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This was an old homework problem that I wasn't able to solve and seem to have lost the solution to. If $p$ is a prime number greater than $3$, then the legendre symbol $(\frac{a}{p})$ as a function of $a$ is the unique Dirichlet character of order $2$. That's easy enough to show, but I've been unable to figure out the next part.

The problem is to show that, if $\chi$ is any nontrivial character on $\mathbb{Z}_p$, then $\sum\limits_{t \in \mathbb{Z}_p}\chi(1-t^2) = \sum\limits_{a+b=1}\chi(a)(\frac{b}{p})$, where $a, b \in \mathbb{Z}_p$.

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up vote 4 down vote accepted

In the first summation $$\sum_{t \in \mathbb Z/p\mathbb Z} \chi(1-t^2) = \sum_{t \in \mathbb Z/p\mathbb Z}_{a = 1 - t^2} \chi(a)$$ every value occurs twice (because $t^2 = (-t)^2$) except $\chi(1)$. Thus is it equal to $$\sum_{a \in \mathbb Z/p\mathbb Z} s(a) \chi(a)$$ where $s(a)$ counts the number of solutions (0, 1 or 2) of $1 - a = t^2$ for some $t$. Clearly $s(a) = 1 + \left(\frac{1-a}{p}\right)$ therefore we multiply out to get $$\sum_{t \in \mathbb Z/p\mathbb Z} \chi(1-t^2) = \sum_{a \in \mathbb Z/p\mathbb Z} \chi(a) + \sum_{a \in \mathbb Z/p\mathbb Z} \left(\frac{1-a}{p}\right) \chi(a)$$ and the first character sum in the RHS is zero (for nontrivial characters) so we have the result.

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thank you so much! –  Shankman Dec 6 '12 at 20:52

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