how does $p | (a^2 + b^2)$ force $p | b^2$ 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 $p|b^2$.

Recall that if $p \mid x$ and $p \mid y$ then $p \mid x-y$. So 
$$p \mid a^2+b^2-a^2=b^2$$
