### Monday, June 30, 2008

## JSH: Measuring newsgroup impact

Now that I'm winding down from new math research I'm focusing more on side details like measuring the impact of posts on the newsgroups on things like statistics of hit counts on my math blog, and I thought it worth sharing some of the information.

For instance, in the past 6 months according to Google Analytics my math blog has hits from 80 countries, though the bulk of the hits come from Great Britain, with the city giving the most hits being London.

According to Google Analytics, the hits coming from what I'll call hostile sources, being the crank.net website and newsgroups through groups.google.com are 10.12% of the total, where crank.net has 6.02% and groups.google.com has 4.10% which is invaluable information to me because it has shown me that hostile posts here have nominal impact in terms of actually driving people from links on negative sources to my math blog.

Which is why I'm posting as I think it's a hoot.

Hostile posters who I've seen at times making it a point to link to negative sites have had a nominal impact based on the evidence I have available, and thankfully, crank.net is a non-force in terms of hit counts.

The hate page is not getting the attention I'm sure it's creator intended, so now it's more of a humorous thing to me to note that and consider what it means about the Internet and how interest is drawn or not drawn by particular tactics.

On a side note, I continue to use search results from the major search engines to gain some sense of my overall worldwide influence, and as it has every year, it is growing.

I now consider the academic journal process to be irrelevant to my needs and it is no longer necessary for me to even consider publication in any math journal at any level.

I can simply go direct, publishing as I choose, what I choose, when I choose.

[A reply to someone who wrote that everybody should be allowed to publish trash, and that James provides the perfect example.]

So does a lot of math society…

You people don't get it yet, eh?

I go around you, and over time prove to people I'm right, then it's not just fun and games in math academia any more.

And now I can go around you.

That's the message.

Yes, it's taking a little time, but if in a couple of years it's over, then what?

When you can't publish the same old stuff through the same old channels. When the Internet is chasing down every bit of wrong math that mathematicians push—no matter what their degree or their status in the community.

When the funding, stops.

Remember, my end goal is the same: remove your public financing.

Then you can come here and chatter all you want, but you will not be paid with taxpayer dollars and I don't care if you believe it or not, but the message to undergrads is, you may get your degree, get into a great program and find you do not have any money.

Some of you may graduate just in time to have no money available.

For instance, in the past 6 months according to Google Analytics my math blog has hits from 80 countries, though the bulk of the hits come from Great Britain, with the city giving the most hits being London.

According to Google Analytics, the hits coming from what I'll call hostile sources, being the crank.net website and newsgroups through groups.google.com are 10.12% of the total, where crank.net has 6.02% and groups.google.com has 4.10% which is invaluable information to me because it has shown me that hostile posts here have nominal impact in terms of actually driving people from links on negative sources to my math blog.

Which is why I'm posting as I think it's a hoot.

Hostile posters who I've seen at times making it a point to link to negative sites have had a nominal impact based on the evidence I have available, and thankfully, crank.net is a non-force in terms of hit counts.

The hate page is not getting the attention I'm sure it's creator intended, so now it's more of a humorous thing to me to note that and consider what it means about the Internet and how interest is drawn or not drawn by particular tactics.

On a side note, I continue to use search results from the major search engines to gain some sense of my overall worldwide influence, and as it has every year, it is growing.

I now consider the academic journal process to be irrelevant to my needs and it is no longer necessary for me to even consider publication in any math journal at any level.

I can simply go direct, publishing as I choose, what I choose, when I choose.

[A reply to someone who wrote that everybody should be allowed to publish trash, and that James provides the perfect example.]

So does a lot of math society…

You people don't get it yet, eh?

I go around you, and over time prove to people I'm right, then it's not just fun and games in math academia any more.

And now I can go around you.

That's the message.

Yes, it's taking a little time, but if in a couple of years it's over, then what?

When you can't publish the same old stuff through the same old channels. When the Internet is chasing down every bit of wrong math that mathematicians push—no matter what their degree or their status in the community.

When the funding, stops.

Remember, my end goal is the same: remove your public financing.

Then you can come here and chatter all you want, but you will not be paid with taxpayer dollars and I don't care if you believe it or not, but the message to undergrads is, you may get your degree, get into a great program and find you do not have any money.

Some of you may graduate just in time to have no money available.

### Sunday, June 22, 2008

## JSH: Probabilistic quadratic residue solving

An off-shoot of my surrogate factoring research is a probabilistic method to solve quadratic residues, as given

k^2 = q mod p

when p is an odd prime, and q is a quadratic residue modulo p, you find k.

The technique requires introduction of a few additional variables starting with T, where

T = 2q + np

where n is an odd natural number so you have T = 2q mod p, but T - 2q is also forced to have p as a factor.

Next you find z, where with integer factors f_1 and f_2 where f_1*f_2 = T:

z = (f_1 + f_2)/2

and now finally you try for an answer for k, with

k = 3^{-1}(2z) mod p.

The method is probabilistic because if I've got the analysis right you have a 50% probability of getting the right k, for each z that you try. Checking is done by looking at k^2 mod p, to see if you get q.

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

Here T=21 does not work, but T = 55 = 5(11), so z = 8 and the answer then from

3k = 2(8) mod 17, is k = 11 mod 17.

Is there any use for such a technique?

k^2 = q mod p

when p is an odd prime, and q is a quadratic residue modulo p, you find k.

The technique requires introduction of a few additional variables starting with T, where

T = 2q + np

where n is an odd natural number so you have T = 2q mod p, but T - 2q is also forced to have p as a factor.

Next you find z, where with integer factors f_1 and f_2 where f_1*f_2 = T:

z = (f_1 + f_2)/2

and now finally you try for an answer for k, with

k = 3^{-1}(2z) mod p.

The method is probabilistic because if I've got the analysis right you have a 50% probability of getting the right k, for each z that you try. Checking is done by looking at k^2 mod p, to see if you get q.

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

Here T=21 does not work, but T = 55 = 5(11), so z = 8 and the answer then from

3k = 2(8) mod 17, is k = 11 mod 17.

Is there any use for such a technique?

### Sunday, June 15, 2008

## JSH: Diminishing concerns on factoring

For years I've had a major fear that if I had a major breakthrough with integer factorization it might have a very bad, negative impact on the global economy but now as I finish out the foundation layer of that research it is clear that my research is not impacting the world. Yeah, the world is having problems but those are its own.

Ok, so what do I mean about the foundation layer?

Well the simplest way to consider what my factoring research does is that while mathematicians have traditionally focused on one type of congruence I use two:

A little while back rather than consider S as a whole I started focusing on its prime factors, which is why I have a lot of equations with

x^2 = y^2 mod p

as p is an odd prime factor of S.

That was a HUGE breakthrough to move from concentrating on S itself, where I had control variables I called n, alpha and k, and would puzzle over things like what was the way to pick k, to figuring out how to get p.

And with the fundamental equations now known that completes the foundation layer of surrogate factoring, and removes the need for further brainstorming at this point, so there is no need for me to discuss further.

Succinctly, the basic research phase on the primary layer is complete.

World seems ok, as my worst fears haven't realized, so I'm kicking back, relaxing and moving to other problems as a MAJOR issue right now is monetization of the web, or so-called Web 2.0 as it's coming up as a bigger and bigger problem to get revenue flowing for creative people around the world, and my current focus is on the biggest player, so YouTube is the focus of problem solving efforts and that will take up most of my time indefinitely.

Math research is done for this iteration.

Ok, so what do I mean about the foundation layer?

Well the simplest way to consider what my factoring research does is that while mathematicians have traditionally focused on one type of congruence I use two:

- x^2 = y^2 mod S
- z^2 = y^2 mod T

A little while back rather than consider S as a whole I started focusing on its prime factors, which is why I have a lot of equations with

x^2 = y^2 mod p

as p is an odd prime factor of S.

That was a HUGE breakthrough to move from concentrating on S itself, where I had control variables I called n, alpha and k, and would puzzle over things like what was the way to pick k, to figuring out how to get p.

And with the fundamental equations now known that completes the foundation layer of surrogate factoring, and removes the need for further brainstorming at this point, so there is no need for me to discuss further.

Succinctly, the basic research phase on the primary layer is complete.

World seems ok, as my worst fears haven't realized, so I'm kicking back, relaxing and moving to other problems as a MAJOR issue right now is monetization of the web, or so-called Web 2.0 as it's coming up as a bigger and bigger problem to get revenue flowing for creative people around the world, and my current focus is on the biggest player, so YouTube is the focus of problem solving efforts and that will take up most of my time indefinitely.

Math research is done for this iteration.

### Monday, June 09, 2008

## JSH: Wow, cool factoring relations

There are times that I just wonder why I'm the only person who seems to really get it that it is just so incredible to try and figure these things out and it's not just about some people supposedly being the only ones with the training or whatever. Like check these out:

With a target composite T, I find 'a' by the use of k, where

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

with p an odd prime, and c is a natural number of arbitrary size.

Then, incredibly, I have that with factors f_1 and f_2, where f_1*f_2 = T:

f_1 = ak mod p^c

and

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

One of the reasons they work is that

a^2 = f_1(f_2 - f_1)^{-1} mod p^c

and in fact, no matter what, if you get an 'a' for a k, then you have a factorization of T, but you can get non-rational ones, which is where I can end up in arguments with people who'd still like to call the relations useless.

But there is so much simple beauty in them, how (T - k^2) has to be a quadratic residue. How 'a' is related to the factors through a prime.

Knowledge unknown to the world before now though it's not new. It's always been there.

With a target composite T, I find 'a' by the use of k, where

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

with p an odd prime, and c is a natural number of arbitrary size.

Then, incredibly, I have that with factors f_1 and f_2, where f_1*f_2 = T:

f_1 = ak mod p^c

and

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

One of the reasons they work is that

a^2 = f_1(f_2 - f_1)^{-1} mod p^c

and in fact, no matter what, if you get an 'a' for a k, then you have a factorization of T, but you can get non-rational ones, which is where I can end up in arguments with people who'd still like to call the relations useless.

But there is so much simple beauty in them, how (T - k^2) has to be a quadratic residue. How 'a' is related to the factors through a prime.

Knowledge unknown to the world before now though it's not new. It's always been there.

### Sunday, June 08, 2008

## JSH: Newest factoring algorithm

After my latest miss-steps I went back, looked everything over again and finally noticed something that was bizarrely, simply obvious and staring me in the face.

Here's the factoring algorithm that results.

Given an odd target composite T, for a test odd prime p, iterate through non-zero integers k until

T - k^2 mod p

is a quadratic residue. If no k will work, discard that prime p and try another. If one does work then get á from

á^2 = (T - k^2)(k^2)^{-1} mod p.

Then you have residues of your factors modulo p from

f_1 = ák mod p

and

f_2 = á^{-1}(1 + á^2)k mod p

where f_1*f_2 = T.

Trivial factorizations if they are given can be discarded by noting that your residues is the same as T, though one caveat is that for a given prime p, one of your factors may equal the residue of the target composite T.

Also if with r^2 = T - k^2 mod p, r^2 = k^2, then that result should be discarded, though it may indicate that one of your factors equals the residue of T for that particular prime.

The sign of the correct k and á that work seems to correlate with the sign of the factors as long as neither factor is less than p.

The simplification occurred to me after noticing a few obvious things. It should work well though I guess maybe I'm still missing something so out it goes to see if someone will test it.

If it does work well then factoring a target composite T is just about getting enough primes that will work.

For example, if 80 primes are found then a target at least as large as 80! would topple even if you took primes starting at 3 and working your way up, so assuming about 50% would work then using only primes under 1000, you could factor a number greater than 7.1569457046263802294811533723187e+118.

Here's the factoring algorithm that results.

Given an odd target composite T, for a test odd prime p, iterate through non-zero integers k until

T - k^2 mod p

is a quadratic residue. If no k will work, discard that prime p and try another. If one does work then get á from

á^2 = (T - k^2)(k^2)^{-1} mod p.

Then you have residues of your factors modulo p from

f_1 = ák mod p

and

f_2 = á^{-1}(1 + á^2)k mod p

where f_1*f_2 = T.

Trivial factorizations if they are given can be discarded by noting that your residues is the same as T, though one caveat is that for a given prime p, one of your factors may equal the residue of the target composite T.

Also if with r^2 = T - k^2 mod p, r^2 = k^2, then that result should be discarded, though it may indicate that one of your factors equals the residue of T for that particular prime.

The sign of the correct k and á that work seems to correlate with the sign of the factors as long as neither factor is less than p.

The simplification occurred to me after noticing a few obvious things. It should work well though I guess maybe I'm still missing something so out it goes to see if someone will test it.

If it does work well then factoring a target composite T is just about getting enough primes that will work.

For example, if 80 primes are found then a target at least as large as 80! would topple even if you took primes starting at 3 and working your way up, so assuming about 50% would work then using only primes under 1000, you could factor a number greater than 7.1569457046263802294811533723187e+118.

### Saturday, June 07, 2008

## JSH: Some basics about factoring

It occurs to me that some of you may not understand how my research changes the landscape with factoring so I want to explain simply and give some factoring basics.

First off, if you have z^2 = y^2 + nT

where T is your target to factor, and you also have z mod p_1, z mod p_2, and z mod p_3, where z is for a non-trivial factorization and those are odd primes with a minimum value greater than 2sqrt(nT)/3, then necessarily you can calculate z exactly and with z you can factor nT from

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

That is just an absolute in terms of basic algebra.

So it IS a big deal for me to present equations that allow you to just calculate z mod p.

Most modern factoring methods in some way or another use an equation like z^2 = y^2 + nT, or more familiarly you often have x^2 = y^2 mod N, where N is the target composite, so it's just about variable names and if you know any math at all you know that shifting letters is not a big deal.

Even the Number Field Sieve is a lot about using x^2 = y^2 mod N, as in trying to find x and y (I think it uses two congruences of that type), so the result I have has implications for the most advanced factoring techniques known.

But, you may then naturally wonder, if it's such a big deal to find z mod p, then how can it be something argued out on newsgroups without experts in the field caring?

One simple possible answer to that question is that I must be wrong. REMEMBER, if p_1, p_2 and p_3 can be found in the size range necessary then it is an ABSOLUTE that you can factor non-trivially.

I claim to have a method that gives z mod p, so if that claim is correct and you can get z mod p in the necessary range for just three prime numbers then ABSOLUTELY you will factor non-trivially.

So theory says one thing, absolutely. Where notice I still haven't answered the question of whether or not I must be wrong.

Well, there's the derivation which you can look over, and there is doing examples and you might wonder if maybe with a big target composite T, maybe it IS really hard to find odd primes p that will work, and you can muddle along with those questions believing there must be something wrong somewhere or top people in the field would acknowledge this result!

I think this situation for some of you is a test of your trust in people versus your trust in mathematics and it's probably not fair, but I think some of you wrongly believe that you have mathematical ability, when you do not.

Short of it is that how the newsgroups react doesn't matter. No matter what if the research is viable that will be known and probably in a rather short amount of time as we have a world today that consumes information. But what you cannot forget later, or I don't want you to forget it, is if you couldn't resolve the issue on your own despite the algebra being easy and the problem being hugely significant.

As if you cannot evaluate easy algebra and get the right answer when it's handed to you because you're waiting on some other people or trusting that someone else out there has the judgment you need, then you are NOT a mathematician, no matter what you tell yourself when you look in the mirror.

You are then a social person who relies on other people who really do know mathematics to tell you what is true or not.

First off, if you have z^2 = y^2 + nT

where T is your target to factor, and you also have z mod p_1, z mod p_2, and z mod p_3, where z is for a non-trivial factorization and those are odd primes with a minimum value greater than 2sqrt(nT)/3, then necessarily you can calculate z exactly and with z you can factor nT from

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

That is just an absolute in terms of basic algebra.

So it IS a big deal for me to present equations that allow you to just calculate z mod p.

Most modern factoring methods in some way or another use an equation like z^2 = y^2 + nT, or more familiarly you often have x^2 = y^2 mod N, where N is the target composite, so it's just about variable names and if you know any math at all you know that shifting letters is not a big deal.

Even the Number Field Sieve is a lot about using x^2 = y^2 mod N, as in trying to find x and y (I think it uses two congruences of that type), so the result I have has implications for the most advanced factoring techniques known.

But, you may then naturally wonder, if it's such a big deal to find z mod p, then how can it be something argued out on newsgroups without experts in the field caring?

One simple possible answer to that question is that I must be wrong. REMEMBER, if p_1, p_2 and p_3 can be found in the size range necessary then it is an ABSOLUTE that you can factor non-trivially.

I claim to have a method that gives z mod p, so if that claim is correct and you can get z mod p in the necessary range for just three prime numbers then ABSOLUTELY you will factor non-trivially.

So theory says one thing, absolutely. Where notice I still haven't answered the question of whether or not I must be wrong.

Well, there's the derivation which you can look over, and there is doing examples and you might wonder if maybe with a big target composite T, maybe it IS really hard to find odd primes p that will work, and you can muddle along with those questions believing there must be something wrong somewhere or top people in the field would acknowledge this result!

I think this situation for some of you is a test of your trust in people versus your trust in mathematics and it's probably not fair, but I think some of you wrongly believe that you have mathematical ability, when you do not.

Short of it is that how the newsgroups react doesn't matter. No matter what if the research is viable that will be known and probably in a rather short amount of time as we have a world today that consumes information. But what you cannot forget later, or I don't want you to forget it, is if you couldn't resolve the issue on your own despite the algebra being easy and the problem being hugely significant.

As if you cannot evaluate easy algebra and get the right answer when it's handed to you because you're waiting on some other people or trusting that someone else out there has the judgment you need, then you are NOT a mathematician, no matter what you tell yourself when you look in the mirror.

You are then a social person who relies on other people who really do know mathematics to tell you what is true or not.

### Friday, June 06, 2008

## JSH: Iterative process, factoring

To me it's just such a remarkable journey to do basic mathematical research where you start with a concept and look to figure out the mathematics that says the concept is viable or it's not. And today we have advantages that just were not on this planet before. Modern problem solving techniques are just so much more powerful than what was known in the past when researchers were often more like hobbyists like Fermat or introducing new systems like Newton, whereas today, the science and art of problem solving is fully matured.

It only makes sense that people using the best techniques of our modern era can figure out extraordinary things.

I've been trained to use iterative processes and brainstorming so the oddity of the posting style that I use is easily explained as throwing ideas out there as part of brainstorming, and there is usually a progression through a fairly structured process to each iteration.

With factoring after considering what was previously known I asked myself, can you maybe factor one number by using the factorization of some other number?

And that was just a question, which became a concept I called surrogate factoring and that start was over 4 years ago.

I've talked to people who are amazed that you can work on a math problem for years, but for readers on these forums it should not, of course, sound all that remarkable. What is I think remarkable though is the simple answer that years of iterations, with lots of brainstorming, so many posts, so many false starts and so much analysis has brought:

To solve for positive integer x, when x^2 = y^2 mod N, where N is an odd target composite to be factored you can use helper primes to solve the explicit equation:

x^2 = y^2 + cN

where c is an integer chosen to force cN mod 3 = 2, where for each prime that works you will have

x^2 = 8^{-1}(9cN) mod p

where p is an odd prime that is less than 2x/3 for which the quadratic residue exists, but further it must be true that the residue given by a prime in that range for which a quadratic residue exists must give an x near a value I call x_0 where x_0 is the largest positive integer divisible by 3 such that

abs(cN - 8x_0^2/9)

is a minimum, and x must be within x_0/(3p) positive steps from x_0.

That the question of whether or not one factorization can lead to another then has a surprising answer which represents the outcome of over four years of basic research.

Weird thing is that the derivation of

x^2 = 8^{-1}(9cN) mod p

is easy to the point of trivial so I've presented it a lot (see other threads), while the other rules are harder to explain so you'd have to dig a bit to understand where they come from, where that last part about x_0 is rather remarkable in terms of the why of the constraint.

I will say a few words about how I do problem solving which typically does include some insults in a post, often lambasting the mathematical community, which has a lot to do with motivating replies to my posts so that I can get other people looking over mathematical ideas. Years of experience on newsgroups has taught me that conflict draws readers and without it, you don't get readers at the same level.

Maybe unfortunate but it is the reality that I've found to be true in my experiments with what works and what does not as that has been an iterative process and a lot about problem solving as well.

If some other technique worked better, I'd use it.

[A reply to someone who suggested that James should write some code.]

I've written code before that factors numbers though not very large as I think it was usually right in the 40 bit range. For people who wonder what 40 bits is, it's numbers larger than 2^40 or 1099511627776. Those are considered small for factoring.

My experience is that posters deny any and all evidence or just keep raising the bar until the request is to demonstrate factoring better than anything else previously known.

The behavior as I've noted before is like saying that unless Albert Einstein built a working atomic bomb, and exploded it, his research was garbage.

The theory guy doesn't have to be the implementing guy or team.

Sometimes you just need to open the door and reasonable people can see value in your research effort, as we live in a very connected world. If there is anything to this research I'm sure there will be people building on it, which frees me up to consider theoretical issues.

[A reply to someone who told James that Einstein never claimed that he could build an atomic bomb.]

Scaling a solution can present challenges that have nothing to do with whether or not the theoretical solution is correct.

Mathematical theory says I've solved the factoring problem, like an interpretation of Einstein's research (along with that of others) indicated there were nuclear applications available.

But producing the practical product from theory can be a different enterprise.

The mathematical theory is about mathematical proof, which says the problem is solved.

Disbelief in mathematical proof can lead to a request for demonstration as if only seeing is believing.

Real mathematicians, however, know that mathematical proof means that the thing can be done.

You people could do something more useful and attack the claim of proof versus asking for a demonstration as if only an example can prove a proof, which displays a basic ignorance of mathematics, as examples can be nice but proof is proof.

[A reply to someone who explained James that no barrier prevents him from implementing of a solution to the factoring problem.]

I don't have to factor anything.

Of one thing there is little doubt in my mind it's if I'm right then someone will do the grunt work.

Now if you deny that then you're insane.

Why argue?

If I have solved the factoring problem and I say theory proves it, and I'm not implementing then if I am correct someone in the world will do the rest, so why bother arguing?

It's just stupid.

[A reply to someone who asked James how long will it be with nobody implementing his solution before he realises that he is wrong again.]

I'm not wrong. The derivation is trivial. All your arguing is equivalent to claiming that given

z^2 = y^2 + nT

x^2 = y^2 mod p

2x = k and z = x + k

that it's meaningless that

z^2 = 8^{-1} (9nT) mod p

which is a DERIVED result. If the derivation is useless then the algebra that gives it is nonsensical and lost, like a human being.

Issues around finding p as an odd prime are separate from the mathematical reality that those 4 equations give that result, which is about loving mathematics for what it tells us, not denying it for human reasons.

[A reply to someone who asked James how often does he reconsider his assertions that

Daily. Or better yet, several times a day.

People, including myself, are notable for easily being complete fools, idiots and total morons for things they want to believe. And are quite capable of doing so to unseemly limits, like blowing themselves up.

We as a species are remarkably bonkers in general.

I fully believe that I am quite capable of being totally and completely wrong, and believing I'm right indefinitely, and have reached a point where I find I trust little if anything, especially myself.

Sometimes I believe that our planet is best simply described as world of fools and I may be one of the greatest of all time.

It only makes sense that people using the best techniques of our modern era can figure out extraordinary things.

I've been trained to use iterative processes and brainstorming so the oddity of the posting style that I use is easily explained as throwing ideas out there as part of brainstorming, and there is usually a progression through a fairly structured process to each iteration.

With factoring after considering what was previously known I asked myself, can you maybe factor one number by using the factorization of some other number?

And that was just a question, which became a concept I called surrogate factoring and that start was over 4 years ago.

I've talked to people who are amazed that you can work on a math problem for years, but for readers on these forums it should not, of course, sound all that remarkable. What is I think remarkable though is the simple answer that years of iterations, with lots of brainstorming, so many posts, so many false starts and so much analysis has brought:

To solve for positive integer x, when x^2 = y^2 mod N, where N is an odd target composite to be factored you can use helper primes to solve the explicit equation:

x^2 = y^2 + cN

where c is an integer chosen to force cN mod 3 = 2, where for each prime that works you will have

x^2 = 8^{-1}(9cN) mod p

where p is an odd prime that is less than 2x/3 for which the quadratic residue exists, but further it must be true that the residue given by a prime in that range for which a quadratic residue exists must give an x near a value I call x_0 where x_0 is the largest positive integer divisible by 3 such that

abs(cN - 8x_0^2/9)

is a minimum, and x must be within x_0/(3p) positive steps from x_0.

That the question of whether or not one factorization can lead to another then has a surprising answer which represents the outcome of over four years of basic research.

Weird thing is that the derivation of

x^2 = 8^{-1}(9cN) mod p

is easy to the point of trivial so I've presented it a lot (see other threads), while the other rules are harder to explain so you'd have to dig a bit to understand where they come from, where that last part about x_0 is rather remarkable in terms of the why of the constraint.

I will say a few words about how I do problem solving which typically does include some insults in a post, often lambasting the mathematical community, which has a lot to do with motivating replies to my posts so that I can get other people looking over mathematical ideas. Years of experience on newsgroups has taught me that conflict draws readers and without it, you don't get readers at the same level.

Maybe unfortunate but it is the reality that I've found to be true in my experiments with what works and what does not as that has been an iterative process and a lot about problem solving as well.

If some other technique worked better, I'd use it.

[A reply to someone who suggested that James should write some code.]

I've written code before that factors numbers though not very large as I think it was usually right in the 40 bit range. For people who wonder what 40 bits is, it's numbers larger than 2^40 or 1099511627776. Those are considered small for factoring.

My experience is that posters deny any and all evidence or just keep raising the bar until the request is to demonstrate factoring better than anything else previously known.

The behavior as I've noted before is like saying that unless Albert Einstein built a working atomic bomb, and exploded it, his research was garbage.

The theory guy doesn't have to be the implementing guy or team.

Sometimes you just need to open the door and reasonable people can see value in your research effort, as we live in a very connected world. If there is anything to this research I'm sure there will be people building on it, which frees me up to consider theoretical issues.

[A reply to someone who told James that Einstein never claimed that he could build an atomic bomb.]

Scaling a solution can present challenges that have nothing to do with whether or not the theoretical solution is correct.

Mathematical theory says I've solved the factoring problem, like an interpretation of Einstein's research (along with that of others) indicated there were nuclear applications available.

But producing the practical product from theory can be a different enterprise.

The mathematical theory is about mathematical proof, which says the problem is solved.

Disbelief in mathematical proof can lead to a request for demonstration as if only seeing is believing.

Real mathematicians, however, know that mathematical proof means that the thing can be done.

You people could do something more useful and attack the claim of proof versus asking for a demonstration as if only an example can prove a proof, which displays a basic ignorance of mathematics, as examples can be nice but proof is proof.

[A reply to someone who explained James that no barrier prevents him from implementing of a solution to the factoring problem.]

I don't have to factor anything.

Of one thing there is little doubt in my mind it's if I'm right then someone will do the grunt work.

Now if you deny that then you're insane.

Why argue?

If I have solved the factoring problem and I say theory proves it, and I'm not implementing then if I am correct someone in the world will do the rest, so why bother arguing?

It's just stupid.

[A reply to someone who asked James how long will it be with nobody implementing his solution before he realises that he is wrong again.]

I'm not wrong. The derivation is trivial. All your arguing is equivalent to claiming that given

z^2 = y^2 + nT

x^2 = y^2 mod p

2x = k and z = x + k

that it's meaningless that

z^2 = 8^{-1} (9nT) mod p

which is a DERIVED result. If the derivation is useless then the algebra that gives it is nonsensical and lost, like a human being.

Issues around finding p as an odd prime are separate from the mathematical reality that those 4 equations give that result, which is about loving mathematics for what it tells us, not denying it for human reasons.

[A reply to someone who asked James how often does he reconsider his assertions that

- If he is right, then someone will factor big numbers using his method.
- He is right.]

Daily. Or better yet, several times a day.

People, including myself, are notable for easily being complete fools, idiots and total morons for things they want to believe. And are quite capable of doing so to unseemly limits, like blowing themselves up.

We as a species are remarkably bonkers in general.

I fully believe that I am quite capable of being totally and completely wrong, and believing I'm right indefinitely, and have reached a point where I find I trust little if anything, especially myself.

Sometimes I believe that our planet is best simply described as world of fools and I may be one of the greatest of all time.

### Wednesday, June 04, 2008

## Re-cap, why factoring is done

I'm going to shift variables to give you all something more familiar with factoring and explain the significance of a remarkable and simple congruence result:

Given

x^2 = y^2 + cN

where N is a target odd composite to factor, not divisible by 3, and c is a control variable that equals 1 if N mod 3 = 2 and 5 if N mod 3 = 1, it can be trivially proven that

x^2 = 8^{-1] (9cN) mod p

when 8^{-1] (9cN) is a quadratic residue for an odd prime p, where p < 2x/3.

For some of you it may be an extremely counter-intuitive result or terribly hard to understand, so I'll give an example and then talk about why it's significant.

Let N = 299 = 13(23), and since 299 mod 3 = 2, c=1. The only primes available (I'll leave the proof to the reader) are 3, 5, 7, 11 and 13 but since I know 13 is a factor I'll try, 11.

With p = 11, I check to see if the quadratic residue exists:

x^2 = 8^{-1} (9(299)) mod 11 = 7(7) mod 11, so that's easy and x = 7 mod 11 or -7 mod 11.

For those who wonder about the modular inverse 8^{-1} mod 11 = 7 because 7(8) = 1 mod 11.

Just knowing that x needs to be greater than p, you could try 11+7 = 18, and

y = sqrt(18^2 - 299) = sqrt(25) = 5, as I only care about the positive solution.

So I have (18+5)(18-5) = (23)(13) = 299, and a factorization with this approach.

The result is significant for two reasons:

x mod p_1*p_2*p_3

and if p_1*p_2*p_3 > x then, of course, you have x exactly.

So the short of it is, if it had been known early on that you could get x mod p then factoring would NOT have been thought to be a hard problem, so it's that significant.

From a socio-political aspect the result is significant because it's easy to derive, so easy to prove and has extraordinary social impact when developed into a fast factoring method so there is no way that a healthy mathematical society would not inform the world about it (and start celebrating, cheering and being generally happy that such a remarkable and beautiful result exists).

That isn't happening yet and my guess is that part of it is that mathematicians hate me and they're political animals so they are playing politics with the result. Ha ha.

I laugh because they can only get away with that for so long and then their funding can be taken away—all of it—in "pure math" areas so these people will be looking for new jobs soon enough. Ha ha.

Oh, so how do you derive the result? Easy. I've given the derivation in other posts with different variables but wanted to use more familiar variables to help some of you understand.

I have other mathematical research having proven Fermat's Last Theorem, found the prime counting function, delivered a prime gap equation, and if you Google "definition of mathematical proof" you can find my definition in the top 10.

Mathematicians hate me because I say many of them routinely lie because they do lie and they do it for money and prestige and because doing real mathematical research is harder than lying about your research and playing pretend. Ha ha.

Part of the purpose of postings like this one is to remove their naive student support, which will hurt their feelings very much because they are actors, playing at being mathematicians so it will bug them for you to look at them like they're crappy human beings, which they are.

And to me that actually is such a fun thing to contemplate that I have a sense of satisfaction already.

After all, Ribet or Andrew Wiles really just want butt-kissers which so many of you have been for so many years so the shock of that tragic look in your eyes will devastate the con artists, which they deserve.

Merry day!

Given

x^2 = y^2 + cN

where N is a target odd composite to factor, not divisible by 3, and c is a control variable that equals 1 if N mod 3 = 2 and 5 if N mod 3 = 1, it can be trivially proven that

x^2 = 8^{-1] (9cN) mod p

when 8^{-1] (9cN) is a quadratic residue for an odd prime p, where p < 2x/3.

For some of you it may be an extremely counter-intuitive result or terribly hard to understand, so I'll give an example and then talk about why it's significant.

Let N = 299 = 13(23), and since 299 mod 3 = 2, c=1. The only primes available (I'll leave the proof to the reader) are 3, 5, 7, 11 and 13 but since I know 13 is a factor I'll try, 11.

With p = 11, I check to see if the quadratic residue exists:

x^2 = 8^{-1} (9(299)) mod 11 = 7(7) mod 11, so that's easy and x = 7 mod 11 or -7 mod 11.

For those who wonder about the modular inverse 8^{-1} mod 11 = 7 because 7(8) = 1 mod 11.

Just knowing that x needs to be greater than p, you could try 11+7 = 18, and

y = sqrt(18^2 - 299) = sqrt(25) = 5, as I only care about the positive solution.

So I have (18+5)(18-5) = (23)(13) = 299, and a factorization with this approach.

The result is significant for two reasons:

- No one knew in mathematical history that given x^2 = y^2 + cT, you could get x mod p, where p is an odd prime less than 2x/3, when x is forced to be divisible by 3, and it is a "pure math" result with amazing significance there because you have these prime numbers coming in out of the blue in this remarkable way.

And prime numbers are cool. - Given x mod p with successive primes, you can get x exactly, which means you can factor T.

x mod p_1*p_2*p_3

and if p_1*p_2*p_3 > x then, of course, you have x exactly.

So the short of it is, if it had been known early on that you could get x mod p then factoring would NOT have been thought to be a hard problem, so it's that significant.

From a socio-political aspect the result is significant because it's easy to derive, so easy to prove and has extraordinary social impact when developed into a fast factoring method so there is no way that a healthy mathematical society would not inform the world about it (and start celebrating, cheering and being generally happy that such a remarkable and beautiful result exists).

That isn't happening yet and my guess is that part of it is that mathematicians hate me and they're political animals so they are playing politics with the result. Ha ha.

I laugh because they can only get away with that for so long and then their funding can be taken away—all of it—in "pure math" areas so these people will be looking for new jobs soon enough. Ha ha.

Oh, so how do you derive the result? Easy. I've given the derivation in other posts with different variables but wanted to use more familiar variables to help some of you understand.

I have other mathematical research having proven Fermat's Last Theorem, found the prime counting function, delivered a prime gap equation, and if you Google "definition of mathematical proof" you can find my definition in the top 10.

Mathematicians hate me because I say many of them routinely lie because they do lie and they do it for money and prestige and because doing real mathematical research is harder than lying about your research and playing pretend. Ha ha.

Part of the purpose of postings like this one is to remove their naive student support, which will hurt their feelings very much because they are actors, playing at being mathematicians so it will bug them for you to look at them like they're crappy human beings, which they are.

And to me that actually is such a fun thing to contemplate that I have a sense of satisfaction already.

After all, Ribet or Andrew Wiles really just want butt-kissers which so many of you have been for so many years so the shock of that tragic look in your eyes will devastate the con artists, which they deserve.

Merry day!

### Tuesday, June 03, 2008

## JSH: Tested limits

Mathematicians tested my limits and forced me to look at the world objectively enough to figure out what was going on and solve the problem, but questions still remain in my mind as I begin yet again to wait and comprehend how a people, a group of people really, around the world could have acted in this way.

One part of me sees it as destiny meant to force me to do what I would not have otherwise done, requiring me to solve problems I might otherwise not have attempted, as our world heads towards challenges most of you still clearly do not comprehend as intelligence got our species here, and without intelligence our species will not long

survive.

But life and death can seem so distant to comfortable people so today our world is increasingly uncomfortable so more of you will understand.

That is just the warning.

Reality has no problems with death and misery on a scale many of you will simply decide can't happen just like you decided you could ignore mathematical proofs because you decided that this day wouldn't be here.

For a long time I've seen this battle as a decision point for the species that most didn't realize was there when humanity would decide if it would die in misery on this planet or if it would have the courage and strength to develop the mathematics and science to leave this planet and explore the stars.

Many of you have fought for the death of the species, others have fought for life.

The balance is where the answer lies and as we sit in a real life equation calculating its way out before us and as the fate of humanity and the fate of life on this world is decided, it occurs to me that maybe some of you really don't know what limits are.

One of my predecessors Sir Isaac Newton fully understood them, and another, Archimedes comprehended their meaning.

But tests today are lost on some who have never truly been tested. Who do not know the value of intelligence.

Who do not understand what reality is.

Mathematics is the language of the Universe. It is the force that binds and holds reality together.

It is the only true power in reality.

And with it, all that is possible is realized and outside of it there is nothing.

You do not understand the true power of mathematics, or the meaning beyond mysticism, or you would never have set yourselves up in a battle that you could only win, if humanity died.

So you sealed your fates as the equations balance.

One part of me sees it as destiny meant to force me to do what I would not have otherwise done, requiring me to solve problems I might otherwise not have attempted, as our world heads towards challenges most of you still clearly do not comprehend as intelligence got our species here, and without intelligence our species will not long

survive.

But life and death can seem so distant to comfortable people so today our world is increasingly uncomfortable so more of you will understand.

That is just the warning.

Reality has no problems with death and misery on a scale many of you will simply decide can't happen just like you decided you could ignore mathematical proofs because you decided that this day wouldn't be here.

For a long time I've seen this battle as a decision point for the species that most didn't realize was there when humanity would decide if it would die in misery on this planet or if it would have the courage and strength to develop the mathematics and science to leave this planet and explore the stars.

Many of you have fought for the death of the species, others have fought for life.

The balance is where the answer lies and as we sit in a real life equation calculating its way out before us and as the fate of humanity and the fate of life on this world is decided, it occurs to me that maybe some of you really don't know what limits are.

One of my predecessors Sir Isaac Newton fully understood them, and another, Archimedes comprehended their meaning.

But tests today are lost on some who have never truly been tested. Who do not know the value of intelligence.

Who do not understand what reality is.

Mathematics is the language of the Universe. It is the force that binds and holds reality together.

It is the only true power in reality.

And with it, all that is possible is realized and outside of it there is nothing.

You do not understand the true power of mathematics, or the meaning beyond mysticism, or you would never have set yourselves up in a battle that you could only win, if humanity died.

So you sealed your fates as the equations balance.

## But how easy? Fundamental factoring congruence

Given

z^2 = y^2 + nT

where z is a positive integer divisible by 3, and T is a target composite to be factored, it can be shown that

z^2 = 8^{-1} (9nT) mod p

where p is an odd prime less than 2z/3, for which the quadratic residue exists, and n is a control variable chosen such that z has 3 as a factor so it's n = 1 if T mod 3 = 2 and, can be n = 5, if T mod 3 = 1, to force that situation.

If true that relation is one of the most fundamental relations in number theory showing a relationship between prime numbers and every integer factorization.

Yet it turns out it is trivially derived using a technique of adding variables that I first used in order to find a proof of Fermat's Last Theorem, as you can derive that relation by using;

x^2 = y^2 mod p

2x = k and z = x+ k

so you just multiply both sides by k with 2x = k to get 2xk = k^2 and add it to x^2 = y^2 mod p to get

x^2 + 2xk = y^2 + k^2 mod p

and then add k^2 to both sides to get

x^2 + 2xk + k^2 = y^2 + 2k^2 mod p

and now you have

(x+k)^2 = y^2 + 2k^2 mod p

so with T = 2k^2 mod p, and the other equations you are back to

z^2 = y^2 + nT

and the circle is complete.

But now you know that k^2 = 2^{-1}(nT) mod p, and doing the other substitutions you can get to

z^2 = 8^{-1} (9nT) mod p

and with slightly more complicated algebra you can prove that p is an odd prime less than k, so finally you have the p is an odd prime less than 2z/3, where the minimum z if z is a positive integer is sqrt(nT).

Easy math. Trivial. But oh what an impact.

The door is now open to solving for z modulo a succession of odd primes p, and in doing so, factoring nT, and in so doing, factoring T, non-trivially with easy algebra and basic math.

What a revolution.

But how easy is it to understand?

That question is what will decide the fate of mathematicians around the world where if it is very easy and the current delay in acknowledgment continues then mathematicians could face sharp questions about why they sat quiet.

My position as I have multiple major discoveries going back over 6 years is that mathematicians keep quiet because they're fakes fearing that knowledge of this will kill their cash cows and force them to work for a living like most people versus stealing from a public that pays many of them to do worthless research that they must know is false as I have proven it's false.

You lied thinking you'd never get caught. You betrayed humanity thinking the truth didn't matter.

You tried to stop the progress of mathematics which is why you lost.

The only question now is, how big of a price will you pay?

Judgment Day.

z^2 = y^2 + nT

where z is a positive integer divisible by 3, and T is a target composite to be factored, it can be shown that

z^2 = 8^{-1} (9nT) mod p

where p is an odd prime less than 2z/3, for which the quadratic residue exists, and n is a control variable chosen such that z has 3 as a factor so it's n = 1 if T mod 3 = 2 and, can be n = 5, if T mod 3 = 1, to force that situation.

If true that relation is one of the most fundamental relations in number theory showing a relationship between prime numbers and every integer factorization.

Yet it turns out it is trivially derived using a technique of adding variables that I first used in order to find a proof of Fermat's Last Theorem, as you can derive that relation by using;

x^2 = y^2 mod p

2x = k and z = x+ k

so you just multiply both sides by k with 2x = k to get 2xk = k^2 and add it to x^2 = y^2 mod p to get

x^2 + 2xk = y^2 + k^2 mod p

and then add k^2 to both sides to get

x^2 + 2xk + k^2 = y^2 + 2k^2 mod p

and now you have

(x+k)^2 = y^2 + 2k^2 mod p

so with T = 2k^2 mod p, and the other equations you are back to

z^2 = y^2 + nT

and the circle is complete.

But now you know that k^2 = 2^{-1}(nT) mod p, and doing the other substitutions you can get to

z^2 = 8^{-1} (9nT) mod p

and with slightly more complicated algebra you can prove that p is an odd prime less than k, so finally you have the p is an odd prime less than 2z/3, where the minimum z if z is a positive integer is sqrt(nT).

Easy math. Trivial. But oh what an impact.

The door is now open to solving for z modulo a succession of odd primes p, and in doing so, factoring nT, and in so doing, factoring T, non-trivially with easy algebra and basic math.

What a revolution.

But how easy is it to understand?

That question is what will decide the fate of mathematicians around the world where if it is very easy and the current delay in acknowledgment continues then mathematicians could face sharp questions about why they sat quiet.

My position as I have multiple major discoveries going back over 6 years is that mathematicians keep quiet because they're fakes fearing that knowledge of this will kill their cash cows and force them to work for a living like most people versus stealing from a public that pays many of them to do worthless research that they must know is false as I have proven it's false.

You lied thinking you'd never get caught. You betrayed humanity thinking the truth didn't matter.

You tried to stop the progress of mathematics which is why you lost.

The only question now is, how big of a price will you pay?

Judgment Day.

### Monday, June 02, 2008

## JSH: Extra variables and the factoring problem

It is remarkable to me that a technique I first introduced to attack Fermat's Last Theorem has now been key to solving the factoring problem, which is a technique of adding extra variables.

So with FLT I ended up adding a new variable v, and to solve the factoring problem I needed two new variables, k and p, where p is an odd prime.

What you see now in my talking about the solution is an extensive refinement on techniques I've figured out over years where the odd thing now is that there is so much room to simplify so I can give the solution to the factoring problem very succinctly so I will do so again now.

Prior attempts in the math field had the start right:

z^2 = y^2 + nT, or z^2 = y^2 mod T

but failed to realize that adding a few variables blew the problem away as in solved it very quickly:

x^2 = y^2 mod p

2x = k, and z = x+k

I actually came across this approach by deliberately looking for ways to factor one number with another, an approach I christened surrogate factoring, and I've deliberately looked for methods that had completion of the square for some time but did not realize how simple it all could be.

So you go from 2x = k, to 2xk = k^2, and add it to x^2 = y^2 mod p to get

x^2 + 2xk = y^2 + k^2 mod p

and then complete the square by adding k^2 to both sides to get

(x+k)^2 = y^2 + 2k^2 mod p

and now simply enough with z = x+k, you have that

nT = 2k^2 mod p

and doing the substitutions you also have that z = 2k/3, so everything is connected trivially and you have a way to find z modulo p just like that.

For years I've worked on the question of how to find k, or how to pick k as in previous research I'd often set it, but now realize that k is determined and you look for it using

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

where next the issue came up of how to pick p, which I now know is an odd prime.

It took some time and basic research involving theory plus experiments to finally figure out that there was a limit on the size of p, where I erroneously for some time thought that the size of factors f_1 and f_2 where f_1*f_2 = nT determined the size of p, but finally realized that additional congruence relations gave the answer as

f_1 = k mod p

and

f_2 = 2k mod p

so easily enough, k must be greater than p—I keep k a positive integer—as otherwise there would be a contradiction as k would be a factor of nT and of z when z^2 = y^2 + nT.

Given those rules it's then trivial to come up with what is probably the best factoring method possible, which involves picking a large prime p near the limit in size which is at the minimum k possible which would only happen with nT a perfect square, which means a prime near 2sqrt(nT)/3.

Since you need a quadratic residue of that prime, there is a 50% probability that any given odd prime p will work.

Oh yeah, about the variable n, it is one other addition I forgot to mention which is just there to make z divisible by 3, so if T mod 3 = 2 then n=1, but it can, equal, say 5 if T mod 3 = 1, so its size impact is nominal.

Remarkably as you search for k, which would mean picking the first even k near the minimum that has the residue modulo p required by your prime and then iterating by 2p, you can, of course, pick additional primes as well as you increase the size of your test k's, since the rule is that p be less than k.

And THEN of course you can loop modulo more primes as if you have

k mod p_1 and k mod p_2

then of course you can get k mod p_1*p_2

which means that with theory alone you the reader now know this method will be faster than anything else previously known.

Since I know that I previously found a proof of Fermat's Last Theorem using an advanced technique which math society ignored, though I managed briefly to get a key technique published, and I have my prime counting function discovery with its remarkably unique features which math society ignored, and my prime gap equation which math society ignored, plus my work on logic which math society ignored I am

comfortable in saying that the modern math field is completely corrupted.

For instance, supposed great research works are proven by my research to be flawed where I'm still amazed by the many ways I could refute the work of Andrew Wiles where his supposed proof of FLT fails many ways including the use of a simple logical fallacy called cum hoc, ergo propter hoc.

The best explanation for what has happened to the math field is that in esoteric "pure math" areas, people have simply done fake research and claimed it was otherwise, feeling safe and secure in that because the research had no practical value. They clearly believed that the charade could be maintained indefinitely.

But they had two problems to handle:

factoring problem.

That problem is now solved as explained in this post.

The current delay in acceptance is all about the corruption of the modern math field.

My take on the situation is that most modern number theorists rarely if ever tell the truth, especially about math.

So with FLT I ended up adding a new variable v, and to solve the factoring problem I needed two new variables, k and p, where p is an odd prime.

What you see now in my talking about the solution is an extensive refinement on techniques I've figured out over years where the odd thing now is that there is so much room to simplify so I can give the solution to the factoring problem very succinctly so I will do so again now.

Prior attempts in the math field had the start right:

z^2 = y^2 + nT, or z^2 = y^2 mod T

but failed to realize that adding a few variables blew the problem away as in solved it very quickly:

x^2 = y^2 mod p

2x = k, and z = x+k

I actually came across this approach by deliberately looking for ways to factor one number with another, an approach I christened surrogate factoring, and I've deliberately looked for methods that had completion of the square for some time but did not realize how simple it all could be.

So you go from 2x = k, to 2xk = k^2, and add it to x^2 = y^2 mod p to get

x^2 + 2xk = y^2 + k^2 mod p

and then complete the square by adding k^2 to both sides to get

(x+k)^2 = y^2 + 2k^2 mod p

and now simply enough with z = x+k, you have that

nT = 2k^2 mod p

and doing the substitutions you also have that z = 2k/3, so everything is connected trivially and you have a way to find z modulo p just like that.

For years I've worked on the question of how to find k, or how to pick k as in previous research I'd often set it, but now realize that k is determined and you look for it using

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

where next the issue came up of how to pick p, which I now know is an odd prime.

It took some time and basic research involving theory plus experiments to finally figure out that there was a limit on the size of p, where I erroneously for some time thought that the size of factors f_1 and f_2 where f_1*f_2 = nT determined the size of p, but finally realized that additional congruence relations gave the answer as

f_1 = k mod p

and

f_2 = 2k mod p

so easily enough, k must be greater than p—I keep k a positive integer—as otherwise there would be a contradiction as k would be a factor of nT and of z when z^2 = y^2 + nT.

Given those rules it's then trivial to come up with what is probably the best factoring method possible, which involves picking a large prime p near the limit in size which is at the minimum k possible which would only happen with nT a perfect square, which means a prime near 2sqrt(nT)/3.

Since you need a quadratic residue of that prime, there is a 50% probability that any given odd prime p will work.

Oh yeah, about the variable n, it is one other addition I forgot to mention which is just there to make z divisible by 3, so if T mod 3 = 2 then n=1, but it can, equal, say 5 if T mod 3 = 1, so its size impact is nominal.

Remarkably as you search for k, which would mean picking the first even k near the minimum that has the residue modulo p required by your prime and then iterating by 2p, you can, of course, pick additional primes as well as you increase the size of your test k's, since the rule is that p be less than k.

And THEN of course you can loop modulo more primes as if you have

k mod p_1 and k mod p_2

then of course you can get k mod p_1*p_2

which means that with theory alone you the reader now know this method will be faster than anything else previously known.

Since I know that I previously found a proof of Fermat's Last Theorem using an advanced technique which math society ignored, though I managed briefly to get a key technique published, and I have my prime counting function discovery with its remarkably unique features which math society ignored, and my prime gap equation which math society ignored, plus my work on logic which math society ignored I am

comfortable in saying that the modern math field is completely corrupted.

For instance, supposed great research works are proven by my research to be flawed where I'm still amazed by the many ways I could refute the work of Andrew Wiles where his supposed proof of FLT fails many ways including the use of a simple logical fallacy called cum hoc, ergo propter hoc.

The best explanation for what has happened to the math field is that in esoteric "pure math" areas, people have simply done fake research and claimed it was otherwise, feeling safe and secure in that because the research had no practical value. They clearly believed that the charade could be maintained indefinitely.

But they had two problems to handle:

- Computer science advances could long ago have introduced computerized checking of most or all claims of mathematical proof.
- The emergence of a major researcher who could point out numerous errors.
- So they handled each problem in turn in various way—attacking the notion of computers being able to check math proofs and building a system that would allow them to just ignore major research findings.

factoring problem.

That problem is now solved as explained in this post.

The current delay in acceptance is all about the corruption of the modern math field.

My take on the situation is that most modern number theorists rarely if ever tell the truth, especially about math.