I have the following problem: Denoting $S(q,a,\chi ) = \sum_{x=1}^q \chi (x) e(ax/q)$, where $\chi $ is an arbitrary character modulo $q$, I have to prove $$\sum_{a=1}^q \vert S(q,a,\chi ) \vert ^2 = q\phi (q) .$$
I already know for $(a,q)=1$ that $S(q,a,\chi ) = \overline{\chi }(a)\sum_{x=1}^q \chi (x)e(x/q)$, but I don't see how this fact could help.
The short answer is "that you just do it." So let's do it.
We have
$$ \begin{align} \sum_{a \leq q} \lvert S(q,a,\chi)\rvert^2 &= \sum_a \sum_{m,n \leq q} \chi(m)\overline{\chi}(n)e\left( \frac{a(m - n)}{q}\right) \\ &= \sum_{m,n} \chi(m)\overline{\chi}(n) \sum_a e\left( \frac{a(m - n)}{q}\right) \\ &= q\sum_{m \leq q} \chi(m)^2 \\ &= q\phi(q) \end{align},$$
where I used that $\displaystyle\sum_a e(x/q) = \begin{cases} 0 & \text{if } x \neq 0\\ q & \text{else} \end{cases}$.