<s>[INST] Proof that squares are divisible by 3 when their sum is In this proof, they write $3|a^2+b^2 \implies 3|a$, $3|b$. I tried using the same proof used to prove $3|a^2 \implies 3|a$, where $3$ being prime and writing $a^2 = a\cdot a$ suggests that $a$ is divisible by $3$. I'm not sure how to prove the $3|a^2+b^2$ case, though.

E9. There is no quadruple of positive integers $(x, y, z, u)$ satisfying $$x^2 + y^2 = 3(z^2 + u^2).$$
Solution. Suppose there is such a quadruple. We choose the solution with the smallest $x^2 + y^2$. Let $(a, b, c, d)$ be the chosen solution. Then $$a^2 + b^2 = 3(c^2 + d^2) \implies 3|a^2 + b^2 \implies 3|a, 3|b \implies a = 3a_1, b = 3b_1,\\a^2 + b^2 = 9(a^2_1 + b^2_1) = 3(c^2 + d^2) \implies c^2 + d^2 = 3(a^2_1 + b^2_1).$$
We have found a new solution $(c, d, a_1, b_1)$ with $c^2 + d^2 \lt a^2 + b^2$. Contradiction.
We have used the fact that $3|a^2 + b^2 \implies 3|a, 3|b$. Show this yourself. We will return to similar examples when treating infinite descent.

 [/INST] It probably isn't the best solution, but you could try using congruence.
Since 3 is a pretty small number, you can test each case for 
$$a,b\equiv 0,1,2\pmod 3$$
And for each one check if 
$$a² + b² \equiv 0 \pmod 3  $$
It gives you (all results given modulo 3):
$$
    \begin{matrix}
    a & b & a^2+b² \\
    0 & 0 & 0 \\
    0 & 1 & 1 \\
    0 & 2 & 1 \\
    1 & 0 & 1 \\
    1 & 1 & 2 \\
    1 & 2 & 2 \\
    2 & 0 & 1 \\
    2 & 1 & 2 \\
    2 & 2 & 2 \\
    \end{matrix}
$$
As you can see:
$$a² + b² \equiv 0 \pmod 3 $$ iff $$a \equiv 0\pmod 3 \land b\equiv 0\pmod 3 $$
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