# Proving the Jacobi sum $J(\chi,\rho)=\sum_{t\in F_p}\chi(1-t^2)$ when $\rho$ has order $2$?

Suppose $\chi$ is a nontrivial character on $F_p$, and $\rho$ is the character of order $2$.

How can I show $J(\chi,\rho)=\sum_t \chi(1-t^2)$ where the sum is over all $t\in F_p$?

Here $J(\chi,\rho)$ is the Jacobi sum. Since $\rho$ is the only character of order $2$, if follows that $N(x^2=a)=1+\rho(a)$, where $N(x^2=a)$ is the number of solutions to $x^2=a$. I calculated $$J(\chi,\rho)=\sum_{a+b=1}\chi(a)\rho(b)=\sum_{b}\chi(1-b)(N(x^2=b)-1)=\sum_{b}\chi(1-b)N(x^2=b)$$ since I think $\sum_b \chi(1-b)=0$ since $\chi$ is nontrivial. I can't figure out how to show $\sum_{b}\chi(1-b)N(x^2=b)=\sum_b\chi(1-b^2)$. How can one finish? Thanks.

Source: I&R #8.3

$N(x^2 = b) = 2$ if $b$ is a nonzero square, = 1 if $b$ = 0, and 0 otherwise. So your sum is really $$\chi(1) + 2 \sum_{squares \neq 0} \chi (1-b)$$ For each square $b \neq 0$, there are two square roots, both nonzero - if $x^2 = b$, so is $(-x)^2 = b$. So $$2 \sum_{squares \neq 0} \chi (1-b) = \sum_{t \neq 0} \chi (1-t^2)$$ Now, $$\chi(1) + 2 \sum_{squares \neq 0} \chi (1-b) = \chi(1) + \sum_{t \neq 0} \chi (1-t^2) = \sum_t \chi(1-t^2)$$