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?
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.