If $p$ is an odd prime, show that $\sum_{j=1}^{p-1}({\frac{j}{p}})=0$ with $(\frac{j}{p})$ the Legendre Symbol.
I'm not sure if it's enough to say that, because of the Euler's Criteria, there are exactly $\frac{p-1}{2}$ values for which $(\frac{j}{p})=1$ so the other values are $-1$, hence it's proved.
I somehow feel like it's not a good proof.
I've also seen that if $p\equiv 3 (4)$ it's true since $(\frac{-j}{p})=-(\frac{j}{p})$, so the terms cancel out, but following this for $p\equiv 1 (4)$ I don't achieve anything.
Thanks
It's perfectly good. Another way: the nonzero squares mod $p$ are $1^2 \pmod p$, $2^2 \pmod p$,..., $(p-1)^2 \pmod p$. That would be $p-1$, but they're double-counted since $x^2 \equiv (p-x)^2$, so there are $(p-1)/2$ squares mod $p$.