quasi
Posts:
9,903
Registered:
7/15/05


Re: A quicker way?
Posted:
Feb 4, 2013 12:04 PM


luttgma@gmail.com wrote: >quasi wrote: >> 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. > >Thank you, but then you must find u and v, which doesn't >seem quicker than trying a=1,2,3, etc.
Sometimes it's a lot quicker.
For example, try to find all pairs (n,a) of positive integers such that n^2 = 2^100 + a^2.
Using the method a = 1,2,3, ... how long do you think it will take to find even one such pair?
quasi

