How to show that if $9n^2=a^2+b^2$, $a$ and $b$ are multiples of $3$ To be honest, I don't know where to start with this problem:
Let $n\in \mathbb{N}$. Prove that if $9n^2$ is the sum of two perfect squares $(a^2,b^2)$, then $a$ and $b$ are multiples of $3$.

Logically speaking, $$9n^2 = a^2 + b^2 \iff \left({a\over3}\right)^2 + \left({b\over3}\right)^2 = n^2.$$ It is clear that $n^2 \in \mathbb Z$, but because $a^2, b^2$ are perfect squares, then $a,b,n\in\mathbb Z$. Can you see where to take it from here?  
