Re: Quadratic residues and factoring

jstevh@msn.com wrote:
> Wouldn't you know I go to factoring, yet again, as, for one thing, just
> about all of my research relates to factoring one way or another, and
> for another, I think that the factoring problem has a big enough
> economic importance that I focus on it as a way to break the impasse.
>
> This result might be a way to prove Goldbach's Conjecture, but forget
> that, I prove that and mathematicians can just lie about that result
> like everything else.
>
> So I don't see value in working to prove Goldbach's Conjecture, as it
> would really depress me to prove it, and still be arguing with sci.math
> people, like with all my other research results.
>
> Mathematicians can ignore a proof of Goldbach's Conjecture, but I get a
> factoring result, and well...
>
> So what is this thing?
>
> The result is that given odd naturals summed to get a composite:
>
> n_1 + n_2 = C
>
> it must be true that, if k is an odd difference of factors of 2*C that
>
> 8n_2 + k^2 must be a quadratic residue of p, where p is a prime factor
> of n_1, or
>
> (8n_2 + k^2) mod p, where p is a prime factor of n_1,
>
> must be quadratic residue of n_1, or to state it the way I've been told
> is correct terminology,
>
> 8n_2 + k^2 must be a square modulo p, where p is a prime factor of n_1,
> or
>
> (8n_2 + k^2) mod p must be a square modulo p.
>
> So how can that be used to factor?
>
> Well, if n_1 is a composite with two primes as factors, then maybe you
> can find their individual quadratic residues.
>
> For example, using n_1 = 55, I will let n_2 = 17, so I have 2*C = 144,
> and picking the factorization 16(9), I have k = 7.
>
> Then
>
> 8*17 + 49 = 185
>
> and 185 mod 55 = 20.
>
> And 20 is 0 mod 5 and 9 mod 11, so they are squares, but if you didn't
> know that, what would you know?
>
> I don't know. You could check the 0 square modulo p possibility by
> factoring 20, which would give you 5. But hey, you can use another k,
> so let's use the factorization
>
> 48(3), which gives k = 45, then
>
> 8*17 + 45^2 = 2161, and 2161 mod 55 = 16.
>
> And I don't know what that tells me, if anything. Oh well, I'll try
> n_2 = 7, which gives
>
> C = 62, and 2*C = 124, and I'll use the factorization 31(4), so k = 27,
> and I have
>
> 8*7 + 27^2 = 785, and 785 mod 55 = 15.
>
> And I don't know what that gives me either. But, kowing that 11 and 5
> are factors, I do see the 0 square modulo 5, and 4 is, of course a
> square.
>
> Maybe I'm using primes that are too small, and I wonder how often the 0
> case pops up?
>
> Maybe about 2/(p + 1) where p is the smallest prime, so that's no help.
>
> So far with these I have residues with C that are bigger than either
> prime factor, but if you get a residue less than sqrt(C) then maybe
> that could tell you something?
>
> Lots of speculation here, or is there gold in there somewhere?
>
> If so, have I just given away a route to factoring composites made up
> of large primes?
>
> Or is this approach completely useless when it comes to practicality?
>
>
> James Harris
>

I just realised the suggestion I gave someone else for an unrelated
problem is ideal, from your point of view, for your factoring problem.

Patent it and do it straight away.

This gives you:

a legal document so no-one will ever be able to claim they invented your
method. At the moment anyone can make a small change and claim it as
their invention.

control over the technology for a maximum of 17 (or 20?) years. If you
want to factor numbers, fine, but if someone else starts to use it, you
sue them. Money for you and a lesson for the infringer.

responsible handling of a dangerous technology. You can call the tune
as to when the world will have to transition away from RSA, just make a
public announcement of when you will let the patent laps.

In all a much better situation.
Received on Mon May 1 02:06:04 2006