Here $(\cdot)$ denotes the legendre symbol and $p$ be an odd prime number. All terms are of norm 1. So one bound is $p$. can one better bound be given to the sum?
2026-03-25 06:13:04.1774419184
Can a bound be given to $\sum_{a=1}^{p-1}\chi(a)\left(\frac{a^2-1}{p}\right)$, which is smaller than $p$, where $\chi$ is a dirichlet character?
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