# Thread: 2p^2 - q^2 and 2q^2 - p^2 as perfect squares simultaneously

1. ## 2p^2 - q^2 and 2q^2 - p^2 as perfect squares simultaneously

If I can show that

$2p^2-q^2$ and $2q^2-p^2$

can only be perfect squares simultaneously when

$p\ =\ m^2+n^2\ =\ u^2-2uv-v^2$

and

$q\ =\ m^2-2mn-n^2\ =\ u^2+v^2$

is this enough to say that p = q ?

Forgot to say that p and q are integers > 0

Thanks