### Wednesday, June 30, 2010

## JSH: Such a surreal thing

I think it fascinating that I can have these results and the world not seem to notice while math people rip on them and me. Now THAT is interesting.

This latest result is kind of cool because of the way the math handles the issue of large m with k^m = q mod N. It's like, the algebra couldn't care less. It can handle that issue like nothing, but it IS algebra. Human beings are just these puny little things that come for a while and then die.

It is fascinating to me the issue of how some people think, especially how they do certain things like plan for the future when they refuse to acknowledge important information! It's so contradictory!

A few years back because of my surrogate factoring research I started thinking P=NP, and came up with some research in that area which bugged me. I still don't like it but I keep it up, as I can't just throw it away. But let's just say, you don't understand reality if you don't appreciate the implications of that result.

But why should you?

Did it ever occur to any of you that reality might be smarter than you are? And that just MAYBE you'd be a lot more boring if you were smarter? It has occurred to me. Reality is clearly much smarter than I am.

But many of you seem to think that YOU are smarter which means you can be more interesting than otherwise, but I watch you and wish I were like you in that you can do those things, like plan for the future, have expectations. BELIEVE in so many things.

These mathematical results may unfortunately give me knowledge I don't wish to have, but I have you to insult me regardless. Don't you wish you knew what happens to you later?

I wonder how you do so many of the things you do, but realize it's because you have no clue.

THAT is the most amazing thing. You plan for your futures because you have no idea what's going to happen to you.

I find that both bizarre and fascinating. How is it possible?

This latest result is kind of cool because of the way the math handles the issue of large m with k^m = q mod N. It's like, the algebra couldn't care less. It can handle that issue like nothing, but it IS algebra. Human beings are just these puny little things that come for a while and then die.

It is fascinating to me the issue of how some people think, especially how they do certain things like plan for the future when they refuse to acknowledge important information! It's so contradictory!

A few years back because of my surrogate factoring research I started thinking P=NP, and came up with some research in that area which bugged me. I still don't like it but I keep it up, as I can't just throw it away. But let's just say, you don't understand reality if you don't appreciate the implications of that result.

But why should you?

Did it ever occur to any of you that reality might be smarter than you are? And that just MAYBE you'd be a lot more boring if you were smarter? It has occurred to me. Reality is clearly much smarter than I am.

But many of you seem to think that YOU are smarter which means you can be more interesting than otherwise, but I watch you and wish I were like you in that you can do those things, like plan for the future, have expectations. BELIEVE in so many things.

These mathematical results may unfortunately give me knowledge I don't wish to have, but I have you to insult me regardless. Don't you wish you knew what happens to you later?

I wonder how you do so many of the things you do, but realize it's because you have no clue.

THAT is the most amazing thing. You plan for your futures because you have no idea what's going to happen to you.

I find that both bizarre and fascinating. How is it possible?

## Discrete logs result, sum of the factors upper bound?

I've noted a way to solve discrete logarithms through factoring:

q^2 mod N

where with

f_1*...*f_m = q^2 mod N

you can find m simply enough from:

(k-1)*m = (f_1+...+f_c - c) mod N

where c is the count of factors in your factorization of q^2 mod N, where prior posts explain from where the equations derive.

In a prior post I try to set an upper bound for q^2 mod N, which at approximately 4N^2 is rather high, but there may be many reasons why solutions will not approach that upper bound where I noticed one while playing with small numbers, which is that the sum of the factors of q^2 mod N, can be a constraining factor.

Consider 2^m = 60 mod 107. That gives f_1*...*f_m = q^2 mod N = 69 mod 107, and I've been adding N for my first try at a solution so started with 176, which factors as 2*2*2*2*11, and those sum to give 19 but minus 5, I get 14, which may be close, but it's not right. So I kept going, and quickly noticed that the sum of the factors were not getting closer but

Worrying a bit but knowing the underlying mathematics is solid, I went back to 69, and finally noticed: 6+9-2 = 13.

2^13 = 60 mod 107

Intriguingly, with a larger prime relative to the answer, the factors grew rapidly, so that the sum of the factors was way beyond m fairly quickly, which forced early solution?

But still I'd think there'd be other answer with small primes. But of course the math also has the ability to use a large m, which still works, though it's not the smallest one that I'd prefer.

q^2 mod N

where with

f_1*...*f_m = q^2 mod N

you can find m simply enough from:

(k-1)*m = (f_1+...+f_c - c) mod N

where c is the count of factors in your factorization of q^2 mod N, where prior posts explain from where the equations derive.

In a prior post I try to set an upper bound for q^2 mod N, which at approximately 4N^2 is rather high, but there may be many reasons why solutions will not approach that upper bound where I noticed one while playing with small numbers, which is that the sum of the factors of q^2 mod N, can be a constraining factor.

Consider 2^m = 60 mod 107. That gives f_1*...*f_m = q^2 mod N = 69 mod 107, and I've been adding N for my first try at a solution so started with 176, which factors as 2*2*2*2*11, and those sum to give 19 but minus 5, I get 14, which may be close, but it's not right. So I kept going, and quickly noticed that the sum of the factors were not getting closer but

**further**away.Worrying a bit but knowing the underlying mathematics is solid, I went back to 69, and finally noticed: 6+9-2 = 13.

2^13 = 60 mod 107

Intriguingly, with a larger prime relative to the answer, the factors grew rapidly, so that the sum of the factors was way beyond m fairly quickly, which forced early solution?

But still I'd think there'd be other answer with small primes. But of course the math also has the ability to use a large m, which still works, though it's not the smallest one that I'd prefer.

## Discrete logs result and degrees of freedom

I've noted a way to solve for m, when k^m = q mod N, through integer factorization, which is then an approach to solving discrete logarithms in a prior post. In this post I'll explain when the equations MUST work, where a simple analysis can be done trivially using methods familiar to those who've solved simultaneous equations in regular algebra.

Here are relevant equations without the complete detail explaining them all of the prior post which should be read for reference:

Everything follows from use of a simple system of equations:

f_1 = a_1*k mod N thru f_m = a_m*k mod N

Two important constraining equations:

a_1*...*a_m = q mod N

and

a_1+...+a_m = m mod N

Resultant equations:

f_1*...*f_m = q^2 mod N

and

f_1+...+f_m = mk mod N

(These are arbitrary constraints that I used. There may be others that are of practical use.)

Now assume that for some unknown number m-c of the f's that the a's are simply the modular inverse of k, then for that number the f's simply equal 1, which allows me to solve for m with:

(k-1)*m = (f_1+...+f_c - c) mod N

If k-1 is coprime to N, you can simply use the modular inverse to get m. Otherwise you'd to divide off common factors from both sides and then use the modular inverse with what remained.

All of which was given in my prior post, but notice I can now go back to the constraining equations for the a's with the information that some of the a's have been set to the modular inverse of k:

a_1*...*a_c*k^{-(m-c)} = q mod N, so a_1*...*a_c= q*k^{-(m-c)}

and

a_1+...+a_c + (m-c)k^{-1} = m mod N, so a_1+...+a_c = -(m-c)k^{-1} + m mod N

which means there are two simultaneous congruence equations with c unknowns:

a_1*...*a_c= q*k^{-(m-c)}

and

a_1+...+a_c = -(m-c)k^{-1} + m mod N

Using the first to substitute out a_1 into the second and simplifying slightly gives:

q*k^{-(m-c)} + a_2^2*a_3*...*a_c + a_2*a_3^2*...*a_c...+a_2*...*a_c^2 = a_2*...*a_c[-(m-c)k^{-1} + m] mod N

Notice then that with any c-2 variables set arbitrarily, the existence of the remaining one is set if a quadratic residue exists.

So for instance if c=4, then you can have a_3 and a_4 completely free variables, as long as quadratic residues exist to allow for a_2, which would indicate a 50% probability in that case if N is prime.

However, you can also further constrain one more of the a's to remove squares, for instance let a_3 = a_2^{-1} mod N, and you have:

q*k^{-(m-c)} + a_2*a_4*...*a_c + a_3*a_4*...*a_c...+ a_4*...*a_c^2 = a_4*...*a_c[-(m-c)k^{-1} + m] mod N

which allows a solution for a_2 to always exist. Which would leave with c=4, a_4 completely free.

Assuming human nature will be to look for smaller values to factor to actually use the algorithm, one can assume that

f_1*...*f_m = q^2 mod N

will be with smaller values of q^2 mod N, based on human preference, so if a_4 is completely free, and is non-unit, it would likely in many cases mean that f_4 will be 2.

If a_3 and a_4 are free then one would expect f_3 and f_4 to be 2.

With c = 4 or greater then the area to consider variability that cannot just be arbitrarily set to a convenience value is with a_1 and a_2 which could make f_1 and f_2 just about any residue mod N. Worst case they are both near N, so you'd have a size of approximately 4N^2.

So algorithms based on this method should exit within q^2 mod N less than 4N^2.

Here's the example given in my prior post, which should make more sense given the information above:

Solve for m, where:

2^m = 13 mod 23

so k = 2, and f_1*...*f_m = q^2 mod N = 13^2 mod 23 = 8 mod 23.

And I found a solution with f_1*...*f_m = 54, and

f_1 = 2, f_2 = 3, f_3 = 3, f_4 = 3

so c=4, and

m = (k-1)^{-1}(f_1 +...+ f_c - c) mod N = 2+3+3+3 - 4 = 7

And 2^7 = 128 = 13 mod 23.

Notice that 2 is a factor as are small primes. The method will try to fit small primes if you are using small values which is about human preference. The algebra tries to give you what you want based on the size of the numbers you are factoring.

The value of c is dynamically set by the factorization of q^2 mod N. But the results above indicate that the algebra must give you a factorization within that range which will work.

And that is what's found with a first blush basic analysis. It's not clear at this time what further information might result from more basic research. My aim at this point is to answer criticism against this approach.

Routinely posters reply requesting I demonstrate by breaking current encryption. Well, if I could do that I wouldn't need to bother posting on newsgroups now would I?

It's basic research. Early stages.

Here are relevant equations without the complete detail explaining them all of the prior post which should be read for reference:

Everything follows from use of a simple system of equations:

f_1 = a_1*k mod N thru f_m = a_m*k mod N

Two important constraining equations:

a_1*...*a_m = q mod N

and

a_1+...+a_m = m mod N

Resultant equations:

f_1*...*f_m = q^2 mod N

and

f_1+...+f_m = mk mod N

(These are arbitrary constraints that I used. There may be others that are of practical use.)

Now assume that for some unknown number m-c of the f's that the a's are simply the modular inverse of k, then for that number the f's simply equal 1, which allows me to solve for m with:

(k-1)*m = (f_1+...+f_c - c) mod N

If k-1 is coprime to N, you can simply use the modular inverse to get m. Otherwise you'd to divide off common factors from both sides and then use the modular inverse with what remained.

All of which was given in my prior post, but notice I can now go back to the constraining equations for the a's with the information that some of the a's have been set to the modular inverse of k:

a_1*...*a_c*k^{-(m-c)} = q mod N, so a_1*...*a_c= q*k^{-(m-c)}

and

a_1+...+a_c + (m-c)k^{-1} = m mod N, so a_1+...+a_c = -(m-c)k^{-1} + m mod N

which means there are two simultaneous congruence equations with c unknowns:

a_1*...*a_c= q*k^{-(m-c)}

and

a_1+...+a_c = -(m-c)k^{-1} + m mod N

Using the first to substitute out a_1 into the second and simplifying slightly gives:

q*k^{-(m-c)} + a_2^2*a_3*...*a_c + a_2*a_3^2*...*a_c...+a_2*...*a_c^2 = a_2*...*a_c[-(m-c)k^{-1} + m] mod N

Notice then that with any c-2 variables set arbitrarily, the existence of the remaining one is set if a quadratic residue exists.

So for instance if c=4, then you can have a_3 and a_4 completely free variables, as long as quadratic residues exist to allow for a_2, which would indicate a 50% probability in that case if N is prime.

However, you can also further constrain one more of the a's to remove squares, for instance let a_3 = a_2^{-1} mod N, and you have:

q*k^{-(m-c)} + a_2*a_4*...*a_c + a_3*a_4*...*a_c...+ a_4*...*a_c^2 = a_4*...*a_c[-(m-c)k^{-1} + m] mod N

which allows a solution for a_2 to always exist. Which would leave with c=4, a_4 completely free.

Assuming human nature will be to look for smaller values to factor to actually use the algorithm, one can assume that

f_1*...*f_m = q^2 mod N

will be with smaller values of q^2 mod N, based on human preference, so if a_4 is completely free, and is non-unit, it would likely in many cases mean that f_4 will be 2.

If a_3 and a_4 are free then one would expect f_3 and f_4 to be 2.

With c = 4 or greater then the area to consider variability that cannot just be arbitrarily set to a convenience value is with a_1 and a_2 which could make f_1 and f_2 just about any residue mod N. Worst case they are both near N, so you'd have a size of approximately 4N^2.

So algorithms based on this method should exit within q^2 mod N less than 4N^2.

Here's the example given in my prior post, which should make more sense given the information above:

Solve for m, where:

2^m = 13 mod 23

so k = 2, and f_1*...*f_m = q^2 mod N = 13^2 mod 23 = 8 mod 23.

And I found a solution with f_1*...*f_m = 54, and

f_1 = 2, f_2 = 3, f_3 = 3, f_4 = 3

so c=4, and

m = (k-1)^{-1}(f_1 +...+ f_c - c) mod N = 2+3+3+3 - 4 = 7

And 2^7 = 128 = 13 mod 23.

Notice that 2 is a factor as are small primes. The method will try to fit small primes if you are using small values which is about human preference. The algebra tries to give you what you want based on the size of the numbers you are factoring.

The value of c is dynamically set by the factorization of q^2 mod N. But the results above indicate that the algebra must give you a factorization within that range which will work.

And that is what's found with a first blush basic analysis. It's not clear at this time what further information might result from more basic research. My aim at this point is to answer criticism against this approach.

Routinely posters reply requesting I demonstrate by breaking current encryption. Well, if I could do that I wouldn't need to bother posting on newsgroups now would I?

It's basic research. Early stages.

### Tuesday, June 29, 2010

## JSH: Integer factorization IS the short-cut

Intriguingly enough my research results studying the simple system of:

f_1 = a_1*k mod N thru f_m = a_m*k mod N

indicate that integer factorization IS itself the key to modular arithmetic, so what I called surrogate factoring is in retrospect the way modular arithmetic operates.

The result I recently posted showing a way to solve discrete logarithms is amazing for its compactness but also for what it may indicate about the power of knowledge: not knowing how modular arithmetic works at a deep level, certain things that may be easy, appear hard.

That result actually may eliminate the value of a large m, with k^m = q mod N, allowing one to easily handle it by simply cancelling out much of m with a simple technique equivalent to having a certain number of the a's equal to k^{-1} mod N.

I'll be looking at replies with interest. One of the problems with knowledge is when people reject it because, well, because they don't like it!!!

That creates huge problems on a world-wide scale.

Quite simply, people refuse solutions to problems, bad things happen, people get upset, but, they refuse solutions to problems!

It's a HUGE issue. So far the problem has been intractable.

Human beings seem to love misery. I'm not sure why. But make no mistake, the human animal often works very hard to NOT solve its problems, preferring often instead to whine about them, but refusing to solve them.

That may be built into the human genome. The reasons are complex.

f_1 = a_1*k mod N thru f_m = a_m*k mod N

indicate that integer factorization IS itself the key to modular arithmetic, so what I called surrogate factoring is in retrospect the way modular arithmetic operates.

The result I recently posted showing a way to solve discrete logarithms is amazing for its compactness but also for what it may indicate about the power of knowledge: not knowing how modular arithmetic works at a deep level, certain things that may be easy, appear hard.

That result actually may eliminate the value of a large m, with k^m = q mod N, allowing one to easily handle it by simply cancelling out much of m with a simple technique equivalent to having a certain number of the a's equal to k^{-1} mod N.

I'll be looking at replies with interest. One of the problems with knowledge is when people reject it because, well, because they don't like it!!!

That creates huge problems on a world-wide scale.

Quite simply, people refuse solutions to problems, bad things happen, people get upset, but, they refuse solutions to problems!

It's a HUGE issue. So far the problem has been intractable.

Human beings seem to love misery. I'm not sure why. But make no mistake, the human animal often works very hard to NOT solve its problems, preferring often instead to whine about them, but refusing to solve them.

That may be built into the human genome. The reasons are complex.

## JSH: Considering discrete logs

I've noted a fundamental result in modular arithmetic around studying the simple system of equations:

f_1 = a_1*k mod N thru f_m = a_m*k mod N

Specifically by noting that multiplying them together gives:

f_1*...*f_m = a_1*...*a_m*k^m mod N, but ADDING them together gives:

f_1+...+f_m = (a_1 + ...+ a_m)k mod N

Note that the a's are what I call control variables, which can be set to ANY non-zero value.

Remarkably enough that simple result allows one to handle k^m = q mod N through integer factorization.

You can solve for k, when m, q and N are known, or for m, when k, q and N are known, and it is the latter I'll consider in more detail in this post.

Since the a's are control variables their value can be set to some extent, with m degrees of freedom. To handle discrete logarithms entirely, over any m, I need to remove less than m of those degrees of freedom and do so with the following equations:

a_1*...*a_m = q mod N

and

a_1+...+a_m = m mod N

Then k^m = q mod N, and f_1*...*f_m = a_1*...*a_m*k^m mod N, with substitutions gives me:

f_1*...*f_m = q^2 mod N

and f_1+...+f_m = (a_1 + ...+ a_m)k mod N with substitution gives:

f_1+...+f_m = mk mod N

But so far I've removed only 2 degrees of freedom from the a's, so assume that for some unknown number m-c of the f's that the a's are simply the modular inverse of k, then for that number the f's simply equal 1, giving me:

f_1+...+f_c + m - c = mk mod N

Which allows me to solve for m, with:

m = (k-1)^{-1}(f_1+...+f_c - c) mod N

(If k-1 is not coprime to N, then factors in common would be divided off before the modular inverse operation, which actually gives another check for finding useable factors.)

Notice that finding factors then becomes just a matter of considering solutions:

f_1*...*f_m = q^2 mod N

where m-c of the factors have been set to 1. So the reasons for that substitution is clear.

Notice that c is found dynamically from the count of non-unit factors of q^2 mod N.

Here's an example, solve for m, where:

2^m = 13 mod 23

so k = 2, and f_1*...*f_m = q^2 mod N = 13^2 mod 23 = 8 mod 23.

And I found a solution with f_1*...*f_m = 54, and

f_1 = 2, f_2 = 3, f_3 = 3, f_4 = 3

so c=4, and

m = (k-1)^{-1}(f_1 +...+ f_c - c) mod N = 2+3+3+3 - 4 = 7

And 2^7 = 128 = 13 mod 23.

Notice that the method then gives you the number c from factoring numbers q^2 mod N. The means mathematically that c of the a's have been set by the algebra itself to cancel out k, with the modular inverse, which is how this approach eliminates the advantage of a large m—the math simply handles it for you.

And that is the basic research showing how a simple set of modular equations can lead to an approach to solving discrete logarithms through integer factorization. It is not clear at this time how practical it can be made to be.

However, I would assume that it is a method that cannot simply be ignored as the idiot mutterings of some "crackpot". But if any of you do, I hope you pay the appropriate consequences if it turns out to be important. Paying as high a price as your world requires of you. No matter what the price might be.

f_1 = a_1*k mod N thru f_m = a_m*k mod N

Specifically by noting that multiplying them together gives:

f_1*...*f_m = a_1*...*a_m*k^m mod N, but ADDING them together gives:

f_1+...+f_m = (a_1 + ...+ a_m)k mod N

Note that the a's are what I call control variables, which can be set to ANY non-zero value.

Remarkably enough that simple result allows one to handle k^m = q mod N through integer factorization.

You can solve for k, when m, q and N are known, or for m, when k, q and N are known, and it is the latter I'll consider in more detail in this post.

Since the a's are control variables their value can be set to some extent, with m degrees of freedom. To handle discrete logarithms entirely, over any m, I need to remove less than m of those degrees of freedom and do so with the following equations:

a_1*...*a_m = q mod N

and

a_1+...+a_m = m mod N

Then k^m = q mod N, and f_1*...*f_m = a_1*...*a_m*k^m mod N, with substitutions gives me:

f_1*...*f_m = q^2 mod N

and f_1+...+f_m = (a_1 + ...+ a_m)k mod N with substitution gives:

f_1+...+f_m = mk mod N

But so far I've removed only 2 degrees of freedom from the a's, so assume that for some unknown number m-c of the f's that the a's are simply the modular inverse of k, then for that number the f's simply equal 1, giving me:

f_1+...+f_c + m - c = mk mod N

Which allows me to solve for m, with:

m = (k-1)^{-1}(f_1+...+f_c - c) mod N

(If k-1 is not coprime to N, then factors in common would be divided off before the modular inverse operation, which actually gives another check for finding useable factors.)

Notice that finding factors then becomes just a matter of considering solutions:

f_1*...*f_m = q^2 mod N

where m-c of the factors have been set to 1. So the reasons for that substitution is clear.

Notice that c is found dynamically from the count of non-unit factors of q^2 mod N.

Here's an example, solve for m, where:

2^m = 13 mod 23

so k = 2, and f_1*...*f_m = q^2 mod N = 13^2 mod 23 = 8 mod 23.

And I found a solution with f_1*...*f_m = 54, and

f_1 = 2, f_2 = 3, f_3 = 3, f_4 = 3

so c=4, and

m = (k-1)^{-1}(f_1 +...+ f_c - c) mod N = 2+3+3+3 - 4 = 7

And 2^7 = 128 = 13 mod 23.

Notice that the method then gives you the number c from factoring numbers q^2 mod N. The means mathematically that c of the a's have been set by the algebra itself to cancel out k, with the modular inverse, which is how this approach eliminates the advantage of a large m—the math simply handles it for you.

And that is the basic research showing how a simple set of modular equations can lead to an approach to solving discrete logarithms through integer factorization. It is not clear at this time how practical it can be made to be.

However, I would assume that it is a method that cannot simply be ignored as the idiot mutterings of some "crackpot". But if any of you do, I hope you pay the appropriate consequences if it turns out to be important. Paying as high a price as your world requires of you. No matter what the price might be.

## JSH: World of insecurity?

When back in 2004 I realized that my research showing a conflict between algebraic integers and the field of complex numbers would not be accepted by mathematicians who would rather be in error, I decided I needed to find something that could not just be ignored, and I thought the factoring problem was that thing.

Here we are 6 years later and on a convoluted path I've found methods for handling k^m = q mod N through integer factorization, something I designated surrogate factoring, and I was wrong, as it is, it seems, being ignored!

And I realize that years can go by yet again.

There is nothing that will work. There is no discovery that will change these people. Nothing.

But the world should be ok. Years ago I'd worry a lot but from what I've seen, no matter how much "experts" screw-up, people just keep going about their business, if they can. Like now with the oil spill in the gulf.

Experts have no accountability in the modern world.

You don't have to be right!

You can be complete idiots as long as there is a gang of you, and you say things that aren't true, make claims that aren't true, and risk other people's lives because you know it will not matter for you in the long run. No one really cares.

Oddly enough the world can remain quite secure no matter how powerful the mathematical ideas I've found may be because BELIEF controls people to such a degree that no one may bother to exploit the information.

Sitting here I realize that knowledge can be a curse in a world where knowledge is not needed to be correct as you look at idiots do well, ripping on people who are right, but it doesn't matter, because most people go with the idiots, so even when disaster strikes, they stick with them, for reasons maybe God only truly knows.

But that's the real world. I'll quietly celebrate my own findings like I usually do in a world that does not care, and not worry. I wish I knew years ago what I know now: fools are often safe because of their own stupidity.

I'm not worried about the world.

Here we are 6 years later and on a convoluted path I've found methods for handling k^m = q mod N through integer factorization, something I designated surrogate factoring, and I was wrong, as it is, it seems, being ignored!

And I realize that years can go by yet again.

There is nothing that will work. There is no discovery that will change these people. Nothing.

But the world should be ok. Years ago I'd worry a lot but from what I've seen, no matter how much "experts" screw-up, people just keep going about their business, if they can. Like now with the oil spill in the gulf.

Experts have no accountability in the modern world.

You don't have to be right!

You can be complete idiots as long as there is a gang of you, and you say things that aren't true, make claims that aren't true, and risk other people's lives because you know it will not matter for you in the long run. No one really cares.

Oddly enough the world can remain quite secure no matter how powerful the mathematical ideas I've found may be because BELIEF controls people to such a degree that no one may bother to exploit the information.

Sitting here I realize that knowledge can be a curse in a world where knowledge is not needed to be correct as you look at idiots do well, ripping on people who are right, but it doesn't matter, because most people go with the idiots, so even when disaster strikes, they stick with them, for reasons maybe God only truly knows.

But that's the real world. I'll quietly celebrate my own findings like I usually do in a world that does not care, and not worry. I wish I knew years ago what I know now: fools are often safe because of their own stupidity.

I'm not worried about the world.

### Saturday, June 19, 2010

## JSH: Judging difficulty

Confident people can be VERY convincing, and over the years I've more than once ran into the issue of, if so many people say you're wrong, how can you be right? And that has bugged me as well, which goes to the real reason I talked a lot about Google searches.

Sure I can put forward a mathematical argument and say I'm right, say I've traced it out carefully and that in reply a poster just deleted it all out, or simply refuses to address the actual argument and instead changes the subject or just makes false statements, but it can still seem like just one person's word against another's, and the crowd has the greater impact.

And one thing I've noticed over the years is that a group of maybe even only 6 posters will post with a lot of energy in reply to a person and will often use phrases like "everyone disagrees with you" or "no one likes you", or global statements to that effect as clearly they understand the power of presenting a situation that supposedly is about one person refusing to accept the truth from a large number of people.

So Google searches are this unique way to watch their behavior with something much bigger than they are!

Intriguingly though they simply demeaned Google.

Suddenly according to them, Google searches were "meaningless", and getting high search rankings was a trivial thing that anyone could do, though I don't put "anyone" in quotes as that at least isn't something I've seen actually stated.

Oh, and in facing high country counts of hits to my math blog as reported to Google Analytics, some might simply claim I was lying about them, while one poster confidently proclaimed the hits to be robot programs and forcefully argued that point.

One exercise for people contemplating the actual difficulty of something is to imagine doing it themselves, where Google is global enough that you can just go do searches on your own activities, or your ideas, or imagine putting an idea out there and seeing where it ranks.

Amazingly enough with that exercise you may only then be able to contemplate the feat of my having the definition of mathematical proof, by imagining doing it yourself.

So think of sitting down, writing up your own personal definition of mathematical proof—you can actually try this exercise if you wish—putting it out there, and taking over the #1 spot over much of the world, beating dictionaries, MathWorld, and the Wikipedia.

Now do the search in Google (has to be Google): definition of mathematical proof

For me it was surreal when I finally realized I was getting highly ranked for that definition and I learned it from search strings people were using to get to my math blog (um, I have a LOT of data from multiple sources).

Confidence oozing from posters can be mistaken for skill.

Readers can believe that no one would actually just stalk another person claiming they're wrong without having factual or mathematical basis, but I have watched that behavior for years. Confidence does not mean a person is right. Personally I don't trust confidence with mathematics and prefer when someone is honest about worries about a particular mathematical argument. Or concerns that they missed something, or are just plain wrong!

But human nature is to TRUST confidence. So my own admissions of error rather than helping me, are used by confident posters to reinforce the notion that I must be wrong.

It IS a competitive world.

And to me it has been amazing watching posters claim otherwise, as if anyone can have the definition of mathematical proof in Google, or "everyone" can have hits from 120+ countries on a yearly basis to their blog, if they just want it or something? I guess?

But at the end of it all of course the ultimate refuge for readers is the realization that it's really about mainstream mathematicians acknowledging or not acknowledging important mathematical research anyway, so why would I need to talk about Google search results if I were actually right?

Because the Internet as the new thing seems to be upending an old guard which has been informed of a massive error within number theory, which they may see as more easily ignored than addressed.

Google: algebraic integers vs complex numbers

If you even imagine that the ring of algebraic integers conflicts with the field of complex numbers you realize it's a defunct ring, but it's such a huge thing, and then for supposedly top mathematicians to pretend the problem doesn't exist?

How is such a thing possible?

Well as BP continues to try and stop an oil spill. And as the world continues to try and recover from a near financial collapse. And while the Catholic Church has to keep addressing issues of pedophile priests, it's not so hard to understand.

Judging difficulty can guide you well here. I note things all the time which indicate some measure by which you can consider, and when posters in reply will say that Google gives worthless search rankings—because I rank highly—then it's time to re-think confidence in the crowd and confidence in people who say such things with total confidence.

The math world will eventually confront its major problem, I'm sure. My confidence in that regard though is something I ponder myself, as the years have gone by, but the results from the Internet are telling in my opinion.

Such a massive problem requires massive forces to correct it.

It may not have been correctable without the age of the Internet given the resistance that is STILL being shown!!!

So this problem may have waited for the evolution of the Internet to be solved.

These math professors will go about their business as best they can until they are stopped. With Usenet posters fighting for them in hostile posts that will go after anyone and anything to preserve the error, even going after Google.

THAT is how difficult the situation is. Short answer: it's about as hard as it could possibly get. A near impossible task.

Sure I can put forward a mathematical argument and say I'm right, say I've traced it out carefully and that in reply a poster just deleted it all out, or simply refuses to address the actual argument and instead changes the subject or just makes false statements, but it can still seem like just one person's word against another's, and the crowd has the greater impact.

And one thing I've noticed over the years is that a group of maybe even only 6 posters will post with a lot of energy in reply to a person and will often use phrases like "everyone disagrees with you" or "no one likes you", or global statements to that effect as clearly they understand the power of presenting a situation that supposedly is about one person refusing to accept the truth from a large number of people.

So Google searches are this unique way to watch their behavior with something much bigger than they are!

Intriguingly though they simply demeaned Google.

Suddenly according to them, Google searches were "meaningless", and getting high search rankings was a trivial thing that anyone could do, though I don't put "anyone" in quotes as that at least isn't something I've seen actually stated.

Oh, and in facing high country counts of hits to my math blog as reported to Google Analytics, some might simply claim I was lying about them, while one poster confidently proclaimed the hits to be robot programs and forcefully argued that point.

One exercise for people contemplating the actual difficulty of something is to imagine doing it themselves, where Google is global enough that you can just go do searches on your own activities, or your ideas, or imagine putting an idea out there and seeing where it ranks.

Amazingly enough with that exercise you may only then be able to contemplate the feat of my having the definition of mathematical proof, by imagining doing it yourself.

So think of sitting down, writing up your own personal definition of mathematical proof—you can actually try this exercise if you wish—putting it out there, and taking over the #1 spot over much of the world, beating dictionaries, MathWorld, and the Wikipedia.

Now do the search in Google (has to be Google): definition of mathematical proof

For me it was surreal when I finally realized I was getting highly ranked for that definition and I learned it from search strings people were using to get to my math blog (um, I have a LOT of data from multiple sources).

Confidence oozing from posters can be mistaken for skill.

Readers can believe that no one would actually just stalk another person claiming they're wrong without having factual or mathematical basis, but I have watched that behavior for years. Confidence does not mean a person is right. Personally I don't trust confidence with mathematics and prefer when someone is honest about worries about a particular mathematical argument. Or concerns that they missed something, or are just plain wrong!

But human nature is to TRUST confidence. So my own admissions of error rather than helping me, are used by confident posters to reinforce the notion that I must be wrong.

It IS a competitive world.

And to me it has been amazing watching posters claim otherwise, as if anyone can have the definition of mathematical proof in Google, or "everyone" can have hits from 120+ countries on a yearly basis to their blog, if they just want it or something? I guess?

But at the end of it all of course the ultimate refuge for readers is the realization that it's really about mainstream mathematicians acknowledging or not acknowledging important mathematical research anyway, so why would I need to talk about Google search results if I were actually right?

Because the Internet as the new thing seems to be upending an old guard which has been informed of a massive error within number theory, which they may see as more easily ignored than addressed.

Google: algebraic integers vs complex numbers

If you even imagine that the ring of algebraic integers conflicts with the field of complex numbers you realize it's a defunct ring, but it's such a huge thing, and then for supposedly top mathematicians to pretend the problem doesn't exist?

How is such a thing possible?

Well as BP continues to try and stop an oil spill. And as the world continues to try and recover from a near financial collapse. And while the Catholic Church has to keep addressing issues of pedophile priests, it's not so hard to understand.

Judging difficulty can guide you well here. I note things all the time which indicate some measure by which you can consider, and when posters in reply will say that Google gives worthless search rankings—because I rank highly—then it's time to re-think confidence in the crowd and confidence in people who say such things with total confidence.

The math world will eventually confront its major problem, I'm sure. My confidence in that regard though is something I ponder myself, as the years have gone by, but the results from the Internet are telling in my opinion.

Such a massive problem requires massive forces to correct it.

It may not have been correctable without the age of the Internet given the resistance that is STILL being shown!!!

So this problem may have waited for the evolution of the Internet to be solved.

These math professors will go about their business as best they can until they are stopped. With Usenet posters fighting for them in hostile posts that will go after anyone and anything to preserve the error, even going after Google.

THAT is how difficult the situation is. Short answer: it's about as hard as it could possibly get. A near impossible task.

### Thursday, June 17, 2010

## JSH: Say anything

Arguing on Usenet is not about right or wrong. It's about people who like to argue. I admit that and I like to argue.

Actually I could just as easily attack my own research, and have done so in the past and that didn't matter either!

Posters take positions antagonistic to me as THAT'S WHAT YOU DO on Usenet, or you can't argue.

Arguing IS the point.

And it's not like any of us have to do anything anyway. If I'm right my math will win out regardless. If I'm wrong, it never will, regardless.

Easy. But arguing is FUN. So that's what happens, but it's not this thing of cosmic significance or anything.

Usenet is a fringe area of mathematics. It's where the math rejects go so that SOMEONE may actually listen to them, or at least they can play pretend that they matter. But they don't.

IF they mattered, then they could go somewhere else.

It's the gutter system of the mathematical world.

Actually I could just as easily attack my own research, and have done so in the past and that didn't matter either!

Posters take positions antagonistic to me as THAT'S WHAT YOU DO on Usenet, or you can't argue.

Arguing IS the point.

And it's not like any of us have to do anything anyway. If I'm right my math will win out regardless. If I'm wrong, it never will, regardless.

Easy. But arguing is FUN. So that's what happens, but it's not this thing of cosmic significance or anything.

Usenet is a fringe area of mathematics. It's where the math rejects go so that SOMEONE may actually listen to them, or at least they can play pretend that they matter. But they don't.

IF they mattered, then they could go somewhere else.

It's the gutter system of the mathematical world.

### Sunday, June 13, 2010

## JSH: Yeah fame is weird

About a decade or so ago some poster noted in reply to me that I was famous and I replied back that no, I'm infamous. But I guess it really is about how you define "fame" and everybody seems to have their own personal definition.

But, I am read in somewhere around 120 countries that I can verify just by hits to my math blog as reported by Google Analytics. Search strings around my research tend to be in the top 10 at a level that indicates people driving them there from all over the world.

But you may say, you KNOW fame, and there's no way that any of that matters as you don't see me on television, don't read newspaper articles about me. I'm not on Youtube even! How can I be famous?

Well it turns out that I'm read in about 120 countries according to Google Analytics just for hits to my math blog, on a yearly basis.

A lot of people narrowly define fame around celebrity, and ten around the most visible celebrities, or around some vague notion of it that makes sense to them.

But for a while now I've been probably the most influential single human being in the math field on the planet by far.

I've actually been hoping to avoid celebrity. Seems it can be kind of annoying.

So I've been impacting the math field for a while now and noting that impact, as I try to adjust it to the implications of some of my results and hopefully minimize the damage, especially the collateral damage. It's a scary task. Quite simply I've been re-working the mathematical field worldwide, slowly and steadily, for years now.

But, I am read in somewhere around 120 countries that I can verify just by hits to my math blog as reported by Google Analytics. Search strings around my research tend to be in the top 10 at a level that indicates people driving them there from all over the world.

But you may say, you KNOW fame, and there's no way that any of that matters as you don't see me on television, don't read newspaper articles about me. I'm not on Youtube even! How can I be famous?

Well it turns out that I'm read in about 120 countries according to Google Analytics just for hits to my math blog, on a yearly basis.

A lot of people narrowly define fame around celebrity, and ten around the most visible celebrities, or around some vague notion of it that makes sense to them.

But for a while now I've been probably the most influential single human being in the math field on the planet by far.

I've actually been hoping to avoid celebrity. Seems it can be kind of annoying.

So I've been impacting the math field for a while now and noting that impact, as I try to adjust it to the implications of some of my results and hopefully minimize the damage, especially the collateral damage. It's a scary task. Quite simply I've been re-working the mathematical field worldwide, slowly and steadily, for years now.

## JSH: Why being right isn't enough

People wonder how you can argue on Usenet, or in other places, endlessly over subjects like math, and you can look over my recent Usenet postings to see how, as if other people refuse to accept something because they find it distasteful or repugnant, then they can just refuse to accept it.

There is no other known general method other than brute force for finding k, when k^m = q mod N, besides what I found.

Now a rational person learning that a new way has been discovered would find that exciting—if it came from established sources.

But the fact coming from me gets a lot of denial in response as that changes the status quo. And people are social animals.

So yeah, I've seen this behavior for years. Human beings are not rational creatures they are social creatures.

Of course cryptology people will proclaim they'd accept a major new find from any source, reality is, ask yourself: is there any other known general method OF ANY TYPE besides brute force for finding k, when k^m = q mod N besides what I found?

If the answer is, no. Then you're risking national security by ignoring this result.

And no matter what your social gut tells you now, if any of you have security clearances in ANY COUNTRY around the globe, at a minimum those will be stripped from you later, and you will become persona non grata within the security community.

As hindsight is 20-20 and cruel. No government will trust you later no matter what explanation you try to give.

There is no other known general method other than brute force for finding k, when k^m = q mod N, besides what I found.

Now a rational person learning that a new way has been discovered would find that exciting—if it came from established sources.

But the fact coming from me gets a lot of denial in response as that changes the status quo. And people are social animals.

So yeah, I've seen this behavior for years. Human beings are not rational creatures they are social creatures.

Of course cryptology people will proclaim they'd accept a major new find from any source, reality is, ask yourself: is there any other known general method OF ANY TYPE besides brute force for finding k, when k^m = q mod N besides what I found?

If the answer is, no. Then you're risking national security by ignoring this result.

And no matter what your social gut tells you now, if any of you have security clearances in ANY COUNTRY around the globe, at a minimum those will be stripped from you later, and you will become persona non grata within the security community.

As hindsight is 20-20 and cruel. No government will trust you later no matter what explanation you try to give.

### Saturday, June 12, 2010

## General factoring result for k^m = q mod N

Oddly enough a fairly simple general result relates finding k, when k^m = q mod N, where m is a natural number to factoring.

Given an mth residue where m is a natural number, q mod N, to be solved one can find k, where

k^m = q mod N, from

k = (a_1+a_2+...+a_m)^{-1} (f_1 +...+ f_m) mod N

where f_1*...*f_m = T, and T = a_1*...*a_m*q mod N

and the a's are free variables as long as they are non-zero and their sum is coprime to N.

So you get some T, such that T = a_1*...*a_m*q mod N, factor it, and you may have k using its factors with that simple relation.

It's a general result, which may have been known to Gauss and simply didn't get written down, or maybe he did and no one noticed. It's not the sort of thing that had the importance in the past that it MAY have in our modern age of computers and systems based on factoring as a hard problem.

I actually generalized to the full result about a month ago, having in the past on this newsgroup mentioned a simpler quadratic result that I noticed first!

I think some posters derided it as too simple and they moved on. I puzzled over it a few more months and realized that I could generalize to m, a Natural number.

As it is a general result it's hard to say much about how it works, especially with m greater than 2. It has intriguing behavior though even with the quadratic case.

I'll admit is is a sobering find for me, as it's an incredibly simple result to prove, is general for residues connecting them to integer factorization in a deep way, and looks like the kind of result one would expect to be in the front of a number theory textbook on modular arithmetic.

Yet I'm the one who found it, over 200 years since Gauss introduced "mod" in 1801.

Ok, I'll stop there. I won't be surprised to see a lot of hateful and hostile replies in response. I've been talking it out on sci.math and have been ripped on continuously. The insult-fest never ends on Usenet though. That's part of what defines Usenet.

No matter what, someone insults you.

Given an mth residue where m is a natural number, q mod N, to be solved one can find k, where

k^m = q mod N, from

k = (a_1+a_2+...+a_m)^{-1} (f_1 +...+ f_m) mod N

where f_1*...*f_m = T, and T = a_1*...*a_m*q mod N

and the a's are free variables as long as they are non-zero and their sum is coprime to N.

So you get some T, such that T = a_1*...*a_m*q mod N, factor it, and you may have k using its factors with that simple relation.

It's a general result, which may have been known to Gauss and simply didn't get written down, or maybe he did and no one noticed. It's not the sort of thing that had the importance in the past that it MAY have in our modern age of computers and systems based on factoring as a hard problem.

I actually generalized to the full result about a month ago, having in the past on this newsgroup mentioned a simpler quadratic result that I noticed first!

I think some posters derided it as too simple and they moved on. I puzzled over it a few more months and realized that I could generalize to m, a Natural number.

As it is a general result it's hard to say much about how it works, especially with m greater than 2. It has intriguing behavior though even with the quadratic case.

I'll admit is is a sobering find for me, as it's an incredibly simple result to prove, is general for residues connecting them to integer factorization in a deep way, and looks like the kind of result one would expect to be in the front of a number theory textbook on modular arithmetic.

Yet I'm the one who found it, over 200 years since Gauss introduced "mod" in 1801.

Ok, I'll stop there. I won't be surprised to see a lot of hateful and hostile replies in response. I've been talking it out on sci.math and have been ripped on continuously. The insult-fest never ends on Usenet though. That's part of what defines Usenet.

No matter what, someone insults you.

### Friday, June 11, 2010

## JSH: When your math is wrong

The real underlying problem is that years ago I found a foundational error in number theory. So to number theorists thoroughly trained in error there may be a sense that there is no upside in the truth!!!

And you know? From their perspective maybe mathematics betrayed them. How many "learning experiences" do they have? With homework and tests, along with years of "research", long nights? Long discussions with other mathematicians?

Only to find there is mathematics that says that was all junk?

I still can't get over the mathematicians who have simply left the country on sabbatical when confronted with my math. It's like they just have to get away from a while, but they DO come back, and apparently just go back to the familiar. Back to error.

I've speculated that the math error actually selects out certain types of people who tolerate error—despite their denial. As it seems reasonable that people with a mathematical intuition that sensed something was wrong, might go to another field or focus on applied mathematics exclusively—thereby avoiding the error.

IF that speculation is correct, then the people who gained prominence in number theory, paradoxically, are the WORST math people, possibly worse than the average, as they were instead ATTRACTED to error, and found success with it.

To such people mathematical truth may seem to now be the enemy. Having broken brains anyway, they can easily ignore even dramatic evidence proving they are wrong, and just keep doing what they are doing!!!

An intractable situation which has gone on for years now. These people destroyed an entire mathematical journal to protect their error.

Living in error IS their way of life.

And you know? From their perspective maybe mathematics betrayed them. How many "learning experiences" do they have? With homework and tests, along with years of "research", long nights? Long discussions with other mathematicians?

Only to find there is mathematics that says that was all junk?

I still can't get over the mathematicians who have simply left the country on sabbatical when confronted with my math. It's like they just have to get away from a while, but they DO come back, and apparently just go back to the familiar. Back to error.

I've speculated that the math error actually selects out certain types of people who tolerate error—despite their denial. As it seems reasonable that people with a mathematical intuition that sensed something was wrong, might go to another field or focus on applied mathematics exclusively—thereby avoiding the error.

IF that speculation is correct, then the people who gained prominence in number theory, paradoxically, are the WORST math people, possibly worse than the average, as they were instead ATTRACTED to error, and found success with it.

To such people mathematical truth may seem to now be the enemy. Having broken brains anyway, they can easily ignore even dramatic evidence proving they are wrong, and just keep doing what they are doing!!!

An intractable situation which has gone on for years now. These people destroyed an entire mathematical journal to protect their error.

Living in error IS their way of life.

## JSH: Posting realities, residues result

I've been fascinated by the response to my posting of a general result for solving quadratic residues because of the elephant in the room NOT mentioned, which is that it is mod N, whereas mathematicians usually MUST use mod p, where p is a prime.

That's important for factoring as if N is a composite then with k^2 = q mod N, you have multiple non-trivial solutions for k, where I mean, not just k, and N - k, whereas if N=p, then there is only one.

And in all the discussion that erupted around my postings on this result you may have noticed I noted that the approach tends to prefer large k, though all the details of how that works out are not clear even to me.

So imagine N = p_1*p_2, so there are

So it's a way to deliberately probe for factors of N, picking k_1 in order to try and get k_2, which is a technique not available with any other known method, because no other general way to solve for residues is known!

So why would posters argue endlessly with me on other subjects and miss this amazing thing that the approach is mod N and not mod p? Because they hate me. Also they want me to be wrong!!! No matter what as it's personal with them.

Mathematics is a sideline to their actual activity. It doesn't rank in importance to what they really want.

Human beings are quirky creatures. They can do the damndest things. So if you're a Usenet poster arguing with people on Usenet the LAST THING you wish for a target to be, is actually right!

I'll give the result again, and note that it's trivially derived though I won't give the derivation again. I'll also note that the result is basic research so it's not clear how hard it is to get it to work at any level. The devil is in the details.

For instance with nuclear weapons it's a far cry from knowing you just have to slap some plutonium together to actually building a working nuclear weapon.

Now then, back to national security! It IS quite possible that this information could be of interest to governments, oh, all over the world. Failure to disclose of it for some of you could be seen as a LACK OF LOYALTY in your respective countries.

Usenet posters who reply decrying the result could face extraordinary scrutiny in the near future if only from the world press wondering how they could do such a thing. And could find themselves labeled for life.

Post in reply now with care. No matter how little you think of Usenet, you can make a decision in this thread which you can't take back, which will end the life you knew, and move you into a Hell on earth that you will not escape until you die.

Given an mth residue where m is a natural number, q mod N, to be solved one can find k, where

k^m = q mod N, from

k = (a_1+a_2+...+a_m)^{-1} (f_1 +...+ f_m) mod N

where f_1*...*f_m = T, and T = a_1*...*a_m*q mod N

and the a's are free variables as long as they are non-zero and their sum is coprime to N.

It's a general result, which may have been known to Gauss and simply didn't get written down, or maybe he did and no one noticed. It's not the sort of thing that had the importance in the past that it MAY have in our modern age of computers and systems based on factoring as a hard problem.

It is a general result at the heart of modular arithmetic. No one can really say for sure how big it actually may be as general results have that quality. Kind of like differentiation with the calculus. How big is that? Has humanity determined its limits yet?

That's important for factoring as if N is a composite then with k^2 = q mod N, you have multiple non-trivial solutions for k, where I mean, not just k, and N - k, whereas if N=p, then there is only one.

And in all the discussion that erupted around my postings on this result you may have noticed I noted that the approach tends to prefer large k, though all the details of how that works out are not clear even to me.

So imagine N = p_1*p_2, so there are

**TWO**values for k, given k^2 = q mod N. If you**picked**k_1 to find q, such that it will tend to be small, and k_2 is more likely to be selected then the approach I discovered could conceivably give you k_2, then you have p_1 or p_2 from (k_2 - k_1) gcd N.So it's a way to deliberately probe for factors of N, picking k_1 in order to try and get k_2, which is a technique not available with any other known method, because no other general way to solve for residues is known!

So why would posters argue endlessly with me on other subjects and miss this amazing thing that the approach is mod N and not mod p? Because they hate me. Also they want me to be wrong!!! No matter what as it's personal with them.

Mathematics is a sideline to their actual activity. It doesn't rank in importance to what they really want.

Human beings are quirky creatures. They can do the damndest things. So if you're a Usenet poster arguing with people on Usenet the LAST THING you wish for a target to be, is actually right!

I'll give the result again, and note that it's trivially derived though I won't give the derivation again. I'll also note that the result is basic research so it's not clear how hard it is to get it to work at any level. The devil is in the details.

For instance with nuclear weapons it's a far cry from knowing you just have to slap some plutonium together to actually building a working nuclear weapon.

Now then, back to national security! It IS quite possible that this information could be of interest to governments, oh, all over the world. Failure to disclose of it for some of you could be seen as a LACK OF LOYALTY in your respective countries.

Usenet posters who reply decrying the result could face extraordinary scrutiny in the near future if only from the world press wondering how they could do such a thing. And could find themselves labeled for life.

Post in reply now with care. No matter how little you think of Usenet, you can make a decision in this thread which you can't take back, which will end the life you knew, and move you into a Hell on earth that you will not escape until you die.

Given an mth residue where m is a natural number, q mod N, to be solved one can find k, where

k^m = q mod N, from

k = (a_1+a_2+...+a_m)^{-1} (f_1 +...+ f_m) mod N

where f_1*...*f_m = T, and T = a_1*...*a_m*q mod N

and the a's are free variables as long as they are non-zero and their sum is coprime to N.

It's a general result, which may have been known to Gauss and simply didn't get written down, or maybe he did and no one noticed. It's not the sort of thing that had the importance in the past that it MAY have in our modern age of computers and systems based on factoring as a hard problem.

It is a general result at the heart of modular arithmetic. No one can really say for sure how big it actually may be as general results have that quality. Kind of like differentiation with the calculus. How big is that? Has humanity determined its limits yet?

### Monday, June 07, 2010

## JSH: Finally some social crap answers

The great thing about this latest math result of mine is that it's trivial to prove, involves modular arithmetic, so it's in a critical area to information security, is a simple, general result, so it may be one of those rare things in mathematics that should have been discovered in this case, oh, about 200 years ago.

And now I can test math society without any shadows of doubts.

So far, no surprises. The same mathematicians who ignored me before appear to be ignoring me now. I emailed Granville and Ribet DIRECTLY, though their spam filters may have blocked. I've emailed them off and on over the years (not always nicely).

Oh and I emailed Nicely. Oh and I think I emailed Lenstra. Each may have the spam filter excuse.

So what about those mythical grad students that supposedly are there to protect by seizing on a math result even if professors don't? Hiding I guess. I emailed Lagarias and also emailed two of his grad students. They still though all may claim spam filters blocked.

I have the reply from the Annals of Mathematics noting receipt. But don't know whether or not the Princeton staffer actually forwarded to a mathematician or not, but may have and paper could be in a pile somewhere waiting to be considered.

The result is on my math blog of course, and I'm pushing it in several places, but it's hard to say who is maybe reading it from where.

All in all within the muddle some may wonder if I do have answers, but I do. It is revealed that math society moves like a glacier in this situation. And I have responses on this newsgroup.

With this result it's like bas-relief watching replies as I see the same things I've seen for years, but now there is no doubt.

Contrast in the past when mathematicians have had breaking results which echo around the math community and there is a lot of noise with an expectation that the world will follow along. Silly world.

Here's the result again:

Given an mth residue where m is a natural number, q mod N, to be solved one can find k, where

k^m = q mod N, from

k = (a_1+a_2+...+a_m)^{-1} (f_1 +...+ f_m) mod N

where f_1*...*f_m = T, and T = a_1*...*a_m*q mod N

and the a's are free variables as long as they are non-zero and their sum is coprime to N.

It's trivially proven, and shows that integer factorization and residues are very closely related.

To the mathematics, integer factorization may just be about that result.

Simple results in mathematics that are general results tend to have a huge impact, and so are rare, and supposedly all found at this time.

If that result is really big its greatest impact may be known in the future, as if it leads to a new math of integer factorization, if it had been widely known 200 years ago then much of our recent Internet history could not have happened as it did.

Without RSA encryption the history of the Internet at this point would be entirely different.

So maybe powerful forces guaranteed this reality, eh?

I am the latest major discoverer. I have big shoes to fill. There was no way in this modern age I wouldn't be coming in without some seriously big guns. Quite simply I do have to stand on the shoulders of those who came before me, but I still have to top them.

And given the weight of human history that task gets harder and harder with each major discoverer.

So every single one of us has to be a major bad-ass.

And now I can test math society without any shadows of doubts.

So far, no surprises. The same mathematicians who ignored me before appear to be ignoring me now. I emailed Granville and Ribet DIRECTLY, though their spam filters may have blocked. I've emailed them off and on over the years (not always nicely).

Oh and I emailed Nicely. Oh and I think I emailed Lenstra. Each may have the spam filter excuse.

So what about those mythical grad students that supposedly are there to protect by seizing on a math result even if professors don't? Hiding I guess. I emailed Lagarias and also emailed two of his grad students. They still though all may claim spam filters blocked.

I have the reply from the Annals of Mathematics noting receipt. But don't know whether or not the Princeton staffer actually forwarded to a mathematician or not, but may have and paper could be in a pile somewhere waiting to be considered.

The result is on my math blog of course, and I'm pushing it in several places, but it's hard to say who is maybe reading it from where.

All in all within the muddle some may wonder if I do have answers, but I do. It is revealed that math society moves like a glacier in this situation. And I have responses on this newsgroup.

With this result it's like bas-relief watching replies as I see the same things I've seen for years, but now there is no doubt.

Contrast in the past when mathematicians have had breaking results which echo around the math community and there is a lot of noise with an expectation that the world will follow along. Silly world.

Here's the result again:

Given an mth residue where m is a natural number, q mod N, to be solved one can find k, where

k^m = q mod N, from

k = (a_1+a_2+...+a_m)^{-1} (f_1 +...+ f_m) mod N

where f_1*...*f_m = T, and T = a_1*...*a_m*q mod N

and the a's are free variables as long as they are non-zero and their sum is coprime to N.

It's trivially proven, and shows that integer factorization and residues are very closely related.

To the mathematics, integer factorization may just be about that result.

Simple results in mathematics that are general results tend to have a huge impact, and so are rare, and supposedly all found at this time.

If that result is really big its greatest impact may be known in the future, as if it leads to a new math of integer factorization, if it had been widely known 200 years ago then much of our recent Internet history could not have happened as it did.

Without RSA encryption the history of the Internet at this point would be entirely different.

So maybe powerful forces guaranteed this reality, eh?

I am the latest major discoverer. I have big shoes to fill. There was no way in this modern age I wouldn't be coming in without some seriously big guns. Quite simply I do have to stand on the shoulders of those who came before me, but I still have to top them.

And given the weight of human history that task gets harder and harder with each major discoverer.

So every single one of us has to be a major bad-ass.

### Saturday, June 05, 2010

## JSH: Importance of knowledge

I want to emphasize to all of you that the issue is one of a potential technological shift.

If you know your history you should know the importance of technology, from the military tactics of the Roman legions, to the sailing ships of Europeans, and the computers of our modern age, the civilizations with the technological edge, win.

If a country like North Korea or Iran gets through the door on a major exploit then you can wake up in a world where technology has shifted in a way you cannot foresee,

But the future you may live to witness may be one where the dominant country is one you would say can't be dominate that fast, or ever, today.

Also on the darker side what you cannot imagine now can be very clear to people later evaluating behaviors around this result. It is not fun and games if you're facing some people in your home country and they're asking you why you did not do what they would have expected.

Depending on the country what you say then, or what you can say, may mean life or death for you.

So some readers around the world are now in a more delicate position for that reason alone, regardless of anything else.

Because in your country, doing nothing may be seen as a failed test of loyalty. A betrayal of your state. A failure to your country and its leaders.

If you know your history you should know the importance of technology, from the military tactics of the Roman legions, to the sailing ships of Europeans, and the computers of our modern age, the civilizations with the technological edge, win.

If a country like North Korea or Iran gets through the door on a major exploit then you can wake up in a world where technology has shifted in a way you cannot foresee,

But the future you may live to witness may be one where the dominant country is one you would say can't be dominate that fast, or ever, today.

Also on the darker side what you cannot imagine now can be very clear to people later evaluating behaviors around this result. It is not fun and games if you're facing some people in your home country and they're asking you why you did not do what they would have expected.

Depending on the country what you say then, or what you can say, may mean life or death for you.

So some readers around the world are now in a more delicate position for that reason alone, regardless of anything else.

Because in your country, doing nothing may be seen as a failed test of loyalty. A betrayal of your state. A failure to your country and its leaders.

## Solving k^m = q mod N

While I'd prefer to stay away from the hostility, lying, and other misinformation threats of Usenet I'm kind of stuck with a surprising situation to me around my latest major result, a way to solve for k, when k^m = q mod N.

Here's the full result and simple derivation:

Given an mth residue where m is a natural number, q mod N, to be solved one can find k, where

k^m = q mod N, from

k = (a_1+a_2+...+a_m)^{-1} (f_1 +...+ f_m) mod N

where f_1*...*f_m = T, and T = a_1*...*a_m*q mod N

and the a's are free variables as long as they are non-zero and their sum is coprime to N.

It's trivially derived by noting that if you have

f_1 = a_1*k mod N thru f_m = a_m*k mod N

multiplying them together gives f_1*...*f_m = a_1*...*a_m*k^m = a_1*...*a_m*q mod N.

But *adding* them together gives (f_1+...+ f_m) = (a_1+...+a_m)*k mod N, and solving for k is easy enough.

So what I found is a simple general result of modular arithmetic.

But supposedly such results were all found long ago. This result by current mathematical teaching, should not exist as a new result. It should have been discovered nearly 200 years ago, around the time that Gauss introduced modular arithmetic.

And it MAY have been discovered back then, but been considered too trivial to write down…and then it became lost.

But prior mathematicians didn't have a world using integer factorization as a way to secure flows of information and billions upon billions of dollars. To them integer factorization did not have the meaning it has to people today.

I've put the result into a paper and sent to the Annals of Mathematics in Princeton, and got a verification of receipt the next day, but of course I don't know if they're actually looking at it, or I got a polite reply used for "cranks".

I've made several contacts by email, including Granville, Ribet, and others whose names don't come to me right now typing this post.

No replies.

Nothing.

I'm getting little to no feedback on this result. My math blog is showing no up-tick in hits related to it.

The only measure of its impact so far is, yup, Google, where the title of my paper "Solving Residues" is coming up #1.

So I'm desperate enough to come back to Usenet as, um, things aren't supposed to work this way. If integer factorization gets re-written we're on a path to that happening in the worst way, with every major world power possible learning about it occurring by an exploit.

Desperate nations might be the first to find a result like this one. If the general result does lead to another number theory, and another way to factor, and who knows what else, then these people could rapidly gain traction in ways they currently cannot by their current military power or technological infrastructure.

So you could be witnessing the end of the current world order from a bizarre path. It may be destined, but I hope not.

As if that happens, the future you see will not be the United States the dominant country, with Europe a dominant force, but those countries potentially on the BOTTOM, in a technologically advanced future we can no more see today, than people a hundred years ago could see this one.

Being a citizen of the United States and living here quite happily I'd REALLY not like that scenario.

So I know, many of you see yourselves as unimportant. You see what happens in these discussions as of no account as the world doesn't account for it much. In my opinion the dregs of math society end up on Usenet as where else can they go?

You are people who are comfortable with being nobodies being told that the fate of the world requires that you do something, though it's not clear what.

The future will arrive no matter what you do, and if you are later in an upside down world, you unlike so many others who will get dragged along who had no choice, no input, and no way to do anything, will know that you, did.

Here's the full result and simple derivation:

Given an mth residue where m is a natural number, q mod N, to be solved one can find k, where

k^m = q mod N, from

k = (a_1+a_2+...+a_m)^{-1} (f_1 +...+ f_m) mod N

where f_1*...*f_m = T, and T = a_1*...*a_m*q mod N

and the a's are free variables as long as they are non-zero and their sum is coprime to N.

It's trivially derived by noting that if you have

f_1 = a_1*k mod N thru f_m = a_m*k mod N

multiplying them together gives f_1*...*f_m = a_1*...*a_m*k^m = a_1*...*a_m*q mod N.

But *adding* them together gives (f_1+...+ f_m) = (a_1+...+a_m)*k mod N, and solving for k is easy enough.

So what I found is a simple general result of modular arithmetic.

But supposedly such results were all found long ago. This result by current mathematical teaching, should not exist as a new result. It should have been discovered nearly 200 years ago, around the time that Gauss introduced modular arithmetic.

And it MAY have been discovered back then, but been considered too trivial to write down…and then it became lost.

But prior mathematicians didn't have a world using integer factorization as a way to secure flows of information and billions upon billions of dollars. To them integer factorization did not have the meaning it has to people today.

I've put the result into a paper and sent to the Annals of Mathematics in Princeton, and got a verification of receipt the next day, but of course I don't know if they're actually looking at it, or I got a polite reply used for "cranks".

I've made several contacts by email, including Granville, Ribet, and others whose names don't come to me right now typing this post.

No replies.

Nothing.

I'm getting little to no feedback on this result. My math blog is showing no up-tick in hits related to it.

The only measure of its impact so far is, yup, Google, where the title of my paper "Solving Residues" is coming up #1.

So I'm desperate enough to come back to Usenet as, um, things aren't supposed to work this way. If integer factorization gets re-written we're on a path to that happening in the worst way, with every major world power possible learning about it occurring by an exploit.

Desperate nations might be the first to find a result like this one. If the general result does lead to another number theory, and another way to factor, and who knows what else, then these people could rapidly gain traction in ways they currently cannot by their current military power or technological infrastructure.

So you could be witnessing the end of the current world order from a bizarre path. It may be destined, but I hope not.

As if that happens, the future you see will not be the United States the dominant country, with Europe a dominant force, but those countries potentially on the BOTTOM, in a technologically advanced future we can no more see today, than people a hundred years ago could see this one.

Being a citizen of the United States and living here quite happily I'd REALLY not like that scenario.

So I know, many of you see yourselves as unimportant. You see what happens in these discussions as of no account as the world doesn't account for it much. In my opinion the dregs of math society end up on Usenet as where else can they go?

You are people who are comfortable with being nobodies being told that the fate of the world requires that you do something, though it's not clear what.

The future will arrive no matter what you do, and if you are later in an upside down world, you unlike so many others who will get dragged along who had no choice, no input, and no way to do anything, will know that you, did.