Let $\chi$ be a character on $(\mathbb F_q^\times, \cdot)$ (i.e. a homomorphism into $S_1 \subseteq \mathbb C$) with the convention that $\chi$ is defined on the entirety of $\mathbb F_q$ via $\chi(0) = 0$.
Let $u, v \in \mathbb F_q$. I now want to show that:
$\displaystyle \sum_{ a \in \mathbb F_q} \chi(a + u) \overline{\chi}(a + v) = \begin{cases} q-1 & \text{if } u = v \\ -1 & \text{if } u ≠ v \end{cases} $
(where $\overline{\chi}$ denotes the complex conjugated of $\chi$, or equivalently, $\overline{\chi(a)} = \chi(a^{-1})$.)
I mananged to prove the case where $u = v$: in that case, the sum basically reduces to $\sum_{a \in \mathbb F_q, a ≠ -u} \chi((a+u)/(a+u)) = q-1 $, hence giving the desired result. I tried the same method for $u ≠ v$, but wasn't so successful there. In this case, I managed to rewrite the formula as:
$\displaystyle \sum_{a \in \mathbb F_q, a ≠ -v} \chi \left(\frac{a + u}{a + v} \right) $
But don't really know how to go on from there, and if I'm at the right path at all. I know about the orthogonality of group characters, but have little else to work with in terms of criterias or theorems that I can use.
for $v \in \mathbb{F}_q^\times$ : $\chi(u)\overline{\chi(v)} = \chi(u)\chi(v^{-1}) = \chi(uv^{-1})$
if $b \ne 0$ : $a \mapsto ba^{-1}$ is a bijection $\mathbb{F}_q^\times \to \mathbb{F}_q^\times$
$a' \mapsto 1+a'$ is a bijection $\mathbb{F}_q \to \mathbb{F}_q$
And assuming $\chi$ is not the trivial character :
$$\sum_{a\in\mathbb{F}_q} \chi(a+b)\overline{\chi(a)} = \sum_{a\in\mathbb{F}_q^\times} \chi(a+b)\overline{\chi(a)}= \sum_{a\in\mathbb{F}_q^\times} \chi(1+ba^{-1})$$ $$ = \sum_{a'\in\mathbb{F}_q^\times}\chi(1+a') = -\chi(1)+\sum_{a'\in\mathbb{F}_q}\chi(1+a')=-\chi(1)+\sum_{a''\in\mathbb{F}_q} \chi(a'')= -\chi(1) = -1$$