Number Theory proof involving the sum of Legendre symbols

Question

If $b$ is a unit mod $p$, show that $(\frac{b}{p})+(\frac{2b}{p})+\cdot \cdot \cdot +(\frac{(p-1)b}{p})\equiv 0$.

Since $b\in U_p$ then $(\frac{1}{p})+(\frac{2}{p})+\cdot \cdot \cdot +(\frac{(p-1)}{p})$.

(So I'm not sure if I'm stating this part correctly) Therefore for each $(\frac{i}{p})$ for $i=1,...,p-1$, $i$ is either in $Q_p$ or not. Since $|Q_p|=\frac{\phi(p)}{2}=\frac{p-1}{2}$. Half will be in $Q_ p$ and the other half will not. So we get $(\frac{1}{p})+(\frac{2}{p})+\cdot \cdot \cdot +(\frac{(p-1)}{p})=1-1+1-\cdot\cdot\cdot +1\equiv0$.

Is this okay?

@DietrichBurde I just need verification on what i wrote up is okay. — HiPolyEraser, Mar 22 '17 at 20:19
Yes, it is basically OK, and it is exactly the same as the first comment by Arturo given at the duplicate. So you can verify. — Dietrich Burde, Mar 22 '17 at 20:23

score 1 · Answer 1 · answered Mar 22 '17 at 20:23

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Since $b\in U_p$ then $(\frac{1}{p})+(\frac{2}{p})+\cdot \cdot \cdot (\frac{(p-1)}{p})$.

This is not a complete idea. I think you're missing an $=\ldots$ at the end of the expression, probably to assert that it equals the other sum of Legandre symbols. You also lack an explination of how the idea of the last paragraph connects to anything.

The idea of your proof is correct though.

answered Mar 22 '17 at 20:23

Stella Biderman

31,155

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Alright, Thank you. Yea I still have to clean it up a bit and explain a bit more. – HiPolyEraser Mar 22 '17 at 20:25

Number Theory proof involving the sum of Legendre symbols

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