
Re: A quicker way?
Posted:
Feb 4, 2013 11:03 AM


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

