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 $a$.

For any integer $n$, we have $n^2 \equiv 0 \bmod{3}$ or $n^2 \equiv 1 \bmod{3}$.
Since $a^2+b^2 \equiv 0 \bmod{3}$, by the above fact we must have $a^2\equiv 0 \bmod{3}$ and $b^2 \equiv 0 \bmod{3}$.
Since $3$ is a prime dividing $a^2$, $3$ divides $a$.  Similarly $3$ divides $b$.
