Let $p$ be a prime, $q=p^n$, and $\mathbf F_q$ denote the finite field with $q$ elements.
Problem 7 in Section 2.12 of Basic Algebra Vol. 1 by N. Jacobson asks the following:
Let $f(x_1,\ldots,x_n)\in \mathbf F_q[x_1,\ldots,x_n]$ be a polynomial of degree less than $n$. If $f(0,\ldots,0)=0$, then there exists $(a_1,\ldots,a_n)\in \mathbf F_q^n\setminus\{(0,\ldots,0)\}$ such that $f(a_1,\ldots,a_n)=0$ .
Following the hint in the book: Assume on the contrary that the proposition is not true and define $$ g(x_1,\ldots,x_n)=1-f(x_1,\ldots,x_n)^{q-1} $$
Then $g(a_1,\ldots,a_n)=0$ of and only if $(a_1,\ldots,a_n)\neq (0,\ldots,0)$.
Now deifne $$ h(x_1,\ldots,x_n)= (1-x_1^{q-1})\cdots(1-x_n^{q-1}) $$
Note that $g-h$ is evaluztes to zero at all points of $\mathbf F_q^n$.
Thus $g-h\in \langle{x_1^q-x_1,\ldots,x_n^q-x_n\rangle}$.
From here we get $$ 1-f^{q-1}= h+ a_1(x_1^q-x_1)+\cdots+a_n(x_n^q-x_n) $$ for some polynomials $a_1,\ldots, a_n\in \mathbf F_q[x_1,\ldots,x_n]$. Here is where the hypothesis that the degree of $f$ is less than $n$ should come in.
In particular, $f$ has no monomial term which contains all the $x_i$'s.
So if we could somehow show that the term $x_1^{q-1}\cdots x_n^{q-1}$ "survives" in the RHS of the last equation, then we would be done.
I am unable to do this.
Can somebody help?
We want to show that the degree of $h$ is less than that of $g$; you can reduce $g=\sum x_1^{i_1}...x_n^{i_n}$ to $h'=\sum x_1^{i_1\pmod q}\cdots x_n^{i_n \pmod q}$. It is clear that $\forall c_1,\ldots,c_n$, $(h-h')(c_1,\ldots,c_n)=0$ and that the degree of each variable in $h-h'$ is less than $q$. We want to show that $h=h'$. We may use induction on $n$; write $h-h'$ in the form $$h-h'=A_1(x_1,...x_{n-1})+...+A_q(x_1,...,x_{n-1})x^{q-1}$$ For all $c_1,...,c_{n-1}$, $h-h'$ is a polynomial in $x_n$ of degree $q-1$ with $q$ roots so all coefficients are zero and so by the induction hypothesis, $\forall i, A_i(x_1,...,x_{n-1})=0$, so $h-h'=0$ and $h=h'$, and its degree is less than the degree of $g$. So $r(q-1)>n(q-1)$ and $n>q$.