Multiples of 4 as sum or difference of 2 squares Is it true that for any $n \in \mathbb{N}$ we can have $4n = x^{2} + y^{2}$ or $4n = x^{2} - y^{2}$, for $x,y \in \mathbb{N} \cup (0)$?
I was just working out a proof and this turns out to be true from $n=1$ to $n=20$. After that I didn't try, but I would like to see if a counterexample exists for a greater value of $n$.

It is true because
$$ (n+1)^2 - (n-1)^2 = 4n $$
