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.


In $\mathbb{Z}/3$, $a^2=1$,  or $a^2=0$,  $b^2=1$ or $0$  implies that $a^2+b^2=0$ if and only if  $a^2=b^2=0$. 
