$l^2+m^2=n^2$ $\implies$ $lm$ 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.

Here is another approach. The general solution of this equation is:
$l = x^2 - y^2$, $m = 2xy$, $n = x^2 + y^2$. So: $l\cdot m = 2\cdot (x^2 - y^2)\cdot x\cdot y = 2\cdot (x - y)\cdot (x + y) \cdot x\cdot y$. From this, we have some cases to consdier:


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*$3|x$ or $3|y$ then $3|l\cdot m$.

*$3 \not|x$ and $3\not|y$ then if $x \equiv y \pmod 3$, then $3|(x - y)$ and $(x - y)|l\cdot m$, so $3|l\cdot m$, but if $x \neq y \pmod 3$, then $x + y \equiv 0 \pmod 3$, and this implies that $3|l\cdot m$
