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Prove that if $(a,p)=1$ and $p$ is an odd prime, then $\sum^p_{n=1}\left(\frac{an+b}{p}\right)=0$.

In this question we define Legendre symbol $\left(\frac{a}{p}\right)=0$ if $p$ is an odd prime and $p|a$.

I am struggling.

So far, I've tried expanding out the summation to get

$$\sum^p_{n=1}\left(\frac{an+b}{p}\right)=\left( \frac{a+b}{p}\right)+\left( \frac{2a+b}{p}\right)+\dots+\left( \frac{pa+b}{p}\right)$$

By definition of Legendre Symbols, I got

$$\sum^p_{n=1}\left(\frac{an+b}{p}\right)=(a+b)^{(p-1)/2}+(2a+b)^{(p-1)/2}+\dots+(pa+b)^{(p-1)/2} \mod p$$

And then I am stuck.

Yellow Skies
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    Step 1 is to show that there are as many residues mod $p$ as non-residues. Step 2 is to show that the numbers $an + b$ are distinct mod $p$ as $n$ runs from 1 to $p$. Neither is that hard if you know a little group theory, and specifically the structure of the group of integers mod $p$ under multiplication. – Vik78 Sep 16 '16 at 01:49
  • Let me try Step 1: $\sum^p_{n=1}\left(\frac{x}{n}\right)=0$ – Yellow Skies Sep 16 '16 at 01:53
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    Start with the set of residues mod $p$. What's a simple function taking it to the set of non-residues? Show that that function is a bijection and you're done. – Vik78 Sep 16 '16 at 01:58

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