, 2009

We first prove that $k$ is congruent to one of $1, 0, -1$ modulo $(p-1)$. We rephrase the hypothesis in terms of modular arithmetic: If $x \not\equiv a, b, c \pmod p$, then
$$(x - a)^i (x - b)^j \equiv (x - c)^{-k} \pmod{p}.$$
By Fermat's little theorem, there is no harm in shifting $i, j, k$ by multiples of $p-1$, so we may assume $i, j, -k$ are elements of the set $\{0, 1, 2, \dots, p-2\}$. If $i + j \le p-2$, then the condition $i + j + k \equiv 0 \pmod{p-1}$ forces $-k = i + j$. In this case, the polynomial $(x-c)^{-k} - (x-a)^i(x-b)^j$ has degree at most $-k-1$ (since it is the difference of two monic polynomials of degree $-k$), but modulo $p$ it has at least $p-3$ distinct roots (namely the values of $x$ other than $a, b, c$). The only ways to avoid a contradiction are either to have $i+j = -k = p-2$ or to have the polynomial $(x-c)^{-k} - (x-a)^i(x-b)^j$ be identically zero modulo $p$. The former gives $k \equiv 1 \pmod{p-1}$ and we are done. The latter forces $i = j = -k = 0$ by unique factorization of polynomials modulo $p$, so we are again done.

If on the other hand $i + j \ge p - 1$, then we also have
$$(x - a)^{p-1-i} (x - b)^{p-1-j} \equiv (x - c)^{k} \pmod{p}$$
for all $x$ not congruent to $a, b, c$ modulo $p$. If $k \equiv 0 \pmod{(p-1)}$, we are done; otherwise, the fact that $i+j \equiv -k \not\equiv 0 \pmod{(p-1)}$ forces $i+j \ge p$, so $(p-1-i) + (p-1-j) \le p-2$. Thus the previous paragraph implies either $k \equiv 0 \pmod{(p-1)}$ or $k \equiv -1 \pmod{(p-1)}$.

By symmetry, each of $i$, $j$, $k$ is congruent to one of $-1$, $0$, $1$ modulo $(p-1)$. Because $i + j + k \equiv 0$ (mod $(p-1)$) and $p - 1 > 3$, the only possibilities (up to permutations) are $\{i, j, k\} \equiv \{0, 0, 0\}$ (mod $(p-1)$) or $\{i, j, k\} \equiv \{1, 0, -1\}$ (mod $(p-1)$). However, the latter implies that $x - a \equiv x - c$ (mod $p$) for all $x \ne a, b, c$ (mod $p$), which is impossible as soon as there is at least one such $x$ (which is true since $p > 3$). This contradiction leaves only the possibility $\{i, j, k\} \equiv \{0, 0, 0\}$ (mod $(p-1)$), proving the desired result.