Relating a Character sum to a Gauss sum

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Let $q$ be a prime power. Consider the mapping $f:(\mathbb{F}_q)^{\times} \to (\mathbb{F}_q)^{\times},$ where $x\mapsto x^2$. I was interested in sums of the form $$\sum_{t\in \mbox{Im}(f)} \psi_a(t)\quad (a\in \mathbb{F}_q),$$ where $$\psi_a(t)=\exp\Big( \frac{\mbox{tr}(at)}{p}\Big)$$ is an additive character of $\mathbb{F}_q$. I feel it should be possible to relate these sums somehow to the Gauss sums $$\sum_{t\in \mathbb{F}_q} \chi (t) \psi_{\beta}(t),$$ where $\chi$ is a multiplicative character and $\beta\in \mathbb{F}_q$. I can't seem to quite figure out how this should go.

Side Question: Do sums of this type have a name?

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Let $\eta$ be the unique character of order two of $\Bbb{F}_q^*$ (= the Legendre character). Then for all $t\in\Bbb{F}_q^*$ we have $$ 1+\eta(t)=\begin{cases}2,&\text{if $t$ is a square,}\\0,&\text{if $t$ is a non-square.}\end{cases} $$ So your sum takes the form $$ S(a):=\sum_{t\in(\Bbb{F}_q^*)^2}\psi_a(t)=\sum_{t\in\Bbb{F}_q^*}\frac12(1+\eta(t))\psi_a(t). $$ We know that $\sum_{t\in\Bbb{F}_q^*}\psi_a(t)=-1$, so we get $$ S(a)=-\frac12+\frac12\sum_{t\in\Bbb{F}_q^*}\eta(t)\psi_a(t). $$ You will recognize that last sum as a Gauss' sum related to the Legendre character.