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: $$ 1 + \Big(\frac{t}{p}\Big) = \begin{cases} 2 & \text{if $t$ is a quadratic residue,} \\ 0 & \text{if $t$ is a quadratic nonresidue,} \\ 1 & \text{if $t=0$.} \end{cases} $$