Monday, January 14, 2008

 

JSH: Underlying math

The research I've been doing is actually about a Diophantine result, where I've been exploring two sets of equations:

x^2 = y^2 + pr_1

and

z^2 = y^2 + nT

where T is the target composite to be factored, and p is an odd prime, coprime to 3—also I've specified that nT must be odd, coprime to 3, and p.

I discovered you can introduce additional variables, where importantly I let z = x+k. Then I find constraints on what values k can have to give integer solutions, which are congruence results.
Correction: z = x + ak.
Those results are determined by mathematical proof, so they are absolutes.

They represent the conditions under which integer solutions will fit. Challenging the results is not an option as they are proven. The only option is to say that they don't matter and constraints on those solutions are of no use to researchers. But is that the truth?

I repeat, what my latest research is doing is finding the rules whereby there are solutions with all integers for those two equations. That's it. The conclusions I have are supported by an absolute argument.

That is, the support is a mathematical proof.
We shall call them The Rules:

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

where k is key, and k^2 = (a^2+1)^{-1}(nT) mod p.

And now you have the last important helper variable: a. It actually represents a set of solutions such that k exists i.e. (a^2+1)^{-1}(nT).

And, finally, you have

y = (1+2a^2)^{-1} z mod p, or y = -(1+2a^2)^{-1} z mod p.

Those are not just nifty looking equations but are absolutes that determine what integer can fit into

x^2 = y^2 + pr_1

and

z^2 = y^2 + nT

where, you should recall, z = x + ak.

So that's what those equations are: absolute mathematical rules that work over infinity to constrain the set of integer solutions available,

Which means that if you find integers that will fit into those equations they MUST obey ALL THE RULES.

There are a finite number of a's for any given nT and p, which gives a finite number of values for z mod p, y mod p, and therefore, for z-y mod p, and z+y mod p.

If p>sqrt(nT), then one of those values must factor nT. That is a mathematical absolute.

The one issue is, how many of those solutions are there?

The proof says that for every integer solution, there is this equation that defines when the a's exist:

a^2 = 2^{-1}(x^{-1} z - 1) mod p

so you get another quadratic residue. So there are two a's for every possible x mod p and z mod p that can fit. If none do, then there are NO solutions possible for that prime p.

So the a's tell you of the existence of integer solutions, over infinity for every possible (nT) mod p, that can fit with the two underlying equations, and it tells you how many possible residues there are modulo p.

With n=1, T=119, and p=11, note that only a=2 mod 11, and a=5 mod 11 will fit, over infinity.





<< Home

This page is powered by Blogger. Isn't yours?