For characters, $\chi$ and $\rho$, if $k\in \mathbb{F_{p}}$ with $k\neq0$ and $\chi^{2}\neq \epsilon$, then suppose $\sum_{t}\chi{(t(k-t))}=\chi{(k^{2}/2^2)}J(\chi,\rho)$.
How can I show that $g(\chi)^{2}=\chi{2}^{-2}J(\chi,\rho)g(\chi^{2})$?
For characters, $\chi$ and $\rho$, if $k\in \mathbb{F_{p}}$ with $k\neq0$ and $\chi^{2}\neq \epsilon$, then suppose $\sum_{t}\chi{(t(k-t))}=\chi{(k^{2}/2^2)}J(\chi,\rho)$.
How can I show that $g(\chi)^{2}=\chi{2}^{-2}J(\chi,\rho)g(\chi^{2})$?
The $\chi(2)$ and $J(\chi,\rho)$ terms are constant so are more or less irrelevant to the computation.
$\left(\sum_{t} \chi(t)\zeta^t \right)^2 = \sum_{k}\left(\sum_{t+l=k}\chi(t)\chi(l)\right)\zeta^k=\sum_{k}\left(\sum_{t} \chi\big(t(k-t)\big)\right)\zeta^k $
$=\sum_{k} \chi(k^2/4)J(\chi,\rho) \zeta^k=\chi(2)^{-2}J(\chi,\rho)\sum_{k} \chi^2(k)\zeta^k=\frac{J(\chi,\rho)g(\chi^2)}{\chi(2)^2}.$
The only real issue is noting that $\sum_{t+l=k}$ can be replaced with $\sum_t$ as long as we replace $l$ with $k-t$.