# Question on a Gaussian sum and Jacobi sum identity

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 $$(1)$$?

• Why are you using $i^t$ in $g(\chi)$? Shouldn’t they be powers of $e^{2i\pi/5}$ instead? Commented Jul 21, 2023 at 19:23
• @Aphelli, you're right. I've checked with Mathematica that identity (1) holds, so that changes the whole issue. I'll edit the post accordingly. Commented Jul 21, 2023 at 21:11

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 $$\sum_{t=k+1}^{p-1}=0$$.