I have a question that reads:
if $p | a$ and $p|(a^2+b^2)$, then $p | b$.
In the solution menu it reads:
since $p|a$, $p|a^2$. Now $p|a^2$ and $p|(a^2+b^2)$ forces $p|b^2$. we can conclude that $p|b$. I honestly have no idea how $p|(a^2+b^2)$ can give you $p|b^2$.
$p|a^2$ means there exists an integer $n$ such that $np=a^2$. Similarly, $p|(a^2+b^2)$ means there exists an integer $m$ such that $mp=a^2+b^2$. Subtracting the two gives $(m-n)p = b^2$.
Recall that if $p \mid x$ and $p \mid y$ then $p \mid x-y$. So
$$p \mid a^2+b^2-a^2=b^2$$
$p|a$ is equivalent to $a \equiv 0 \mod p$
if $a^2+b^2 \equiv 0 \mod p$ and $a \equiv 0 \mod p$ then $a^2 \equiv 0 \mod p$ hence $b^2 \equiv 0 \mod p$ so $b \equiv 0 \mod p$.
