= Solution
Put $d=\gcd(k,p-1)$ and write $S(a)=\sum_{x\in\mathbb F_p}e(ax^k/p)=pG_{a,p}$. 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 $y\ne0$, <character orthogonality> consequently gives
$$
\#\{x:x^k=y\}=\sum_{\chi^d=1}\chi(y),
$$
where the sum is over its multiplicative <characters>. Thus, for $a\ne0$,
$$
S(a)=1+\sum_{\chi^d=1}\sum_{y\ne0}\chi(y)e(ay/p)=\sum_{\substack{\chi^d=1\\\chi\ne1}}\tau(\chi,a).
$$
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 $y=tz$ gives
$$
|\tau(\chi,a)|^2=\sum_{t\ne0}\chi(t)\sum_{z\ne0}e(a(t-1)z/p).
$$
The inner sum is $p-1$ for $t=1$ and $-1$ otherwise. Since $\sum_{t\ne0}\chi(t)=0$, the result is $p$. The <triangle inequality> now proves the <power Gauss sum over a prime field> bound
$$
\boxed{|G_{a,p}|\le\frac{d-1}{\sqrt p}\le\frac k{\sqrt p}.}
$$
This 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:
$$
R(x)=\#\{(u,v,w):u^k+v^k+w^k=x\}=\frac1p\sum_{a\in\mathbb F_p}S(a)^3e(-ax/p).
$$
The zero-frequency term is $p^2$. Each other term has $|S(a)|\le k\sqrt p$, so
$$
|R(x)-p^2|\le\frac{p-1}{p}k^3p^{3/2}<k^3p^{3/2}.
$$
If $p>k^6$, this is less than $p^2$, giving $R(x)>0$ for every $x$. Hence \b[$p>k^6$ is a sufficient threshold] for the <three power summands over a large prime field> conclusion.
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