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The wikipedia page for Quadratic Gauss Sum says that the fact that $g(a;p)^2 = \left(\frac{-1}{p}\right)p$ is simple, yet I am having trouble coming up with a proof. Are there any references I could be pointed to?

Thanks!

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It is proved in Ireland's and Rosen's book A Classical Introduction to Modern Number Theory. The evaluation of the Gauss sum can be reduced to the case $a = 1$. Basically we transform the double sum with $s\equiv tr$ mod $p$ as follows $$ g(1,p)^2 =\sum_{r=1}^{p-1}\sum_{s=1}^{p-1}\left(\frac{r}{p}\right)\left(\frac{s}{p}\right)e^{\frac{2\pi i(r+s)}{p}} = \sum_{t=1}^{p-1}\left(\frac{t}{p}\right)\sum_{r=1}^{p-1}e^{\frac{2\pi ir(1+t)}{p}}=-\sum_{t=1}^{p-1}\left(\frac{t}{p}\right)+p\cdot \left(\frac{-1}{p}\right), $$ and then use that $\sum_{t=1}^{p-1}\left(\frac{t}{p}\right)=0$, which is also easy to see.

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