 Prime-field character Gauss sum identity

For a nontrivial Dirichlet character modulo a prime $p$ and $a\not\equiv 0(\mathrm{mod}p)$, set $\tau ={\sum }_{y\ne 0}\chi (y)e(ay/p)$. Substituting $y=tz$ in $|\tau {|}^2$ gives ${\sum }_{t\ne 0}\chi (t){\sum }_{z\ne 0}e(a(t-1)z/p)$. The inner sum is $p-1$ at $t=1$ and $-1$ otherwise. Since character orthogonality gives ${\sum }_{t\ne 0}\chi (t)=0$, the result is $p$.