Proof that squares are divisible by 3 when their sum is

Question

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.

Answer 1

For a number $n$ we have

$n\equiv 0,1,2 \mod 3$ so we get $$n^2\equiv 0,1\mod 3$$ For $$a^2+b^2$$ we have

$$a^2+b^2\equiv 0 \mod 3$$ The only possibility is $$a^2=b^2\equiv 0 \mod 3$$

Answer 2

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 $a^2=b^2=0$.

Answer 3

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 $$a \equiv 0\pmod 3 \land b\equiv 0\pmod 3 $$