# Proving $g(\chi\rho)^6=(-1)^{(p-1)/2}p(\overline{\chi(2)J(\chi,\rho)})^4$, from Ireland and Rosen.

Suppose $p\equiv 1\pmod{3}$, $\chi$ is a cubic character, and $\rho$ is the quadratic character on $F_p$.

If $\chi\rho$ is a character of order $6$, why does the Guass sum $g(\chi\rho)^6=(-1)^{(p-1)/2}p\bar{\pi}^4$ where $\pi=\chi(2)J(\chi,\rho)$.

I derived a few formulas, $g(\chi)^3=p\pi$, and if $\chi^2\neq\epsilon$, then $g(\chi)^2=\chi(2)^{-2}J(\chi,\rho)g(\chi^2)$ and $J(\chi,\chi)=\chi(2)^{-2}J(\chi,\rho)$. These are all previous exercises in the book.

Using the last equation, I find $$\pi=\chi(2)\chi(2)^2J(\chi,\chi)=J(\chi,\chi).$$ Also, since $\chi$ is a cubic character, $g(\chi)^3=pJ(\chi,\chi)$.

Since $\chi\rho$ is nontrivial, $$g(\chi\rho)=\frac{g(\chi)g(\rho)}{J(\chi,\rho)}=\frac{g(\chi)g(\rho)}{\chi(2)^2J(\chi,\chi)}$$ so $$g(\chi\rho)^6=\frac{g(\chi)^6g(\rho)^6}{\chi(2)^{12}J(\chi,\chi)^6}=\frac{p^2J(\chi,\chi)^2g(\rho)^6}{J(\chi,\chi)^6}=\frac{p^2g(\rho)^6}{\pi^4}.$$ This looks kind of close since there is at least a power of $\pi^4$. I know that $g(\rho)^6=\pm p^3$, with sign depending on the value of $p\mod 4$, which is unknown. I also know that $\rho(-1)=(-1)^{(p-1)/2}$, which is curious since it shows up in the equation. How can the equality be derived? Thanks.

Source: I&R #8.10.

You've basically got it. The sign in $g(\rho)^6=\pm p^3$ is, as you know, $+$ if $p\equiv1\pmod4$, and $-$ if $p\equiv-1\pmod4$, which is to say, $$g(\rho)^6=(-1)^{(p-1)/2}p^3.$$ I think the only thing you're missing is $\pi\overline{\pi}=p$, because that gives you $${p^2g(\rho)^6\over\pi^4}={p^2(-1)^{(p-1)/2}p^3\over\pi^4}=(-1)^{(p-1)/2}p\overline{\pi}^4,$$ as desired.
So: can you get $p=\pi\overline{\pi}$?