 of 9, as the steps are floor(5/22).

To be 1 step away the k that factored would have had to be greater than or equal to 49.

And that completes a demonstration of the factoring algorithm that follows from the prior results. It's rather remarkable to me to have my own factoring algorithm and also odd that I can't say exactly how well it works!

Nothing succeeds like success.

Antagonism from the mathematical and cryptographic communities to this research is not about its value but about politics and human weakness: guys who think that people like me are beneath them so they refuse to acknowledge research they think puts someone like me higher.

It's all about a pecking order.

Given a target composite T to be factored, first find a prime number p greater than sqrt(T). Then pick a positive integer k = p-1. Make sure k is coprime to T.

Next you calculate positive integer a, with n=1, using

a^2 = (nT)(k^2)^{-1} - 1 mod p

where if it does not exist, you can change n and try again, or decrement k, and try again, where there should be a 50% probability for each try that it will exist.

If you find 'a', you check ak to see if it is greater than p and if not, you decrement k, and try again.

If it is greater than p you check for a factorization into f_1 and f_2,where

f_1*f_2 = nT

with

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p.

If you've succeeded when you take ak modulo p, you will have a factor of T.

If you wish more criteria before you check you can find k such that

abs(nT - (1 + a^2)k^2)

is a minimum, which is what I call k_0. A k that will factor non-trivially must be within (k_0/2p) steps from k_0 where each step is in increments of 2ap, so if you find that you do not have a solution you can check to see how close you were.

The algorithm will factor nT mod p, so the goal is to have it factor nT, where n is a minimum, as the algebra is using n mod p as well, and key to that I'd hypothesize is finding a small 'a' with a large k, where of course ak is greater than p.

So no matter what, if you get to f_1 you are factoring nT mod p, but your hope is to factor nT and it's not completely clear to me how hard it is to get that happening.

Example: Let T=119. Then p=11 is greater than sqrt(119), and trying k=10, with n=1 gives

a^2 = (119)(10^2)^{-1} - 1 mod 11 = 8 mod 11,

so it does not exist. Trying now k=9, gives

a^2 = (119)(9^2)^{-1} - 1 mod 11 = 4 mod 11

so a=2.

And ak = 18, which is greater than 11, so

f_1 = ak = 18 mod 11 = 7 mod 11.

And you have a non-trivial factorization, as 7 is a factor of 119.

Note that k_0, from using

abs(119 - 5k^2)

is k_0 = 5, as that is the value of k that gives the minimum value, and 5 is within 0 steps in increments of 44 of 9, as the steps are floor(5/22).

To be 1 step away the k that factored would have had to be greater than or equal to 49.

And that completes a demonstration of the factoring algorithm that follows from the prior results. It's rather remarkable to me to have my own factoring algorithm and also odd that I can't say exactly how well it works!

Nothing succeeds like success.

Antagonism from the mathematical and cryptographic communities to this research is not about its value but about politics and human weakness: guys who think that people like me are beneath them so they refuse to acknowledge research they think puts someone like me higher.

It's all about a pecking order.

### Saturday, November 22, 2008

## Integer factorization equations

Given a target composite T to be factored remarkably you can simply solve for factors f_1 and f_2, of

f_1*f_2 = nT

where n is defined by

n = (1 + a^2)k^2(T)^{-1} mod p

with

f_1 = ak mod p

and

f_2 = a^{-1}(1 + a^2)k mod p.

Here 'a' and k are chosen at will, but if they can be found such that n=1, for a prime p greater than sqrt(T), you have a factorization of T. Of course n=2, or n=3 or n equal some other known prime will work just as well for you.

But you may then have a trivial factorization.

However, it can be further shown that with all positive integers, for a given 'a' the minimum value for k which I call k_0 which will give a non-trivial factorization is given by finding k such that

abs(nT - (1 + a^2)k^2)

is a minimum, which is just a rather cool and remarkable result, and also amazing to me in that it is the only situation I know of where the algebra cares about trivial versus non-trivial factorizations.

That result is one of my most beautiful research finds. It was sobering when the math community rejected it like so many others and in the months since I first made that discovery I've considered how real mathematicians could ignore a stunningly beautiful result.

I've concluded they could not. Given the reality of the result then, the people who the world sees as its mathematicians, are fakes.

And those equations define integer factorization. They also show what I call deep variables and a helper variable which is the prime number p.

f_1*f_2 = nT

where n is defined by

n = (1 + a^2)k^2(T)^{-1} mod p

with

f_1 = ak mod p

and

f_2 = a^{-1}(1 + a^2)k mod p.

Here 'a' and k are chosen at will, but if they can be found such that n=1, for a prime p greater than sqrt(T), you have a factorization of T. Of course n=2, or n=3 or n equal some other known prime will work just as well for you.

But you may then have a trivial factorization.

However, it can be further shown that with all positive integers, for a given 'a' the minimum value for k which I call k_0 which will give a non-trivial factorization is given by finding k such that

abs(nT - (1 + a^2)k^2)

is a minimum, which is just a rather cool and remarkable result, and also amazing to me in that it is the only situation I know of where the algebra cares about trivial versus non-trivial factorizations.

That result is one of my most beautiful research finds. It was sobering when the math community rejected it like so many others and in the months since I first made that discovery I've considered how real mathematicians could ignore a stunningly beautiful result.

I've concluded they could not. Given the reality of the result then, the people who the world sees as its mathematicians, are fakes.

And those equations define integer factorization. They also show what I call deep variables and a helper variable which is the prime number p.

### Wednesday, November 19, 2008

## Deep variable factoring method

Remarkably the seemingly well-known quadratic equation

z^2 = y^2 + nT

can be analyzed rather thoroughly with a solution for z using some basic algebra. That solution for z is

z = (1 + 2a^2)k/(2a)

where with positive non-zero integers z, y, n and T, with z and y coprime, it is not necessarily the case that k and 'a' are themselves integers, and the result may seem trivial until it's noted that there must exist a prime number p such that

k^2 = (1 + a^2)^{-1}(nT) mod p

where p is what I call a helper prime as it is only relevant to the analysis in order to find z, which is done by finding k and 'a' when they are integers, which is why I call this a deep variable factoring method.

The result STILL may seem trivial until it's noted that for cases where k and 'a' are positive integers, it can be proven that the minimum value for k, which I call k_0 is given by finding k such that

abs(nT - (1 + a^2)k^2)

is a minimum, which is kind of remarkably odd, as given k, you have z, and a factorization of nT, but knowing the minimum k does not necessarily make solving for k easy as an even deeper result is that k will be within 2ap steps of floor(k_0/2p) which brings the helper prime p to the forefront of the question of the usefulness of this approach!

The desired number of steps is 0, and it can be shown that p near sqrt(T), as long as n is small should do the trick! So how do you find the "helper prime" with the appropriate size?

There are two more important equations relevant to that question. If you have positive integers f_1 and f_2 where

f_1*f_2 = nT

then it must be true that

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p

so this approach will GIVE f_1 directly if f_1 is less than p, which is great!!! But notice that if k is less than p as well, then you have that k must be a factor of nT, which is not so great, and in fact, that is a huge damper as if you already knew a factor of nT, you wouldn't be looking for a k to factor it!

(That result has deeper implications for what solutions can be available for a given k and 'a' which is of more "pure math" interest.)

So now finally limiting factors on the helper prime p are clear where it is primary that if f_1 is less than p then ak must be greater, and now finally I can reach the crucial equation which determines the efficacy of this method, as now the variable n proves its worth, as from before I have

k^2 = (1 + a^2)^{-1}(nT) mod p

so I can reverse things and solve for n to get

n = (1 + a^2)k^2(T)^{-1} mod p

where the BEST case is n=1, as then if p > sqrt(T), then I have a factorization from f_1 = ak, so ALL determination of the efficacy of this approach of deep variables rests on how easily one can find k and 'a' such that n is a minimum!

If z does not have integer solutions for k and 'a' when n=1, then that will preclude n=1 since

z = (1 + 2a^2)k/(2a)

but then how about n=2? Or some small n that will still factor? As if you can find a small enough n, then regardless of the size of the number all of the equations above will work and just GIVE you f_1 and a factorization of nT, if the helper prime p is greater than sqrt(nT).

So all issues with the potential of this method boil down to how easily k and 'a' can be picked such that

(1 + a^2)k^2 = m mod p

where m is some arbitrary residue, preferably very small where the best is m=1, when p is a very large prime.

And it took me YEARS to figure out this approach and MONTHS to boil down questions of its efficacy to just that equation which is what I only just accomplished today in terms of realizing its significance!

It is rather exciting to me to be able to add to the prior knowledge about such a seemingly simple equation as

z^2 = y^2 + nT

where it's also of "pure math" interest as you can actually still factor with k and 'a' when those are non-rationals and it is of number theoretic interest how residues behave with them as non-rationals as well.

And the helper prime p is just amazing as a deep variable that is only there for this purpose and vanishes like a ghost when you have an answer, as in actuality so do k and 'a', as once you have z the machinery needed to find it just disappears! Which reminds me of thoughts of deep or hidden variables in physics as this algebra may indicate that they could have a role in our physical reality, where phenomena is less easily explained by the surface variables alone, than by transient variables that briefly exist to help, and then just disappear.

z^2 = y^2 + nT

can be analyzed rather thoroughly with a solution for z using some basic algebra. That solution for z is

z = (1 + 2a^2)k/(2a)

where with positive non-zero integers z, y, n and T, with z and y coprime, it is not necessarily the case that k and 'a' are themselves integers, and the result may seem trivial until it's noted that there must exist a prime number p such that

k^2 = (1 + a^2)^{-1}(nT) mod p

where p is what I call a helper prime as it is only relevant to the analysis in order to find z, which is done by finding k and 'a' when they are integers, which is why I call this a deep variable factoring method.

The result STILL may seem trivial until it's noted that for cases where k and 'a' are positive integers, it can be proven that the minimum value for k, which I call k_0 is given by finding k such that

abs(nT - (1 + a^2)k^2)

is a minimum, which is kind of remarkably odd, as given k, you have z, and a factorization of nT, but knowing the minimum k does not necessarily make solving for k easy as an even deeper result is that k will be within 2ap steps of floor(k_0/2p) which brings the helper prime p to the forefront of the question of the usefulness of this approach!

The desired number of steps is 0, and it can be shown that p near sqrt(T), as long as n is small should do the trick! So how do you find the "helper prime" with the appropriate size?

There are two more important equations relevant to that question. If you have positive integers f_1 and f_2 where

f_1*f_2 = nT

then it must be true that

f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p

so this approach will GIVE f_1 directly if f_1 is less than p, which is great!!! But notice that if k is less than p as well, then you have that k must be a factor of nT, which is not so great, and in fact, that is a huge damper as if you already knew a factor of nT, you wouldn't be looking for a k to factor it!

(That result has deeper implications for what solutions can be available for a given k and 'a' which is of more "pure math" interest.)

So now finally limiting factors on the helper prime p are clear where it is primary that if f_1 is less than p then ak must be greater, and now finally I can reach the crucial equation which determines the efficacy of this method, as now the variable n proves its worth, as from before I have

k^2 = (1 + a^2)^{-1}(nT) mod p

so I can reverse things and solve for n to get

n = (1 + a^2)k^2(T)^{-1} mod p

where the BEST case is n=1, as then if p > sqrt(T), then I have a factorization from f_1 = ak, so ALL determination of the efficacy of this approach of deep variables rests on how easily one can find k and 'a' such that n is a minimum!

If z does not have integer solutions for k and 'a' when n=1, then that will preclude n=1 since

z = (1 + 2a^2)k/(2a)

but then how about n=2? Or some small n that will still factor? As if you can find a small enough n, then regardless of the size of the number all of the equations above will work and just GIVE you f_1 and a factorization of nT, if the helper prime p is greater than sqrt(nT).

So all issues with the potential of this method boil down to how easily k and 'a' can be picked such that

(1 + a^2)k^2 = m mod p

where m is some arbitrary residue, preferably very small where the best is m=1, when p is a very large prime.

And it took me YEARS to figure out this approach and MONTHS to boil down questions of its efficacy to just that equation which is what I only just accomplished today in terms of realizing its significance!

It is rather exciting to me to be able to add to the prior knowledge about such a seemingly simple equation as

z^2 = y^2 + nT

where it's also of "pure math" interest as you can actually still factor with k and 'a' when those are non-rationals and it is of number theoretic interest how residues behave with them as non-rationals as well.

And the helper prime p is just amazing as a deep variable that is only there for this purpose and vanishes like a ghost when you have an answer, as in actuality so do k and 'a', as once you have z the machinery needed to find it just disappears! Which reminds me of thoughts of deep or hidden variables in physics as this algebra may indicate that they could have a role in our physical reality, where phenomena is less easily explained by the surface variables alone, than by transient variables that briefly exist to help, and then just disappear.

### Tuesday, November 18, 2008

## JSH: Settled factoring research, what now?

I ended up bringing factoring back up after I brought up solving for quadratic residues mod N, and yesterday recognized that I'm just tired of this area, and am ready to move on. But as one last parting shot I decided today to give what is my settled factoring research, which is what I derived the method for solving quadratic residues from, and the surprise is that it is mostly straightforward.

Simply enough I worked out that given non-zero integers, z, y and nT, where

z^2 = y^2 + nT

it must be true that

z = (1 + 2a^2)k/(2a)

where k and 'a' are to be determined.

That solution for z is not trivial because it can be shown that

k^2 = (1 + a^2)^{-1}(nT) mod p

for some odd prime p, where p is what I call a helper prime as it is only there for these results! And I ran into trouble trying to define rules for what p's will work where posters kept finding counterexamples to what I thought were the rules.

It's not a minor issue either as, it can be shown that, if f_1*f_2 = nT, then

f_1 = ak mod p

and

f_2 = a^{-1}(1 + a^2)k mod p

so if p is large enough then you have a factorization!!!

ALL issues with the research boiled down then on whether or not you can get a large enough p. To be a very useful technique the method should allow you to use a VERY large p, so that you can easily get f_1 and f_2.

But before I get back to that issue or try to resolve it, there are more equations as it can be shown that with all positive integers, so all the variables are positive integers, the MINIMUM k, which I call k_0 can be found by finding k_0 such that

abs(nT - (1 + a^2)k_0^2)

is a minimum.

And the k that will work to give you z, must be within floor(k_0/(2p)) steps, so clearly you'd want a p large enough so that you'd have the least number of steps! Where 0 steps is the best.

So again, everything with this approach comes down to picking a large helper prime p, and if you can do so, then the theory says you can very quickly factor, but there are problems with picking a very large prime p and they go back to

f_1 = ak mod p

as if f_1 is less than p, because it's so HUGE, and ak is less than p, then f_1 = ak, so that is an explicit equation, which forces k into f_1, which forces k into nT, and means that with

z^2 = y^2 + nT

k is a factor of all, as it is then a factor of z, y and nT, as remember from before:

z = (1 + 2a^2)k/(2a)

so that is the crucial equation and I fumbled around with various rules that were meant to force p to be LESS than f_1, when the actual requirement is clearly seen to be that ak must be less than p, if f_1 is less than p, if not then it doesn't matter.

So p CAN be very large, as long as you do things carefully without the rules I thought were necessary before.

And that is what I realized yesterday, so I updated things and thought to myself I'm sick of this factoring research so I so posted.

But a couple of hours ago, I realized something that worried me enough to put it out there and see if it's not a big deal, as if the issue is getting an ak bigger than p, why not just pick 'a' and k?

The equations do not say you cannot.

So, you pick a large k near the sqrt(T), and then pick a large prime p BIGGER than k, but next you pick an 'a' such that ak is GREATER than p.

Then you go to

k^2 = (1 + a^2)^{-1}(nT) mod p

and REVERSE it to solve for n, which gives

n = (1 + a^2)k^2(T)^{-1} mod p

and now you just solve for z using

z = (1 + 2a^2)k/(2a),

and solve for y using y = sqrt(z^2 - nT), and you have a factorization from

(z-y)(z+y) = nT.

So you have this straightforward easy approach, but I seem to remember coming up with something like this months ago and I'm sure it was shot down by replies from posters.

One immediate issue of course is that the factorization you get may not factor T itself!!!

Also I wonder if I'm right that you solve for z, as maybe you still have to search for k, but part of the point of this approach is that floor(k_0/(2p)) = 0, but if n is HUGE then maybe k_0 is still much bigger than p, and you need a lot of steps, which could be the problem then.

So it's up in the air. But I've covered the settled research with this approach and also tossed in one unsettled piece at the end.

I know, annoying, since I said I was done. But I wish I had everything tied down so that I was sure. And posting is a way to get feedback and critiques.

And it IS kind of remarkable to me at least that you can just solve for z given

z^2 = y^2 + nT

in this way, with such wacky variables as 'a' and k, where I haven't even fully delved into their wackiness as they can be non-rational and STILL help you factor rationals, but the vote is from the mathematical community.

And I discovered what I call the z constraint more than half a year ago, so it seems that community has voted against this research as interesting, which leaves me chattering about it as an afterthought before moving on.

Simply enough I worked out that given non-zero integers, z, y and nT, where

z^2 = y^2 + nT

it must be true that

z = (1 + 2a^2)k/(2a)

where k and 'a' are to be determined.

That solution for z is not trivial because it can be shown that

k^2 = (1 + a^2)^{-1}(nT) mod p

for some odd prime p, where p is what I call a helper prime as it is only there for these results! And I ran into trouble trying to define rules for what p's will work where posters kept finding counterexamples to what I thought were the rules.

It's not a minor issue either as, it can be shown that, if f_1*f_2 = nT, then

f_1 = ak mod p

and

f_2 = a^{-1}(1 + a^2)k mod p

so if p is large enough then you have a factorization!!!

ALL issues with the research boiled down then on whether or not you can get a large enough p. To be a very useful technique the method should allow you to use a VERY large p, so that you can easily get f_1 and f_2.

But before I get back to that issue or try to resolve it, there are more equations as it can be shown that with all positive integers, so all the variables are positive integers, the MINIMUM k, which I call k_0 can be found by finding k_0 such that

abs(nT - (1 + a^2)k_0^2)

is a minimum.

And the k that will work to give you z, must be within floor(k_0/(2p)) steps, so clearly you'd want a p large enough so that you'd have the least number of steps! Where 0 steps is the best.

So again, everything with this approach comes down to picking a large helper prime p, and if you can do so, then the theory says you can very quickly factor, but there are problems with picking a very large prime p and they go back to

f_1 = ak mod p

as if f_1 is less than p, because it's so HUGE, and ak is less than p, then f_1 = ak, so that is an explicit equation, which forces k into f_1, which forces k into nT, and means that with

z^2 = y^2 + nT

k is a factor of all, as it is then a factor of z, y and nT, as remember from before:

z = (1 + 2a^2)k/(2a)

so that is the crucial equation and I fumbled around with various rules that were meant to force p to be LESS than f_1, when the actual requirement is clearly seen to be that ak must be less than p, if f_1 is less than p, if not then it doesn't matter.

So p CAN be very large, as long as you do things carefully without the rules I thought were necessary before.

And that is what I realized yesterday, so I updated things and thought to myself I'm sick of this factoring research so I so posted.

But a couple of hours ago, I realized something that worried me enough to put it out there and see if it's not a big deal, as if the issue is getting an ak bigger than p, why not just pick 'a' and k?

The equations do not say you cannot.

So, you pick a large k near the sqrt(T), and then pick a large prime p BIGGER than k, but next you pick an 'a' such that ak is GREATER than p.

Then you go to

k^2 = (1 + a^2)^{-1}(nT) mod p

and REVERSE it to solve for n, which gives

n = (1 + a^2)k^2(T)^{-1} mod p

and now you just solve for z using

z = (1 + 2a^2)k/(2a),

and solve for y using y = sqrt(z^2 - nT), and you have a factorization from

(z-y)(z+y) = nT.

So you have this straightforward easy approach, but I seem to remember coming up with something like this months ago and I'm sure it was shot down by replies from posters.

One immediate issue of course is that the factorization you get may not factor T itself!!!

Also I wonder if I'm right that you solve for z, as maybe you still have to search for k, but part of the point of this approach is that floor(k_0/(2p)) = 0, but if n is HUGE then maybe k_0 is still much bigger than p, and you need a lot of steps, which could be the problem then.

So it's up in the air. But I've covered the settled research with this approach and also tossed in one unsettled piece at the end.

I know, annoying, since I said I was done. But I wish I had everything tied down so that I was sure. And posting is a way to get feedback and critiques.

And it IS kind of remarkable to me at least that you can just solve for z given

z^2 = y^2 + nT

in this way, with such wacky variables as 'a' and k, where I haven't even fully delved into their wackiness as they can be non-rational and STILL help you factor rationals, but the vote is from the mathematical community.

And I discovered what I call the z constraint more than half a year ago, so it seems that community has voted against this research as interesting, which leaves me chattering about it as an afterthought before moving on.

### Sunday, November 16, 2008

## JSH: Surrogate factoring already achieved perfection

The success rate for what I call surrogate factoring already achieved perfection as in 100% factoring months ago, back in March. What happened then is that posters claimed it was too slow to be useful.

And I wandered off from the subject as I ended up mostly facing taunts that I should factor a very large number which I won't do.

Instead now in looking back over the research I'm focusing on solving quadratic residues mod N, which involves using a result which follows from the other research where instead of using surrogate factoring to find the target T, you use factoring T to solve for what I call k, which gives you a quadratic residue solution as

k^2 = (T/2) mod p

in the original surrogate factoring equations, and I've generalized to k^2 = (T/2) mod N, which they allow.

The underlying mathematics is complex enough, while mostly simple algebra that I simply stay away from it in giving the latest methods, and I've just stayed very far away from talking about surrogate factoring itself, until this post.

It is my most controversial baby. My most conflicted research in terms of how I view it.

And possibly the most misunderstood of my findings.

A completely new way to factor that is too controversial for me to get much traction with it.

And I wandered off from the subject as I ended up mostly facing taunts that I should factor a very large number which I won't do.

Instead now in looking back over the research I'm focusing on solving quadratic residues mod N, which involves using a result which follows from the other research where instead of using surrogate factoring to find the target T, you use factoring T to solve for what I call k, which gives you a quadratic residue solution as

k^2 = (T/2) mod p

in the original surrogate factoring equations, and I've generalized to k^2 = (T/2) mod N, which they allow.

The underlying mathematics is complex enough, while mostly simple algebra that I simply stay away from it in giving the latest methods, and I've just stayed very far away from talking about surrogate factoring itself, until this post.

It is my most controversial baby. My most conflicted research in terms of how I view it.

And possibly the most misunderstood of my findings.

A completely new way to factor that is too controversial for me to get much traction with it.

### Wednesday, November 12, 2008

## JSH: The nightmare I couldn't stop

My fight for so many years has been for your survival.

And I'm almost out of time.

No. I'm not in danger. YOU are.

Forget what you think you know. The world is far more brutal than you can imagine and your societies are running out of time.

I've delayed as long as I can.

There is no more time for you. I have no more mercy left.

And I'm almost out of time.

No. I'm not in danger. YOU are.

Forget what you think you know. The world is far more brutal than you can imagine and your societies are running out of time.

I've delayed as long as I can.

There is no more time for you. I have no more mercy left.

### Tuesday, November 11, 2008

## JSH: Weird situation, explanation attempt

I've said for years that I'm a major mathematical discoverer and something was going wrong with the mathematical community which is why my results weren't recognized, but that didn't seem to convince many people, so for several years I worked on the factoring problem. You know, finding a general way to solve for f_1 and f_2 such that f_1*f_2 = N because it seemed like the kind of result mathematicians couldn't avoid.

Well I progressed at that task for years and last year started piling result on top of result by analyzing the simple system of equations:

x^2 = y^2 + S

and

z^2 = y^2 + T

and as I worked at this I came across something I call the z constraint: fine structure behavior within the seemingly simple second equation which DEFINED z relative to two new variables I called k and alpha.

Well I looked one way to generally solve the factoring problem with this new information and argued as usual with people on math newsgroups, but the issue came up about how fast it was, and they claimed it was slow, and I contemplated getting close to my goal with cold-feet.

Because I was TERRIFIED of crashing the world economy and all along had believed that mathematicians would never let things go too far. It was a game of chicken. A game of chicken with the fate of the world…

But besides I didn't think I was quite done, and was still pondering how to proceed, debating with myself when I thought back to a solution for quadratic residues mod p that I'd come up with earlier. I'd done so because I thought the method of final solution would require finding quadratic residues modulo some prime, and why not see if that could be done with my own method?

Well, I pondered that thing a bit and it kind of occurred to me that maybe it might work mod N, and then I kind of let it drop.

I said I was giving up on factoring research and moving on to other things, which I did.

But let me make an interesting point here: I'd had astounding research finds to my mind which showed I was close to solving the factoring problem and STILL couldn't get anywhere with the math community and no, I wasn't just arguing with people on newsgroups. I was also emailing people like Lenstra, and others.

And part of me thought, hey, I must be wrong, right? No way all these people would just sit by if I'd solved the thing, would they?

And then I really got involved in other things. That was some months ago.

But now, a couple of days ago, I found myself re-visiting the work that was almost a year old, and I posted it again, puzzling over what I'd said before as I thought about it, and noticed that hey, it looked like a general solution to finding a quadratic residue mod N. Wow.

So that's the story.

So the approach has sat out there for around a year.

Um, let's say for the sake of argument that it is a solution to the factoring problem, how did it sit out for so long?

I've done other things to occupy my time, like generally solving binary quadratic Diophantine equations, and no one is paying attention to that either.

But hey, if this solution is right, then your public key might be easily factored. How could THAT have sat out for almost a year?

Oh, and I don't think it did crash the world economy as that's even weirder as it would require a lot of people around the world lying about what was actually going on, and why would they and just leave me babbling out here like in this post?

The situation is befuddling.

Well I progressed at that task for years and last year started piling result on top of result by analyzing the simple system of equations:

x^2 = y^2 + S

and

z^2 = y^2 + T

and as I worked at this I came across something I call the z constraint: fine structure behavior within the seemingly simple second equation which DEFINED z relative to two new variables I called k and alpha.

Well I looked one way to generally solve the factoring problem with this new information and argued as usual with people on math newsgroups, but the issue came up about how fast it was, and they claimed it was slow, and I contemplated getting close to my goal with cold-feet.

Because I was TERRIFIED of crashing the world economy and all along had believed that mathematicians would never let things go too far. It was a game of chicken. A game of chicken with the fate of the world…

But besides I didn't think I was quite done, and was still pondering how to proceed, debating with myself when I thought back to a solution for quadratic residues mod p that I'd come up with earlier. I'd done so because I thought the method of final solution would require finding quadratic residues modulo some prime, and why not see if that could be done with my own method?

Well, I pondered that thing a bit and it kind of occurred to me that maybe it might work mod N, and then I kind of let it drop.

I said I was giving up on factoring research and moving on to other things, which I did.

But let me make an interesting point here: I'd had astounding research finds to my mind which showed I was close to solving the factoring problem and STILL couldn't get anywhere with the math community and no, I wasn't just arguing with people on newsgroups. I was also emailing people like Lenstra, and others.

And part of me thought, hey, I must be wrong, right? No way all these people would just sit by if I'd solved the thing, would they?

And then I really got involved in other things. That was some months ago.

But now, a couple of days ago, I found myself re-visiting the work that was almost a year old, and I posted it again, puzzling over what I'd said before as I thought about it, and noticed that hey, it looked like a general solution to finding a quadratic residue mod N. Wow.

So that's the story.

So the approach has sat out there for around a year.

Um, let's say for the sake of argument that it is a solution to the factoring problem, how did it sit out for so long?

I've done other things to occupy my time, like generally solving binary quadratic Diophantine equations, and no one is paying attention to that either.

But hey, if this solution is right, then your public key might be easily factored. How could THAT have sat out for almost a year?

Oh, and I don't think it did crash the world economy as that's even weirder as it would require a lot of people around the world lying about what was actually going on, and why would they and just leave me babbling out here like in this post?

The situation is befuddling.

## Solving quadratic residues mod N

There is a remarkably simple yet perfect way to generally solve quadratic residues mod a non-unit natural number N, coprime to 3.

Given a quadratic residue q modulo N where N is a non-unit natural number coprime to 3, where you wish to find k, where

k^2 = q mod N

you first find

T = 2q mod N

where T = 2q + jN, and j is a non-zero integer chosen such that T mod 3 = 2.

You next check each factorization of T to find f_1 and f_2 where f_1*f_2 = T, and look for k from

k = (f_1 + f_2)/3

with 100% certainty that you will get k that will work for some factorization of T.

Example: Let q=5, N=19, then T = 10 mod 19, and T = 29, works with f_1=29 and f_2 = 1, giving

3k = 30 mod 19, and k=10.

10^2 = 5 mod 19.

The result follows from some fascinating yet trivial algebra.

Given a quadratic residue q modulo N where N is a non-unit natural number coprime to 3, where you wish to find k, where

k^2 = q mod N

you first find

T = 2q mod N

where T = 2q + jN, and j is a non-zero integer chosen such that T mod 3 = 2.

You next check each factorization of T to find f_1 and f_2 where f_1*f_2 = T, and look for k from

k = (f_1 + f_2)/3

with 100% certainty that you will get k that will work for some factorization of T.

Example: Let q=5, N=19, then T = 10 mod 19, and T = 29, works with f_1=29 and f_2 = 1, giving

3k = 30 mod 19, and k=10.

10^2 = 5 mod 19.

The result follows from some fascinating yet trivial algebra.

## JSH: Just math

Given z^2 = y^2 + T

where T is composite, and y and z are coprime integers, for any factorizations of T into two positive integers f_1 and f_2, where there exists an odd prime p less than both, or if f_1 is the smaller factor p-f_1 is less than f_1, where also exists an alpha such that

k^2 = (1 + alpha^2)^{-1}(T) mod p

then

z = (1 + 2alpha^2)k/(2alpha).

That is just math. It is absolutely perfect math, but it is just math.

The politics though are bigger than mathematical proof when it comes to acceptance.

I remember last year when I had that result, kind of vaguely though as to know when I had it I do searches in Google Groups and they indicate December 2007. It seemed so neat. Just this beautiful little result.

It's a deeper number theoretic structure underlying a familiar equation.

Maybe you thought you knew

z^2 = y^2 + T

but you didn't. With integers there is that deeper structure. I call it the z constraint.

A big deal if number theorists were real researchers, but they're fakes.

I now have all these nice research results. So many I forget what I have, as if it matters.

where T is composite, and y and z are coprime integers, for any factorizations of T into two positive integers f_1 and f_2, where there exists an odd prime p less than both, or if f_1 is the smaller factor p-f_1 is less than f_1, where also exists an alpha such that

k^2 = (1 + alpha^2)^{-1}(T) mod p

then

z = (1 + 2alpha^2)k/(2alpha).

That is just math. It is absolutely perfect math, but it is just math.

The politics though are bigger than mathematical proof when it comes to acceptance.

I remember last year when I had that result, kind of vaguely though as to know when I had it I do searches in Google Groups and they indicate December 2007. It seemed so neat. Just this beautiful little result.

It's a deeper number theoretic structure underlying a familiar equation.

Maybe you thought you knew

z^2 = y^2 + T

but you didn't. With integers there is that deeper structure. I call it the z constraint.

A big deal if number theorists were real researchers, but they're fakes.

I now have all these nice research results. So many I forget what I have, as if it matters.

## JSH: z constraint and solving quadratic residues

I started a thread on solving quadratic residues mod p, which is about a simple technique for finding k, when

k^2 = q mod p.

Remarkably the thread has degenerated so I thought I'd make another one noting some about where the research result comes from, which is what I call the z constraint:

Given z^2 = y^2 + T

where T is composite, and y and z are coprime integers, for any factorizations of T into two positive integers f_1 and f_2, where there exists an odd prime p less than both, or if f_1 is the smaller factor p-f_1 is less than f_1, where also exists an alpha such that

k^2 = (1 + alpha^2)^{-1}(T) mod p

then

z = (1 + 2alpha^2)k/(2alpha)

in general when T mod 3 = 2, and for some reason I had with 50% probability when it doesn't though I'm not sure where I got that from. But when T mod 3 = 2, it is 100%.

So I just creatively used those equations with alpha=1, and T=2q mod p, to get:

Given a quadratic residue q modulo p where p is an odd prime and not 3, where you wish to find k, where

k^2 = q mod p.

you start with T = 2q mod p

where T cannot equal 2q, so you'd start with T = 2q+p or T = 2q - p,

and next you have to factor it:

With f_1 and f_2 where f_1*f_2 = T,

you get k from

k = 3^{-1}(f_1 + f_2) mod p.

Example: Let q=5, p=19, then T = 10 mod 19, and T = 29, works with f_1=29 and f_2 = 1, giving

3k = 30 mod 19, and k=10.

10^2 = 5 mod 19.

The original result follows from research of mine which is often ridiculed that I find it hard to even mention it.

Posters, of course, blame me. When to me it is bizarre how easily the math community can block ANY mathematical research idea. Just look at the nothing happening with my solution to binary quadratic Diophantine equations. No debates about correctness. But it doesn't matter to be correct.

They just slap a nasty label on you and proof doesn't matter and THEY don't give a damn about the importance mathematically of the result.

Don't tell me it's just a matter of putting it out there either. I've had these results for about a year now both posted on newsgroups and on my math blog.

So I'm talking old research.

Just do a search on "z constraint".

It's like what they did to me with Fermat's Last Theorem: they could get points just by ridiculing even the CONCEPT of working on the problem as if it automatically proved I was a crackpot and had to be wrong.

Then, dismissing actual research finds was academic.

Mathematicians can block public knowledge of mathematical discovery, at will.

k^2 = q mod p.

Remarkably the thread has degenerated so I thought I'd make another one noting some about where the research result comes from, which is what I call the z constraint:

Given z^2 = y^2 + T

where T is composite, and y and z are coprime integers, for any factorizations of T into two positive integers f_1 and f_2, where there exists an odd prime p less than both, or if f_1 is the smaller factor p-f_1 is less than f_1, where also exists an alpha such that

k^2 = (1 + alpha^2)^{-1}(T) mod p

then

z = (1 + 2alpha^2)k/(2alpha)

in general when T mod 3 = 2, and for some reason I had with 50% probability when it doesn't though I'm not sure where I got that from. But when T mod 3 = 2, it is 100%.

So I just creatively used those equations with alpha=1, and T=2q mod p, to get:

Given a quadratic residue q modulo p where p is an odd prime and not 3, where you wish to find k, where

k^2 = q mod p.

you start with T = 2q mod p

where T cannot equal 2q, so you'd start with T = 2q+p or T = 2q - p,

and next you have to factor it:

With f_1 and f_2 where f_1*f_2 = T,

you get k from

k = 3^{-1}(f_1 + f_2) mod p.

Example: Let q=5, p=19, then T = 10 mod 19, and T = 29, works with f_1=29 and f_2 = 1, giving

3k = 30 mod 19, and k=10.

10^2 = 5 mod 19.

The original result follows from research of mine which is often ridiculed that I find it hard to even mention it.

Posters, of course, blame me. When to me it is bizarre how easily the math community can block ANY mathematical research idea. Just look at the nothing happening with my solution to binary quadratic Diophantine equations. No debates about correctness. But it doesn't matter to be correct.

They just slap a nasty label on you and proof doesn't matter and THEY don't give a damn about the importance mathematically of the result.

Don't tell me it's just a matter of putting it out there either. I've had these results for about a year now both posted on newsgroups and on my math blog.

So I'm talking old research.

Just do a search on "z constraint".

It's like what they did to me with Fermat's Last Theorem: they could get points just by ridiculing even the CONCEPT of working on the problem as if it automatically proved I was a crackpot and had to be wrong.

Then, dismissing actual research finds was academic.

Mathematicians can block public knowledge of mathematical discovery, at will.

### Saturday, November 08, 2008

## Solving quadratic residues mod p

Given a quadratic residue q modulo p where p is an odd prime and not 3, where you wish to find k, where

k^2 = q mod p.

you start with

T = 2q mod p

where T cannot equal 2q, so you'd start with T = 2q+p or T = 2q - p, and next you have to factor it:

With f_1 and f_2 where f_1*f_2 = T,

you get k from

k = 3^{-1}(f_1 + f_2) mod p

with a 50% probability that you will get the correct k.

Example: Let q=2, p=17 so T = 2(2) mod 17 = 4 mod 17.

I tried T = 21 and that didn't give me a positive k, and nothing did until T = 55, then f_1=5, f_2=11 gives

3k = 16 mod 17, so k = 11.

11^2 = 2 mod 17.

Here's another example, let q=5, p=19, then T = 10 mod 19, and T = 29, works with f_1=29 and f_2 = 1, giving

3k = 30 mod 19, and k=10.

10^2 = 5 mod 19.

So there can be something of a search but it is a smaller search than looping through squaring each residue and checking directly.

And these equations connect solving a quadratic residue modulo p squarely to integer factorization.

k^2 = q mod p.

you start with

T = 2q mod p

where T cannot equal 2q, so you'd start with T = 2q+p or T = 2q - p, and next you have to factor it:

With f_1 and f_2 where f_1*f_2 = T,

you get k from

k = 3^{-1}(f_1 + f_2) mod p

with a 50% probability that you will get the correct k.

Example: Let q=2, p=17 so T = 2(2) mod 17 = 4 mod 17.

I tried T = 21 and that didn't give me a positive k, and nothing did until T = 55, then f_1=5, f_2=11 gives

3k = 16 mod 17, so k = 11.

11^2 = 2 mod 17.

Here's another example, let q=5, p=19, then T = 10 mod 19, and T = 29, works with f_1=29 and f_2 = 1, giving

3k = 30 mod 19, and k=10.

10^2 = 5 mod 19.

So there can be something of a search but it is a smaller search than looping through squaring each residue and checking directly.

And these equations connect solving a quadratic residue modulo p squarely to integer factorization.

### Sunday, November 02, 2008

## JSH: So now you know

So yes, I've had 12 years to contemplate what's wrong with the math field and academia in general.

You can continue to look away, and look away from the approach to the spherical packing problem, and implications for materials science.

Look away from binary quadratic Diophantine equations solution.

Look away from DMESE and a way for people to legally copy their bought DVD's.

But you're no longer scientists.

If you do look away then the corruption has already gone too far and the human species is doomed.

I have a theory about why global warming is progressing so rapidly, it's that maybe our Universe on some level likes winners, and has concluded the human species is just one of Her failures.

So She is beginning the process of cleaning up, maybe starting over.

There may be millions of sentient species out there on worlds throughout just our galaxy where the beings there don't go down a dark tunnel and lose their investigative ability, lose their ability to continue scientific progress.

Worlds worth the effort. Worlds where truth still matters.

I may be the last major discoverer on this planet.

Born maybe to be the one witness who knows and understands, and, sorry, yes, ultimately agrees with the diagnosis.

The patient appears to be terminal. The cancer has progressed too far.

Prognosis for the species is, imminent extinction.

You can continue to look away, and look away from the approach to the spherical packing problem, and implications for materials science.

Look away from binary quadratic Diophantine equations solution.

Look away from DMESE and a way for people to legally copy their bought DVD's.

But you're no longer scientists.

If you do look away then the corruption has already gone too far and the human species is doomed.

I have a theory about why global warming is progressing so rapidly, it's that maybe our Universe on some level likes winners, and has concluded the human species is just one of Her failures.

So She is beginning the process of cleaning up, maybe starting over.

There may be millions of sentient species out there on worlds throughout just our galaxy where the beings there don't go down a dark tunnel and lose their investigative ability, lose their ability to continue scientific progress.

Worlds worth the effort. Worlds where truth still matters.

I may be the last major discoverer on this planet.

Born maybe to be the one witness who knows and understands, and, sorry, yes, ultimately agrees with the diagnosis.

The patient appears to be terminal. The cancer has progressed too far.

Prognosis for the species is, imminent extinction.

## JSH: How village idiots took over number theory

If you've followed the discussion at all where I've been arguing with sci.math'ers where things have now degenerated down to who introduced various terminology when, you may wonder how can such stupid people have killed a freaking math journal and blocked a major research result for years. Well I'm going to raise the stakes for you by showing how my research impacts physics and specifically materials science (and you may understand a wry joke) more directly.

Turns out I worked as a health physicist while in the U.S. Army so I have worked as a physicist though I only had a B.Sc. which is one of those great things about the military, you can work a job for which in the civilian world you'd be required to have a higher educational level. My official title was Nuclear Medical Science Officer.

In any event at one point I wondered if modern problem solving techniques might be useful against hard math problems with enough history that people might be in a rut following what was done before so I wanted to start from scratch. Yes it was arrogant but I was a young hotshot lieutenant. Why shouldn't I be arrogant?

And I picked TWO PROBLEMS. Not one. I picked TWO PROBLEMS.

I picked the spherical packing problem, and Fermat's Last Theorem.

The truly bizarre part of this story is, I solved them both.

I found simpler solutions for both, where I found the spherical packing problem solution, FIRST.

Google, in quotes: "spherical packing problem"

Then you can read my solution, or what I remember of it, as that was back in 1996. Over 12 years ago.

The approach I used could revolutionize materials science allowing scientists to kind of dial-up compounds and substances of whatever structure they wish and mathematically test for stability, and properties, from the computer.

It may be one of the greatest advances in human knowledge, ever.

So what happened to the mathematical field that such research could be blocked as they blocked my spherical packing problem solution FIRST, and instead trumpeted a complex claim of proof to the problem that I think may still be being checked by computer scientists looking over output from a complex computer program meant to solve it?

Here's the best explanation: after Gauss with gaussian integers, mathematicians came across the idea of using monic polynomials with integer coefficients, which they called algebraic integers.

Seemed like a good idea but the ring has a subtle flaw, which if you do not know it exists allows you to convincingly appear to prove things mathematically not true.

Maybe good mathematicians felt something was wrong, but the error allowed MORE people into the field who clearly did not. As time went by, a critical mass developed of people with no real mathematical ability, able to "prove" things completely false because of the subtle error.

Over a hundred years went by.

Today the mathematical field is overtaken by people who do not know how to really do mathematics.

The village idiots took over number theory because of this subtle error, which is why what they call "pure math" has never, and will never, be useful, for physicists.

It's not pure. It's just wrong.

They've moved from the mathematical field to trying to take over the physics field, which is why "string theory" is all the rage. If they succeed there can be a lot more "physicists", but what they do will be wrong, but they will deny it, like the "mathematicians" do today.

The village idiots took over number theory, and they're coming for the physics field next.

Turns out I worked as a health physicist while in the U.S. Army so I have worked as a physicist though I only had a B.Sc. which is one of those great things about the military, you can work a job for which in the civilian world you'd be required to have a higher educational level. My official title was Nuclear Medical Science Officer.

In any event at one point I wondered if modern problem solving techniques might be useful against hard math problems with enough history that people might be in a rut following what was done before so I wanted to start from scratch. Yes it was arrogant but I was a young hotshot lieutenant. Why shouldn't I be arrogant?

And I picked TWO PROBLEMS. Not one. I picked TWO PROBLEMS.

I picked the spherical packing problem, and Fermat's Last Theorem.

The truly bizarre part of this story is, I solved them both.

I found simpler solutions for both, where I found the spherical packing problem solution, FIRST.

Google, in quotes: "spherical packing problem"

Then you can read my solution, or what I remember of it, as that was back in 1996. Over 12 years ago.

The approach I used could revolutionize materials science allowing scientists to kind of dial-up compounds and substances of whatever structure they wish and mathematically test for stability, and properties, from the computer.

It may be one of the greatest advances in human knowledge, ever.

So what happened to the mathematical field that such research could be blocked as they blocked my spherical packing problem solution FIRST, and instead trumpeted a complex claim of proof to the problem that I think may still be being checked by computer scientists looking over output from a complex computer program meant to solve it?

Here's the best explanation: after Gauss with gaussian integers, mathematicians came across the idea of using monic polynomials with integer coefficients, which they called algebraic integers.

Seemed like a good idea but the ring has a subtle flaw, which if you do not know it exists allows you to convincingly appear to prove things mathematically not true.

Maybe good mathematicians felt something was wrong, but the error allowed MORE people into the field who clearly did not. As time went by, a critical mass developed of people with no real mathematical ability, able to "prove" things completely false because of the subtle error.

Over a hundred years went by.

Today the mathematical field is overtaken by people who do not know how to really do mathematics.

The village idiots took over number theory because of this subtle error, which is why what they call "pure math" has never, and will never, be useful, for physicists.

It's not pure. It's just wrong.

They've moved from the mathematical field to trying to take over the physics field, which is why "string theory" is all the rage. If they succeed there can be a lot more "physicists", but what they do will be wrong, but they will deny it, like the "mathematicians" do today.

The village idiots took over number theory, and they're coming for the physics field next.

### Saturday, November 01, 2008

## JSH: Peer reviewed and published

So yeah, I had this result I'm pushing so hard formally peer reviewed and published in the mathematical journal SWJPAM, which stands for Southwest Journal of Pure and Applied Mathematics.

When it happened a poster on sci.physics opined that now you'd never hear the end of it.

Yeah, as if the rules matter.

You people put a label like "crackpot" on someone and then feel free to watch special treatment and special behavior outside the rules no matter what they can prove or how far they get within the rules.

But you're screwing yourselves as well.

My point then has been that the formal peer review system has clearly shown itself to be terribly flawed.

It is fallible to group opinion against a major result that is not liked by the majority.

And academics in other disciplines cannot be trusted to come over into another area even when a case is dramatic to the extreme: a dead math journal with a result that can be simplified to basic algebra which shows an error going back over a hundred years.

I want to break the modern academic system entirely.

The point here is to give me the power of full reform across the board against any discipline, so not only mathematics or even physics but Literature, or Biology, or even Medicine.

I need the full authority to shut down entire programs. End entire journals. And do necessary things like destroy tenure.

THAT is the point of the drama.

THAT is the reason for the energy.

And THAT is what will happen.

You are empowering me to be one of the most powerful men in the history of the human race by your refusal to do the right thing, and I will take that power for the good of the human species.

Thank you very much. And goddamn you to hell because I hate this situation, but I will do what it takes so that it never happens again.

When it happened a poster on sci.physics opined that now you'd never hear the end of it.

Yeah, as if the rules matter.

You people put a label like "crackpot" on someone and then feel free to watch special treatment and special behavior outside the rules no matter what they can prove or how far they get within the rules.

But you're screwing yourselves as well.

My point then has been that the formal peer review system has clearly shown itself to be terribly flawed.

It is fallible to group opinion against a major result that is not liked by the majority.

And academics in other disciplines cannot be trusted to come over into another area even when a case is dramatic to the extreme: a dead math journal with a result that can be simplified to basic algebra which shows an error going back over a hundred years.

I want to break the modern academic system entirely.

The point here is to give me the power of full reform across the board against any discipline, so not only mathematics or even physics but Literature, or Biology, or even Medicine.

I need the full authority to shut down entire programs. End entire journals. And do necessary things like destroy tenure.

THAT is the point of the drama.

THAT is the reason for the energy.

And THAT is what will happen.

You are empowering me to be one of the most powerful men in the history of the human race by your refusal to do the right thing, and I will take that power for the good of the human species.

Thank you very much. And goddamn you to hell because I hate this situation, but I will do what it takes so that it never happens again.

## JSH: Pity the poor math people

So the irony is that I found the problem with my proof of Fermat's Last Theorem. Posters argued with me over the details of that proof so I focused on the key area where the arguments raged, expanded it out and wrote a paper.

The paper got published. The journal retracted after an email campaign by sci.math'ers and the journal died with a rain of abuse hurled at it from the sci.math newsgroup.

But that was just one piece of the full proof of Fermat's Last Theorem.

I kept working at it as I worked to understand the result and I began a steady simplification.

Pity the poor math people.

They were fighting a result that could be simplified and simplified until now I can demonstrate it with a few lines:

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

and get to

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

Pity the poor math people.

They killed one of their own math journals for a few years of grace from the truth, if you call it grace.

For a few more years to teach naive young minds.

Pity the poor math people, and the horror that they wrought upon the world.

Pity them and then pity yourself for they robbed us all.

The paper got published. The journal retracted after an email campaign by sci.math'ers and the journal died with a rain of abuse hurled at it from the sci.math newsgroup.

But that was just one piece of the full proof of Fermat's Last Theorem.

I kept working at it as I worked to understand the result and I began a steady simplification.

Pity the poor math people.

They were fighting a result that could be simplified and simplified until now I can demonstrate it with a few lines:

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

and get to

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

Pity the poor math people.

They killed one of their own math journals for a few years of grace from the truth, if you call it grace.

For a few more years to teach naive young minds.

Pity the poor math people, and the horror that they wrought upon the world.

Pity them and then pity yourself for they robbed us all.

## JSH: Just consider Andrew Wiles

I can easily see mathematicians doggedly walking through a rain of spit to keep teaching the wrong ideas if they can get to their offices, close the door, and continue the fantasy, as what is the alternative?

Can you imagine the horror of the position of say, Andrew Wiles if he accepts what follows from my simple demonstration?

It's almost too cruel to comprehend, a man who spent decades of his life in pursuit of a dream, who had accolades and prestige heaped upon him, to see it all not only yanked away, but shown to never have been based on any real accomplishment.

Why wouldn't he fight a result like this one? (Not saying he personally is fighting it.)

Remember key facts: the paper that killed the journal SWJPAM went by Barry Mazur before. Also, I tried to get it published at a journal where Ralph McKenzie was editor because I saw he was at my alma mater Vanderbilt University. He claimed to not understand it when I was just using email, until I forwarded him Mazur's email, and then he offered that I should explain it in person, so I traveled to Nashville, and explained.

The journal SWJPAM had my paper for nine months. I corrected them in email conversations from calling me "Dr. Harris" as I noted that I was an amateur researcher. After publication one of the editors in response to my profuse gratitude said that all that mattered was the correctness of the result.

Yet they caved a little while later when some sci.math'ers mounted that crafty email campaign.

The denial here has nothing to do with what I can prove mathematically but all about human weakness.

Would you want to be Andrew Wiles now?

If any of you were inspired by his story, can you imagine being him now?

If you were him, would you tell the truth if you read about my simple demonstration showing the problem with the ring of algebraic integers?

Or would you prefer to just go one more day, as many days as you could where you were a hero, known as one of the greatest mathematicians in history? Just one more day? Just one more day? Just one more day?

As, what do you do the day after the truth is known?

How do you pick up the pieces then? How do you even define yourself in this world?

If precedent is followed then this post may be picked up increasingly if people bother searching on "Andrew Wiles".

The world may absorb the rest of the story whatever the mathematical community does to try and block the truth and hold on to the fantasy.

History will catch up to you.

Can you imagine the horror of the position of say, Andrew Wiles if he accepts what follows from my simple demonstration?

It's almost too cruel to comprehend, a man who spent decades of his life in pursuit of a dream, who had accolades and prestige heaped upon him, to see it all not only yanked away, but shown to never have been based on any real accomplishment.

Why wouldn't he fight a result like this one? (Not saying he personally is fighting it.)

Remember key facts: the paper that killed the journal SWJPAM went by Barry Mazur before. Also, I tried to get it published at a journal where Ralph McKenzie was editor because I saw he was at my alma mater Vanderbilt University. He claimed to not understand it when I was just using email, until I forwarded him Mazur's email, and then he offered that I should explain it in person, so I traveled to Nashville, and explained.

The journal SWJPAM had my paper for nine months. I corrected them in email conversations from calling me "Dr. Harris" as I noted that I was an amateur researcher. After publication one of the editors in response to my profuse gratitude said that all that mattered was the correctness of the result.

Yet they caved a little while later when some sci.math'ers mounted that crafty email campaign.

The denial here has nothing to do with what I can prove mathematically but all about human weakness.

Would you want to be Andrew Wiles now?

If any of you were inspired by his story, can you imagine being him now?

If you were him, would you tell the truth if you read about my simple demonstration showing the problem with the ring of algebraic integers?

Or would you prefer to just go one more day, as many days as you could where you were a hero, known as one of the greatest mathematicians in history? Just one more day? Just one more day? Just one more day?

As, what do you do the day after the truth is known?

How do you pick up the pieces then? How do you even define yourself in this world?

If precedent is followed then this post may be picked up increasingly if people bother searching on "Andrew Wiles".

The world may absorb the rest of the story whatever the mathematical community does to try and block the truth and hold on to the fantasy.

History will catch up to you.

## JSH: So yes, it is simple, what now?

I think I've finally stumbled across an irrefutable way to demonstrate that the ring of algebraic integers is severely flawed. You can try to be in that ring for it.

Given

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

Step through getting the factorization:

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

So that you know it CAN be done, algebraically, I'll give you the primary way.

Multiply both sides by 7 in a particular way which is my controversial way:

7*(175x^2 - 15x + 2) = (5b_1(x) + 2)(5*7*b_2(x)+ 7)

and now use the substitutions b_1(x) = a_1(x) + 1, and 7*b_2(x) = a_2(x), to get

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

and you're done. Note that b_1(x) and b_2(x) are given by this method. That is also a requirement for any answers from the mathematical community: give b_1(x) and b_2(x).

Those of you who try to handle the example will find you can't get b_1(x) and b_2(x) in general that are algebraic integers such that

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

AND give b_1(x) and b_2(x), as the algebra cannot let you, as if it could, then you could just step through the route I take, and you'd have

7*b_2(x) = a_2(x)

and a contradiction with the ring of algebraic integers.

The rational explanation is that the definition of algebraic integers as roots of monic polynomials with integer coefficients chops off numbers that NEED to be in the ring to prevent apparent contradiction.

The result above is so devastating that you can use it to prove that 7 is a factor of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

algebraically AND prove that 7 cannot be a factor of either root if you are irreducible over Q IN THE RING OF ALGEBRAIC INTEGERS.

So you can do the mathematically verboten thing of proving two contradictory things, and I showed that several years ago in a paper published in the mathematical journal SWJPAM.

Now I DID pass the paper by Barry Mazur first. He never claimed an error with it, but just asked some general questions. I DID explain non-polynomial factorization in person to Ralph McKenzie at my alma mater Vanderbilt University, having traveled there to be able to speak to him in-person.

So the paper was fully vetted before it went to SWJPAM. And they had it for nine months, and I was in continual contact with two of the editors up until it was published, so it was not mistakenly published.

For years now posters have routinely lied, misrepresented or made up details.

The reason is that the result is so crushing for people established in the field, but less so for undergrads, so what they are doing I think, is trying to pull you into their nightmare: get you to invest years or even decades of your life with the flawed mathematical ideas and be as dedicated as them in preserving them.

They're fighting for their sense of self and dignity. They're fighting to keep believing they're brilliant mathematicians.

And they're trying to sacrifice you to do it.

Don't just believe me. Do the math. Try the demonstration I've mentioned. Read around. Do the research and face the truth.

You are some of the luckiest undergrads ever as if you choose the right path then for the first time in a over a hundred years, math students turned from the dark side with these number theory results.

Life is hard. Mathematics is often harder. Only the best will do. Only the truly best will do the right thing.

Follow the math.

Given

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

Step through getting the factorization:

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

So that you know it CAN be done, algebraically, I'll give you the primary way.

Multiply both sides by 7 in a particular way which is my controversial way:

7*(175x^2 - 15x + 2) = (5b_1(x) + 2)(5*7*b_2(x)+ 7)

and now use the substitutions b_1(x) = a_1(x) + 1, and 7*b_2(x) = a_2(x), to get

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

and you're done. Note that b_1(x) and b_2(x) are given by this method. That is also a requirement for any answers from the mathematical community: give b_1(x) and b_2(x).

Those of you who try to handle the example will find you can't get b_1(x) and b_2(x) in general that are algebraic integers such that

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

AND give b_1(x) and b_2(x), as the algebra cannot let you, as if it could, then you could just step through the route I take, and you'd have

7*b_2(x) = a_2(x)

and a contradiction with the ring of algebraic integers.

The rational explanation is that the definition of algebraic integers as roots of monic polynomials with integer coefficients chops off numbers that NEED to be in the ring to prevent apparent contradiction.

The result above is so devastating that you can use it to prove that 7 is a factor of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

algebraically AND prove that 7 cannot be a factor of either root if you are irreducible over Q IN THE RING OF ALGEBRAIC INTEGERS.

So you can do the mathematically verboten thing of proving two contradictory things, and I showed that several years ago in a paper published in the mathematical journal SWJPAM.

Now I DID pass the paper by Barry Mazur first. He never claimed an error with it, but just asked some general questions. I DID explain non-polynomial factorization in person to Ralph McKenzie at my alma mater Vanderbilt University, having traveled there to be able to speak to him in-person.

So the paper was fully vetted before it went to SWJPAM. And they had it for nine months, and I was in continual contact with two of the editors up until it was published, so it was not mistakenly published.

For years now posters have routinely lied, misrepresented or made up details.

The reason is that the result is so crushing for people established in the field, but less so for undergrads, so what they are doing I think, is trying to pull you into their nightmare: get you to invest years or even decades of your life with the flawed mathematical ideas and be as dedicated as them in preserving them.

They're fighting for their sense of self and dignity. They're fighting to keep believing they're brilliant mathematicians.

And they're trying to sacrifice you to do it.

Don't just believe me. Do the math. Try the demonstration I've mentioned. Read around. Do the research and face the truth.

You are some of the luckiest undergrads ever as if you choose the right path then for the first time in a over a hundred years, math students turned from the dark side with these number theory results.

Life is hard. Mathematics is often harder. Only the best will do. Only the truly best will do the right thing.

Follow the math.