Proof involving characters

I'm self studying from a Classical Introduction to Modern Number Theory by Ireleand and Rosen. In the exercises for the chapter on Gauss and Jacobi sums I came across this question.

Let $\chi$ be a non-trivial multiplicative character of $F_p$ and let $\rho$ be the character of order 2 (the legendre symbol), then if $k \in F_p$ and $k \neq 0$, then $\sum_{t=0}^{p-1} \chi (t(k-t)) = \chi(k^2/4)J(\chi,\rho)$.

Now, $\sum_{t=0}^{p-1} \chi(t(k-t)) = \sum_{c=0}^{p-1}N(t(k-t)=c)\chi(c)$, where $N(t(k-t)=c)$ is the number of solutions to the equations $t(k-t) = c$. Now, $t(k-t) = c$ is solveable iff its discriminant is a quadratic residue and if its solveable there are two solutions. So $N(t(k-t) = c) = N(s^2 = k^2 - 4c) = 1 + \rho(k^2-4c)$

So, the sum becomes $\sum_{c=0}^{p-1}(1 + \rho(k^2-4c))\chi(c) = \sum_{c=0}^{p-1}\rho(4)\rho(k^2/4 -c)\chi(c) = \sum_{c=0}^{p-1}\rho(c)\chi(k^2/4 - c)$

Now, $\chi(k^2/4)J(\chi,\rho) = \chi(k^2/4)\sum_{t=0}^{p-1}\chi(t)\rho(1-t)$

And no matter how I try to manipulate it, I can't get the equations to equal each other. Did I do something wrong along the way?

1 Answer

In $\sum_{c=0}^{p-1}(1 + \rho(k^2-4c))\chi(c)$,

try replacing $c$ with $k^2/4c$ instead of $k^2/4-c$. (To get to the right hand side you want to be able to pull out a $\chi(k^2/4)$, so put yourself in a position to do this.) This should take you to

$\chi(k^2/4)\sum_{t=0}^{p-1}\chi(t)\rho(1-t)$.