quasi
Posts:
12,023
Registered:
7/15/05


Re: A quicker way?
Posted:
Feb 4, 2013 10:36 AM


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

