Question on a Gaussian sum and Jacobi sum identity

Question

I've tried, unsuccessfully, to solve exercise 8.5 from Ireland and Rosen's A Classical Introduction to Modern Number Theory. The exercise asks to prove \begin{align}g(\chi)^2=\frac{J(\chi,\rho)g(\chi^2)}{\chi(2)^2},\tag{1}\end{align} where $\chi:\mathbb{F}_p\longrightarrow \mathbb{C}$ is a Dirichlet character with order not dividing $2$. Moreover, $\rho$ is the Legendre symbol character, $g(\chi)$ is a Gauss sum and $J(\chi,\rho)$ is a Jacobi sum.

A hint is provided for this exercise that asks to expand the product $g(\chi)^2$ and use previous exercise: $$\sum_{t=0}^{p-1}\chi(t(k-t))=\chi\Big(\frac{k^2}{4}\Big)J(\chi,\rho),$$ where $k\in \mathbb{F}_p$ is non-zero. I've tried to follow the hint but with no success: \begin{align*} g(\chi)^2&=\Big(\sum_{k=0}^{p-1}\chi(k)\zeta^k\Big)^2\\&=\sum_{k=0}^{2p-2}\Big(\sum_{t=0}^k\chi(t)\chi(k-t)\Big)\zeta^k, \end{align*} where $\zeta=e^{2\pi i/p}$ is a $p$th root of unity. I'm not able to use the hint because the inner sum in previous expression doesn't go up to $p-1$, but only up to $k$.

This question has been asked before, but I'm not sure the given answer is correct due to the said disparity of sum indexes.

EDIT: the counterexample that I gave of identity $(1)$ was wrong. Given the evidence, how can one prove $(1)$?

Answer 1

The previously given answer to this exercise is correct, but the indexes got me confused. Here's a more detailed solution:

By definition of $g(\chi)$: \begin{align*} g(\chi)^2&=\bigg(\sum_{k=0}^{p-1}\chi(k)\zeta^k\bigg)^2\\ &=\sum_{k=0}^{2p-2}\bigg(\sum_{s+t=k}\chi(s)\chi(t)\bigg)\zeta^k. \end{align*} Now, one has to be careful with the inner sum. We have $$\sum_{s+t=k}\chi(s)\chi(t)=\begin{cases}\displaystyle{\sum_{t=0}^{k}\chi(t)\chi(k-t)}&\text{if }k\leq p-1\\[1ex] \displaystyle{\sum_{t=k-p+1}^{p-1}\chi(t)\chi(k-t)}&\text{if }k>p-1\end{cases}.$$ Moreover, while $k$ goes from $p$ to $2p-2$, the power $\zeta^k$ goes from $\zeta^0$ to $\zeta^{p-2}$, since $\zeta$ is a $p$th root of unity. Also, $\chi(t(p+k-t))=\chi(t(k-t))$, since both arguments are congruent. Thus, \begin{align}g(\chi)^2&=\sum_{k=0}^{p-1}\bigg(\sum_{t=0}^k\chi(t(k-t))+\sum_{t=k+1}^{p-1}\chi(t(k-t))\bigg)\zeta^k\\ &=\sum_{k=0}^{p-1}\bigg(\sum_{t=0}^{p-1}\chi(t(k-t))\bigg)\zeta^k. \end{align} Just a nitpick: for $k=p-1$ we set $\sum_{t=k+1}^{p-1}=0$.

The result now follows easily applying exercise 8.4, as @anon does.