How do I show that $(\sum_{k=0}^{p-1} \omega ^{x^{2}})^2 = p$, where $\omega$ is a primitive pth root of unity of a field $\mathbb{F}$ and $p \equiv 1 \pmod{4}$ is a prime?
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Look up Gauss sum. We may have also covered this on the site already, but I'm not 100%. – Jyrki Lahtonen Sep 28 '19 at 21:32
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Also, the question, while about an interesting fact, could use additional context as described in our guide for new askers. The way it looks like in its present form may raise a suspicion that this is a homework assignment. I know, having thought this on several courses, that this bit is nearly always a theorem in the lecture notes/textbook. That does raise the natural question as to where did you run into this? Explaining that is one acceptable form of context. – Jyrki Lahtonen Sep 28 '19 at 21:39
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For more about Gauss sums on our site see here. In this old answer of mine, I explain what goes wrong when $p=7\not\equiv1\pmod4$. The same technique I describe for $p=7$ actually works for you also, but you need to know a few things. Most notably a result describing in how many ways we can write an element of $\Bbb{Z}/p\Bbb{Z}$ as a sum of two squares. It is helpful to know that $-1$ is a quadratic residue in your case. – Jyrki Lahtonen Sep 28 '19 at 21:50
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The sum is over $k$ but the expression being summed has an $x$ in it instead of a $k$. – Barry Cipra Sep 28 '19 at 21:52