proving that $p$ divides $f(1,2,\ldots, p -1)$ 
Let $p$ be a prime and let $f(x_1,\ldots, x_{p-1})$ be a symmetric polynomial in $p-1$ variables. Suppose $f$ is homogeneous of degree $d$ with $(p - 1) \not | d$. Prove that $p$ divides $f(1,2,\ldots, p -1)$. Assume the coefficients of $f$ are in any ring/field that makes the statement true (e.g. $\mathbb{C}$ or $\mathbb{R}$).

Since $f$ is a symmetric polynomial, it can be written as a polynomial function of elementary symmetric polynomials. Also, since $f$ is homogeneous of degree d, $f(t x_1,\ldots, t x_{p-1}) = t^{p-1}f(x_1,\ldots, x_{p-1})$ for every scalar $t$. I also know that if $a$ is an integer coprime to $p$, then $a, 2a,\ldots, (p-1) a$ forms a complete sequence of nonzero residues modulo p. Is any of this information useful for the proof? How is the fact that $(p-1)\not | d$ useful?
 A: Let $p$ be a prime, and let $f\in\mathbb{Z}[x_1,...,x_{p-1}]$ be symmetric in the variables $x_1,...,x_{p-1}$, and homogeneous of degree $d$, where $(p-1){\,\not\mid\,}d$.

Claim:$\;f\bigl(1,2,3,...,(p-1)\bigr)\equiv 0\;(\text{mod}\;p)$.

Proof:

Assume the hypothesis.

If $t$ is any integer not divisible by $p$, the integers
$$
t,2t,3t,...,(p-1)t
$$
are distinct and nonzero mod $p$, hence the sequence
$$
t,2t,3t,...,(p-1)t
$$
when reduced mod $p$, is just a permutation of the sequence
$$
1,2,3,...,(p-1)
$$
Now let $a$ be a primitive root mod $p$.

Then $a$ has order $p-1$, mod $p$, hence since $(p-1){\,\not\mid\,}d$, it follows that $p{\,\not\mid\,}(a^d-1)$.

Then since $f$ is symmetric we get
$$
f\bigl(a,2a,3a,...,(p-1)a\bigr)\equiv f\bigl(1,2,3,...,(p-1)\bigr)\;(\text{mod}\;p)
$$
and since $f$ is homogeneous of degree $d$ we get
\begin{align*}
&
a^df\bigl(1,2,3,...,(p-1)\bigr)\equiv f\bigl(1,2,3,...,(p-1)\bigr)\;(\text{mod}\;p)
\\[4pt]
\implies\;&
(a^d-1)f\bigl(1,2,3,...,(p-1)\bigr)\equiv 0\;(\text{mod}\;p)
\\[4pt]
\implies\;&
f\bigl(1,2,3,...,(p-1)\bigr)\equiv 0\;(\text{mod}\;p)
\\[4pt]
\end{align*}
as was to be shown.
