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May 20, 2022 at 23:38 vote accept José
May 17, 2022 at 1:01 answer added Mark Lewko timeline score: 3
May 16, 2022 at 23:19 comment added Ofir Gorodetsky 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 $g^p-g$.
May 16, 2022 at 23:10 history edited José CC BY-SA 4.0
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May 16, 2022 at 23:10 comment added José 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.
May 16, 2022 at 23:03 comment added LSpice 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 $x \mapsto e^{2\pi i\operatorname{tr}_{\mathbb F_q/\mathbb F_p}(x)/p}$.
May 16, 2022 at 22:59 history asked José CC BY-SA 4.0