Re: Surrogate factoring, objective look

From: Jim Spriggs (jim.sprigs_at_ANTISPAMbtinternet.com.invalid)
Date: 02/09/05 

Date: Wed, 9 Feb 2005 12:54:04 +0000 (UTC)

jstevh@msn.com wrote:
>
> I have found a factorization method which I claim is new and important.
> Here is an attempt at being objective about my own finding, by
> explaining how it works, and comparing it with congruence of squares.
>
> Congruence of square relies on a simple idea to factor numbers using
> quadratics.
>
> Say you have f_1 f_2 = M. Consider the quadratic
>
> x^2 + cx - M = 0, where you trivially have x(x + c) = M,
>
> so if all all are integer x must be a factor of M, but if it's 1 or M,
> it's a trivial factor.
>
> Solving that quadratic is easy enough using the quadratic formula, and
> you get
>
> x = (-c +/- sqrt(c^2 + 4M))/2
>
> and if that square root is an integer, then you have an integer x.

Why? What if -c +/- sqrt(c^2 + 4M) is odd?

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