Danube Mathematical Competition

Throughout the proof congruences are taken modulo $p$. Begin by ruling out the case $p = 2$. If $p = 2$, then $a^2 + ab + b^2$ and $a^n + b^n + c^n$ are both even, and $a + b + c$ is odd. The first condition forces both $a$ and $b$ even, so $c$ is also even by the second, contradicting the third. Consequently, $p$ must be odd, and the conclusion follows unless $n$ is odd.

Henceforth assume $n$ odd. It is easily seen from the conditions in the statement that $a \not\equiv 0$, so $a^{-1}$ exists modulo $p$, and the hypotheses yield $B^2 + B + 1 \equiv 0$, $B^n + C^n + 1 \equiv 0$ and $B + C + 1 \not\equiv 0$, where $B = a^{-1}b$ and $C = a^{-1}c$. The first congruence yields $B^3 \equiv 1$, where $B \not\equiv 1$. (Otherwise, $3 \equiv 0$, so $p = 3$, $C^n \equiv 1$ and $C \not\equiv 1$ which is impossible since $n$ is odd.) Hence $3$ is a factor of $p-1$; in particular, $p \ge 7$ and $p-1$ is divisible by $6$, so the conclusion follows unless $n = 6m \pm 1$.

Let $n = 6m \pm 1$ and recall that $B^3 \equiv 1$ to deduce that $B^{2n} + B^n + 1 \equiv B^{\pm 2} + B^{\pm 1} + 1 \equiv 0$, so $C^n \equiv B^{2n} \equiv (-B-1)^n$; that is, $(-C(B+1)^{-1})^n \equiv 1$, since $B+1 \not\equiv 0$. The condition $B + C + 1 \not\equiv 0$ shows that $-C(B+1)^{-1} \not\equiv 1$, so the multiplicative order of $-C(B+1)^{-1}$ in $\mathbb{Z}_p^*$ is a divisor $d$ of $n$, greater than $1$. Since $d$ is also a divisor of $p-1$, the conclusion follows.