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How do I know that

$\sum\limits_{x=1}^{p-1}\left(\frac{x}{p}\right) = 0$

(where $\left(\frac{a}{b}\right)$ is the Legendre symbol and $p$ is an odd prime)?

I know that there must be an even number of terms and I guess the solution must be because we have an even number of $1$'s and $-1$'s being added up but how do I know this is the case?

How do I know that if $\left(\frac{a}{b}\right) = 1$ then $\left(\frac{a + 1}{b}\right)$ must be equal to $-1$?

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It's not true that $(\frac{a}{b})=1\Rightarrow (\frac{a+1}{b})=-1$, take for example $a=1,b=7$.

To show that the asked sum is equal to $0$, it's enough to know that exactly $\frac{p-1}{2}$ non-zero residues modulo $p$ are quadratic residues. Once you know this, answer these questions: How many $+1$'s appear in your sum? How many $-1$'s? What is their sum in that case?

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