# Equivalent definitions of the quadratic gauss sum

In Ireland and Rosen, the quadratic Gauss sum of $a$, $g_a$, is defined by $$g_a:=\sum_{t=0}^{p-1}\left(\frac tp\right)\zeta_p^{at}$$ with $\zeta_p$ a $p$th root of unity, $p$ an odd prime and $(\frac\cdot\cdot)$ the legendre symbol. One writes $g_1=g$. Exercise 6.8 asks to prove an equivalent definition:

By evaluating $\sum_{t=0}^{p-1}(1+(\frac tp))\zeta_p^t$ in two different ways, prove that $g=\sum_{t=0}^{p-1}\zeta_p^{t^2}$

Clearly, $$\sum_{t=0}^{p-1}\left(1+\left(\frac tp\right)\right)\zeta_p^t=\sum_{t=0}^{p-1}\zeta^t+\sum_{t=0}^{p-1}\left(\frac tp\right)\zeta^t=0+g_1=g.$$

Unfortunately I don't see an alternative evaluation of the sum... Can someone give me a hint?

Think about how this might be useful: $$1 + \Big(\frac{t}{p}\Big) = \begin{cases} 2 & \text{if t is a quadratic residue,} \\ 0 & \text{if t is a quadratic nonresidue,} \\ 1 & \text{if t=0.} \end{cases}$$
• So $g=2\sum_{t\in QR}\zeta^t$ with $QR=\bigl\{x\in\{1,\dots,p-1\}\bigm|\text{$x$is a quadratic residue mod$p$}\bigr\}$ but how is this linked to $\zeta^{t^2}$? – lvaneesbeeck Mar 30 '14 at 0:08
• If $t$ is a quadratic residue then $t=s^2$ for some $s$, modulo $p$. The sum $\sum_s \zeta^{s^2}$ counts each quadratic residue twice since $s$ and $-s$ both have the same square. Do you see it? – user134824 Mar 30 '14 at 0:19