$l^2+m^2=n^2$ $\implies$ $lm$ is always a multiple of 3 when $l,m,n,$ are positive integers. Let $l,m,n$ be any three positive integers such that $l^2+m^2=n^2$ 
Then prove that $lm$ is always a multiple of 3.

The quadratic resides modulo $3$ are $0$ and $1$. In other words, each of $l^2, m^2$ must be equivalent to either $ 0, 1\pmod3$. 
However, $l^2$ and $m^2$ cannot both be equivalent to $1$ modulo 3, because their sum , $n^2$, a perfect square, would be equivalent to $2\pmod3$, a quadratic non-residue.
Hence, at least one of $l^2, m^2$ must be equivalent to $0\pmod3$. As such $l^2m^2 = (lm)^2 \equiv 0\pmod 3$. By Euclid's lemma, $lm \equiv 0 \pmod 3$.
