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