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This is a poly over $F_p$, $$q(x) = \sum_{k=1}^{p-1} k^{(p-1)/2} x^k$$ We can just get the powers of $k$ by repeated differentiation and multiplication by $x$. $$q(x) = \left(x \frac{d}{dx}\right)^\frac{p-1}{k} \sum_{k=1}^{p-1} x^k$$ $$q(x) = \left(x \frac{d}{dx}\right)^{(p-1)/k} \left( \frac{x^p-1}{x-1} - 1\right)$$ $x^p\equiv x \mod p$ so $$q(x) = \left(x \frac{d}{dx}\right)^{(p-1)/k} \left( \frac{x-1}{x-1} - 1\right) = 0$$
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Fri, 28 Aug 2020 07:56 GMT