Power Gauss sum over a prime field ($G_{a,p}=p^{-1}{\sum }_{x\in {\mathbb{F}}_p}{e}^{2\pi ia{x}^k/p}$)

For $a\ne 0$ in the finite field ${\mathbb{F}}_p$, put $d=\gcd (k,p-1)$. The multiplicative group of a finite field is cyclic, and character orthogonality expresses the number of $k$th roots of $y\ne 0$ as ${\sum }_{{\chi }^d=1}\chi (y)$. The trivial character contributes $-1$, cancelling the $x=0$ term. Each of the remaining $d-1$ Gauss sums of Dirichlet characters has modulus $\sqrt{p}$, by the prime-field character Gauss sum identity. Consequently $|G_{a,p}|\le (d-1)/\sqrt{p}\le k/\sqrt{p}$.

For $S(a)={\sum }_{u}e(a{u}^k/p)$, character orthogonality counts representations of $x$ by $p^{-1}{\sum }_{a}S(a{)}^{3}e(-ax/p)$. Its zero-frequency term is $p^2$. The power Gauss sum over a prime field bounds the other terms in total by $k^3{p}^{3/2}$, strictly less than $p^2$ when $p>k^6$. Every residue class therefore has a representation.