Date: Feb 4, 2013 11:03 AM
Author: mluttgens
Subject: Re: A quicker way?

Le lundi 4 février 2013 16:36:14 UTC+1, quasi a écrit :
> quasi wrote:
>

> >luttgma@gmail.com wrote:
>
> >
>
> >>Let n^2 = N + a^2, where n, N and a are integers.
>
> >>Knowing N, it is of course possible to find n by trying
>
> >> a = 1, 2, 3, 4, etc...
>
> >>But doesn't exist a quicker way ?
>
> >
>
> >if you rewrite the equation in the form
>
> >
>
> > (n - a)(n + a) = N
>
> >
>
> >then for each pair of integers u,v with u*v = N, you can
>
> >solve the equations
>
> >
>
> > n - a = u
>
> > n + a = v
>
> >
>
> >for n and a.
>
>
>
> Also, since n - a and n + a have the same parity (both even
>
> or both odd), you only need to consider pairs of integers u,v
>
> with u*v = N for which u,v are both even or both odd.
>
>
>
> quasi


Thank you, but then you must find u and v, which doesn't seem quicker than trying a=1,2,3, etc.

Marcel