Borevich Shafarevich page 8 problem 3 and 4 Under the assumptions of Warning's theorem, show that the solutions $A_i (i = 1, \dots, s)$ satisfy the congruences $\sum_{i=1}^s a_1^{(i)} \equiv \dots \equiv \sum_{i=1}^s a_n^{(i)} \equiv 0 \pmod{p}$ 
Problem 4 is we generalize to $\sum_{i=1}^s (a_1^{(i)})^k \equiv \dots \equiv \sum_{i=1}^s (a_n^{(i)})^k \equiv 0 \pmod{p}$ for $k = 0,1,\dots,p-2$
Here Warning's theorem is The number of solutions of the congruence $F(x_1,\dots,x_n) \equiv 0 \pmod{p}$ is divisible by $p$ provided the degree of $F$ is less than $n$, $p$ is prime.
And some notation if the congruence $F(x_1,\dots,x_n) \equiv 0 \pmod{p}$ has $s$ solutions we denote the solutions $A_i = (a_1^{(i)},\dots,a_n^{(i)})~~ i=1,\dots,s$
 A: The trick is to adapt the proof of Warning's theorem. I'll go through the solution for 3, and hopefully you'll see how you would adapt it to prove 4.
The proof of Warning's theorem goes like this.
Observe that $$\sum_{x\in\Bbb{F}_q}x^i = 0 \text{ for }0 \le i < q-1.$$
Therefore if $F$ is a polynomial in $n$ variables, and $\deg F < n(q-1)$,
$$\sum_{\mathbf{x}\in\Bbb{F}_q^n} F(\mathbf{x}) = 0,$$
since every monomial will have some variable of degree less than $q-1$ by pigeonhole.
Then we get Warning's theorem by observing that if $f$ has degree less than $n$,
the polynomial $F=1-f^{q-1}$ has degree at most $(n-1)(q-1)$, so the previous statement applies to $F$. However 
$$F(\mathbf{x}) = \begin{cases} 1 &\text{if $f(\mathbf{x})=0$}\\ 0 & \text{otherwise,}\end{cases}$$
so we conclude Warning's theorem. However, if we modify $F$ to be $F'(\mathbf{x}) = (1-f^{q-1}(\mathbf{x}))x_i$, we have
$$F'(\mathbf{x}) = \begin{cases} x_i &\text{if $f(\mathbf{x})=0$}\\ 0 & \text{otherwise,}\end{cases}$$
and $\deg F' \le (n-1)(q-1)+1 = n(q-1) - q + 2 < n(q-1)$, so once again the statement applies, and we get
$$\sum_{\mathbf{x}\in\Bbb{F}_q^n} F'(\mathbf{x}) = \sum_{\mathbf{x} : f(\mathbf{x})=0 } x_i = 0,
$$
as desired.
