Harristotelian Logic

Intriguingly to me, or maybe a curiosity to others my latest factoring research allows you to find the last digits of the factors of a product of two primes with absolute certainty and do so for even a public key size number or a public key itself.

Now what I'm showing is a demonstration to let you know that what I've discovered is like nothing else found before by doing something that no one else can do.

Of course, finding the last digits of the factors does not give you the full factorization, but hey, it's a telling way for me to impress upon you why what I've found is of interest to REAL mathematicians and cryptographers so those who act not interested, are fakes.

Let z^2 = y^2 + T,

where at this point, of course, z and y are unknowns but I'm going to solve for them modulo 5.

You will first find what I call alpha and k, where

k^2 = (a^2 +1)^{-1} T mod 5

so you pick alpha so that the quadratic residue exist.

Now you can solve for z, as

z = k(1+2alpha^2)(2alpha)^{-1) mod 5.

Isn't that incredible!!! Such a surprisingly simple answer!!!

With z you can find y, by solving as a congruence y^2 = z^2 - T, and then you find

f_1 = (z-y) mod 5 and f_2 = (z+y) mod 5.

If the residue is odd then that is the last digit of either factor, if it is even, then adding it to 5 will give the last digit.

And that's it! You can do it on anything, including an RSA public key. But now I'll show you on something much, much easier and smaller.

Here's a demonstration: Let T = 247 = 13*19. Then I find that

k^2 = (a^2 + 1)^{-1} (247) mod 5 = (a^2 + 1)^{-1} (2) mod 5

so alpha = 1 is a solution, which gives k = 1 mod 5 or k = -1 mod 5 as solutions where I give both for a reason which will be clear in a moment, as then

z = (1+2)(2)^{-1) mod 5 = 3(3) mod 5 = 4 mod 5.

Then y^2 = 16 - 247 mod 5 = -1 mod 5, so y = 3 mod 5.

Then I get z-y = 1 mod 5, and z+y=7 mod 5, and notice I have the last digit of the target, so that is the trivial factorization, but I have another solution for k, so using k=-1 mod 5, I get

z = -(1+2)(2)^{-1) mod 5 = -3(3) mod 5 = 1 mod 5.

Then y^2 = 1 - 247 mod 5 = -1 mod 5, so again y = 3 mod 5.

And z+y = 4 mod 5 and z-y = -2 mod 5 = 3 mod 5.

And remembering my rules from above that if the residue is even you add 5, I have the last digits of both factors as one is 9, and the other is 3.

There is a trivial derivation on my math blog, so yes the mathematics is absolute and YES, you can be used on a public key.

You can do the same with an RSA challenge number and determine the last digits of its factors or you can do that with an RSA public key.

The method does not care.

But I need your help. There are powerful people in the mathematical and cryptographic communities who are just politicians. They lie about math because they can't really do it, and seek to block out real mathematicians so that they can make a new world where fakery is all there is and real mathematical truth is lost.

Your challenge is to help me beat them. No real mathematician could dismiss this result.

Remember that—no REAL mathematician could dismiss it. If they claim it's worthless then challenge them to do the same—to match me, with a method of their own.

The challenge is now yours.

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Dumb. I realized after making this post that it doesn't do much modulo 5, so I'm re-doing modulo 11.

You will first find what I call alpha and k, where

k^2 = (a^2 +1)^{-1} T mod 11

so you pick alpha so that the quadratic residue exist.

Now you can solve for z, as

z = k(1+2a^2)(2a)^{-1) mod 11.

Isn't that incredible!!! Such a surprisingly simple answer!!!

With z you can find y, by solving as a congruence y^2 = z^2 - T, and then you find

f_1 = (z-y) mod 11 and f_2 = (z+y) mod 11.

If the residue is odd then its last digit is the last digit of either factor, if it is even, then adding it to 11, take the last digit.

And that's it! You can do it on anything, including an RSA public key. But now I'll show you on something much, much easier and smaller.

Here's a demonstration: Let T = 247 = 13*19. Then I find that

k^2 = (a^2 + 1)^{-1} (247) mod 11 = (a^2 + 1)^{-1} (5) mod 11

so alpha must equal 2, -2, 5 or -5 for k to exist.

Trying the first, I find that with alpha = 2, k=1 mod 11 or -1 mod 11, and trying the first:

z = k(1+2alpha^2)(2alpha)^{-1) mod 11 = (9)(4)^{-1) mod 11 = 5 mod 11.

Then y^2 = 25 - 247 mod 11 = -222 mod 11 = 9 mod 11, so y=3 mod 11 is a solution.

Then I get z-y = 2 mod 11, and z+y= 8 mod 11.

So in both case you add 11, and get 13 and 19, so also in this case you get a factorization.

You can do the same with an RSA challenge number and determine the last digits of its factors or you can do that with an RSA public key.

The method does not care.

But people do, which is why I'm stuck trying to make some kind of demonstration to make the math dance like a trained monkey in a way that enough people around the world know this is something big so that I can't get past the gatekeepers.

Making beautiful mathematics dance like a monkey for people to believe.

Luckily, the math does not care. It is true no matter what.

But remember later that to get the truth out against the people called mathematicians, the math had to be made to dance, and entertain.

It had to dance and entertain like a trained monkey because you had to be convinced.

This modern world HATES ideas and more than it hates ideas it hates people like me who find them, while it says otherwise while exploiting what was found before, while pretenders claim they're still discovering things today.

You hate discovery, but more than you hate discovery, you hate the discoverers.