Prove that if $a^2+b^2$ is a multiple of three, then a and b are multiples of three I have attempted to prove the above. I am uncertain about the correctness of my proof:
Both numbers have to be multiples of three, i.e. $3a+3b=3n$, $\ 3(a+b)=3n$
It is not possible to arrive at an integer that is a multiple of three without adding two integers that are multiples of three.
Assumption: Suppose that $b$ is not a multiple of three, then it can be expressed as $(3v \pm 1)$, therefore we have: 
\begin{align*} a^2+(3v \pm 1)^2=3n\\ a^2= 3n-9v^2 \mp 6v -1\\ a=\sqrt{3(n-3v^2\mp2v-\frac{1}{3})}
\end{align*}
which is not a multiple of three? (or is it).
As mentioned before, $(a+b) \ne 3m \ $ if either $a$ or $b$ is not a multiple of 3, in which case assumption that $b$ is not a multiple of 3 is false. And hence it is a multiple of three, so is $a$.

If $a\ne 0\ (\ mod\ 3)$, then $a^2\equiv 1\ (\ mod\ 3) $
If $a\equiv 0\ (\ mod\ 3)$, then $a^2\equiv 0\ (\ mod\ 3)$
If $b\ne 0\ (\ mod\ 3)$, then $b^2\equiv 1\ (\ mod\ 3) $
If $b\equiv 0\ (\ mod\ 3)$, then $b^2\equiv 0\ (\ mod\ 3)$
So, the only way to get $a^2+b^2\equiv 0\ (\ mod\ 3)$ is $a\equiv b\equiv 0\ (\ mod\ 3)$
