If $3 \mid a^2+b^2$, then show that $9 \mid ab$. 
If $3 \mid a^2+b^2$, then show that $9 \mid ab$.

If $3$ divides $a^2+b^2$ I have that $a^2+b^2 \equiv 0 \pmod{3}$. Looking at the squares it seems that starting from $9$ for every third I’ll get a remainder $0$. Also squares mod $3$ are only $0$ or $1$ so I must have $a^2 \equiv 0 \pmod{3}$ and $b^2 \equiv 0 \pmod{3} $, but still how does this imply the result we’re after?

$a^2\equiv0\pmod3$ means  $3\mid a^2$, which implies  $3\mid a$ (by Euclid's lemma)
;
similarly $3\mid b$, so $9\mid ab$.
