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Let $\mathbb{F}_q$ be a finite field, $\psi$ be a non-trivial additive character over $\mathbb{F}_q$, and $a, b \in \mathbb{F}_q$ constants. Is there any known estimate for the gaussian sum

$$\sum_{x \in \mathbb{F}_q} \psi( a x^m + b x^n),$$

possibly for specific values of $m, n \in \mathbb{Z}_{\ge 2}$, $m \ne n$?

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    $\begingroup$ For $m = n = 2$ and $q$ odd, it is (what I think of as) a classical Gauss sum, whose value is well known: the square is $q\operatorname{sgn}_{\mathbb F_q}(-(a + b))$, and the sign is known in terms of how $\psi$ differs from the "basic" character $x \mapsto e^{2\pi i\operatorname{tr}_{\mathbb F_q/\mathbb F_p}(x)/p}$. $\endgroup$
    – LSpice
    Commented May 16, 2022 at 23:03
  • $\begingroup$ Thank you. However, now I realize that I must also assume $m\ne n$, otherwise it would be just a monomial gaussian sum. I will edit the question to add this. $\endgroup$
    – José
    Commented May 16, 2022 at 23:10
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    $\begingroup$ Weil proved in general that the modulus of $\sum_{x \in \mathbb{F}_q} \psi(f(x))$ is at most $(\deg f - 1)\sqrt{q}$ for any polynomial $f$ not of the shape $g^p-g$. $\endgroup$
    – Ofir Gorodetsky
    Commented May 16, 2022 at 23:19

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As Ofir Gorodetsky notes in the comments, Weil's bound gives that the absolute value of the sum is most order $\max(m,n) \sqrt{q}$. This is non-trivial as long as $m$ and $n$ are $o(\sqrt{q})$, after this the estimate is worse than the trivial bound.

When $m$ and $n$ are large there are known estimates from additive combinatorics, at least when $q$ is prime. There are some obstructions to estimates, particularly when $m$, $n$ or $m-n$ have a large common factor with $q-1$. Otherwise one can obtain cancellation. See: J. Bourgain's paper "Mordell's Exponential Sum Estimate Revisited".

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