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Put $d=\gcd (k,p-1)$ and write $S(a)={\sum }_{x\in {\mathbb{F}}_p}e(a{x}^k/p)=p{G}_{a,p}$. The multiplicative group of a finite field is cyclic, so on its nonzero elements the $k$th-power map has kernel of size $d$. For $y\ne 0$, character orthogonality consequently giveswhere the sum is over its multiplicative characters. Thus, for $a\ne 0$,The term from the trivial character is $-1$ and cancels the initial one.

For a nontrivial character, the prime-field character Gauss sum identity can be proved directly. Since $|\chi (z)|=1$ on nonzero elements, substituting $y=tz$ givesThe inner sum is $p-1$ for $t=1$ and $-1$ otherwise. Since ${\sum }_{t\ne 0}\chi (t)=0$, the result is $p$. The triangle inequality now proves the power Gauss sum over a prime field boundThis also includes $d=1$, when the nonzero-frequency sum vanishes.

For $x\in {\mathbb{F}}_p$, additive character orthogonality counts its representations as three powers:The zero-frequency term is $p^2$. Each other term has $|S(a)|\le k\sqrt{p}$, soIf $p>k^6$, this is less than $p^2$, giving $R(x)>0$ for every $x$. Hence $p>k^6$ is a sufficient threshold for the three power summands over a large prime field conclusion.

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 $|G_{a,p}|\le (d-1)/\sqrt{p}\le k/\sqrt{p}$.