How to solve this system of equations in $\mathbb{Z}_3$? Knowing that the possible values are $\overline{0},\overline{1},\overline{2}$, I get by trial and error that the solution is $\overline{k},\overline{m}=\overline{0} \lor \overline{n},\overline{l}=\overline{0}$, but how to solve this correctly?
$\left\{ \begin{array}{c@{\,}} kl=mn \mod 3\\ kn=-lm \mod 3 \end{array} \right. $
Suppose that $l\not\equiv0\pmod3$. Then $$kl\equiv mn\pmod3\implies kl^2\equiv mln\pmod3\implies kl^2\equiv0\pmod3$$ from the second congruence. Hence $k\equiv0\pmod3$, so $ml\equiv0\pmod3$, giving $m\equiv0\pmod3$
Now let $l\equiv0\pmod3$. Then $mn\equiv0\pmod3$ and $kn\equiv0\pmod3$.