Re: Factoring problem and the SFT
- From: jstevh@xxxxxxx
- Date: 27 Apr 2005 18:26:24 -0700
Tim Peters wrote:
> [Rick Decker]
> [...]
> > Now if we knew A = p * q was the product of two distinct
> > primes and if we had no reason to assume anything about
> > some set of N numbers, then we would expect N/p of them
> > to be divisible by p, N/q to be divisible by q, and
> > N/pq to be divisible by both p and q. So we'd expect
> >
> > N/p + N/q - N/pq
> >
> > useful factors.
>
> Sheesh: you and Nora _both_ lying to James about this, week after
week. No
> wonder he calls Nora "a liar" and you "lying scum" <wink>.
>
They're lying to the public. Their posts are meant to influence
others, not me.
Now I can now prove that the SFT has to work.
It's not even complicated.
If you consider
f_1 = (-(z - 2A^2)+ sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
where z is an integer, you have a finite set of solutions, where at
least one value of z must factor A non-trivially.
Well, it turns out that you can solve the equations that define z, for
rational factorizations of B^2(A^2 - B^2) such that you get an integer
z.
It's not even hard to do the math.
James Harris
.
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