$l^2+m^2=n^2$ $\implies$ $lm$ 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.
 A: We show that $3$ must divde $lm$, by showing that $3$ divides $l$ or $3$ divides $m$. 
Any integer $x$ is either divisible by $3$, or is congruent to $1$ or $-1$ modulo $3$. And iff $x\equiv \pm 1\pmod{3}$, then $x^2\equiv 1 \pmod{3}$. 
Thus if neither $l$ nor $m$ is divisible by $3$, then $l^2+m^2\equiv 2\pmod{3}$. It follows that $l^2+m^2$ cannot be a perfect square. 
A: 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:


*

*$3|x$ 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 $3|l\cdot m$
A: 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, $lm \equiv 0 \pmod 3$.
