, 2024

Let $n = \deg P(x) \ge 1$ and $a \ne 0$ be the leading coefficient of $P$, then comparing the leading coefficients of both sides to get $a = \pm 1$. Note that if $P(x)$ satisfies then so does $-P(x)$; without loss of generality, we assume that $a = 1$.

First, let solve the problem when $n = 1$, consider $P(x) = x + b$. Substituting into the given condition,
$$x^3 - 7x + b = (x - 7 + b)(x - 8 + b)(x - 3 + b).$$
Comparing the coefficient of degree 2, we have $0 = b - 7 + b - 8 + b - 3$ so $b = 6$. Therefore, $P(x) = x + 6$, which is a solution.

Now, for any $n \ge 1$, let $P(x) = (x + 6)^n + Q(x)$ with $\deg Q < n$. If $Q(x) \equiv 0$ then we have $P(x) = (x + 6)^n$, which satisfies since
$$(x^3 - 7x + 6)^n = (x - 1)^n (x - 2)^n (x + 3)^n.$$
Now assume that $Q(x) \ne 0$ and put $\deg Q = m < n$. Substituting in the given condition then we get
$$(x^3 - 7x + 6)^n + Q(x^3 - 7x) = [(x - 1)^n + Q(x - 7)] [(x - 2)^n + Q(x - 8)] [(x + 3)^n + Q(x - 3)]$$
Then expanding and simplifying, we get
$$\begin{align*} Q(x^3 - 7x) &= Q(x - 7)Q(x - 8)Q(x - 3) \\ &\quad + (x - 1)^n Q(x - 8)Q(x - 3) + (x - 2)^n Q(x - 7)Q(x - 3) \\ &\quad + (x + 3)^n Q(x - 7)Q(x - 8) + (x - 1)^n (x - 2)^n Q(x - 3) \\ &\quad + (x - 2)^n (x + 3)^n Q(x - 7) + (x + 3)^n (x - 1)^n Q(x - 8). \end{align*}$$
Comparing the degree of both sides, $3m = 2n + m$ or $m = n$, a contradiction.

From these arguments, one can conclude that all solutions of the given condition are $P(x) = \pm (x + 6)^n$ for all positive integers $n$. $\square$