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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 $g(\chi)^{2}=\chi{2}^{-2}J(\chi,\rho)g(\chi^{2})$?

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This is exactly exercise 5 to Chapter 8 of Ireland & Rosen's Classical Introduction to Modern Number Theory. Why didn't you tell us where the problem is coming from, for context? – KCd Feb 24 '12 at 18:09
up vote 3 down vote accepted

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 $k-t$.

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Thanks! this is the part is was stuck on. – Edison Feb 24 '12 at 20:09

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