TITLE: Sums of powers mod p
QUESTION [20 upvotes]: For prime $p > 7$ with $p-1=rs$, $r>1$, $s>1$, let $A=\{x^r|x \in \mathbb{Z}_p\}$ and $B = \{x^s|x \in \mathbb{Z}_p\}$. If $g$ is a primitive root mod $p$ then $A = \{0\} \cup \{g^{ir}|0 \leq i < s \}$ and $B = \{0\} \cup \{g^{js}|0 \leq j < r \}$.  Is it always true that $\mathbb{Z}_p \neq A + B$?

REPLY [2 votes]: For fixed $r>1$, use of Weil's bound will give the result. 
    We refer to Theorem 5.1 from chapter 2 of W.M.Schmidt, Equations over Finite Fields: An Elementary Approach.
Theorem
    Let $f_1,\cdots, f_n$ be polynomials with coefficients in $\mathbb{F}_q$ and of degree $\leq m$. Put $\delta=\textrm{lcm}(d_1,\cdots,d_n)$, and $d=d_1d_2\cdots d_n$. Let X be a variable and let $\eta_1,\cdots,\eta_n$ be algebraic quantities with
    \begin{equation}
    \eta_1^{d_1}=f_1(X), \cdots, \eta_n^{d_n}=f_n(X).\
    \end{equation}
    Suppose
    \begin{equation}
    [\overline{\mathbb{F}}_q(X, \eta_1,\cdots,\eta_n):\overline{\mathbb{F}}_q(X)]=d.
    \end{equation}
    Then if $q>100\delta^3m^2n^2$, the number $N$ of solutions $(x,y_1,\cdots,y_n)\in\mathbb{F}_q^{n+1}$ of the equations $y_1^{d_1}=f_1(X),\cdots,y_n^{d_n}=f_n(X)$ satisfies
    \begin{equation}
    |N-q|<5mnd\delta^{5/2}q^{1/2}.
    \end{equation} 
    We will use this theorem to solve the problem for fixed $r>1$. We have $q=p$ prime number. Let $g$ be a primitive root modulo $p$. We put $n=r+1$, $m=1$, $d_1=\cdots=d_n=r$, $\delta=r$, $d=r^{r+1}$, $f_i(X)=g(X-g^{is})$ for $0\leq i \leq r-1$, and $f_r(X)=gX$. Then the condition is satisfied, and the number $N$ of solutions to the system $Y_i^r=f_i(X)$ ($0\leq i \leq r$) is 
    \begin{equation}
    |N-p|<5(r+1) r^{r+1+5/2}p^{1/2}.
    \end{equation}
We look for the number $N^{*}$ of solutions with no $Y_i$ being zero. 
Then we have
$$N^{*}\geq p-5(r+1) r^{r+1+5/2}p^{1/2}-(r+1)r^{r+1}.$$ 
This is in fact greater than zero for sufficiently large $p$. For any such solution $(X,Y_0,\cdots, Y_r)\in \mathbb{F}_p^{r+2}$, we obtain that
$$X\in \mathbb{Z}_p-(A+B).$$ 
This proves the result.