Let $p \geq 3$ be prime, and let $n \geq 1$. Let $F \in \mathbb{Z}[x_1,...,x_n].$ Let $d=\deg(F).$ Let $b=\Bigl\lceil\frac {n-d} d\Bigr\rceil$.
Let $N=\operatorname{card}\{x\in \{0,1,...,p-1\}^n : F(x) \equiv 0$ (mod p) $\}$.
I want to prove that $p^b$ divides $N$.
My attempt:
Write $F(x)=a_1x^{e_1}+...+a_mx^{e_m}$, where the $e_i$ are vectors in $\mathbb{Z}^n$, not nessecarily the elementary basis vectors.
Since $(\mathbb{Z}/p^n\mathbb{Z})^\times$ is cyclic of order $p^n-p^{n-1}$, there is a $g\in \mathbb{Z}$ such that $\forall ξ\in (\mathbb{Z}/p^n\mathbb{Z})^\times\exists α\in \mathbb{Z}[$theIntegerEquivalentOf($ξ)\equiv g^α$ (mod p)]
Define the set $S=\{0,g^{0p^n},g^{1p^n},...,g^{(p-2)p^n}\} \subseteq \mathbb{Z}$.
I have shown that the elements of $S$ are distinct, although I won't repeat the proof here.
Define $A=\sum_{x\in S^n}F(x)^{(p-1)p^n}$
We have $N \equiv \sum_{x\in S^n} (1-F(x)^{(p-1)p^n}) \equiv p^n-A \equiv -A$ (mod $p^n$).
Therefore, if we can show that $p^b$ divides $A$, we're done.
We have $A=[\sum_{k_1+...+k_m=(p-1)p^n} [{(p-1)p^n \choose k_1,...,k_m}a_1^{k_1}...a_m^{k_m} [\sum _{x\in S^n} x^{k_1e_1+...+k_me_m}]]$
I have shown that $p^ \frac {σ(k_1)+...+σ(k_m)-(p-1)} {p-1}$ divides ${(p-1)p^n \choose k_1,...,k_m}$, where the notation $σ(k_i)$ denotes the sum of digits in the base-p expansion of $k_i$.
I think the key to finishing the proof will be to show that $\sum _{x\in S^n} x^{k_1e_1+...+k_me_m}$ is a multiple of some certain power of p, call it $p^c$, and then to show that $b \leq \frac {σ(k_1)+...+σ(k_m)-(p-1)} {p-1} + c$. But I'm stuck. Please help!
First you can show that $\sum _{x \in S^n}x^{k_1e_1+...+k_me_m}\equiv p^α(p-1)^β0^γ$ (mod $p^n$) where $α$ is the number of components in the vector $k_1e_1+...+k_me_m$ equal to zero, $β$ is the number of components divisible by $p-1$ yet nonzero, and $γ$ is the number of components not divisible by $p-1$. (I'm following the convention $0^0=0$.)
If $γ\neq0$, we're done, so assume $γ=0$.
Then $\sum _{x \in S^n}x^{k_1e_1+...+k_me_m}$ is clearly divisible by $p^α$.
We can show that $β(p-1) \leq $ (sum of coordinates in the vector $σ(k_1)e_1+...+σ(k_m)e_m$) $\leq d(σ(k_1)+...+σ(k_m))$, and from this inequality deduce the inequality desired by OP (taking $c=α$)