Saudi Arabia booklet 2024

Let $n=\mathrm{deg}P(x)\ge 1$ 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 $\mathrm{deg}Q<n$. 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 $\mathrm{deg}Q=m<n$. 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 $$\begin{array}{rl}Q(x^3-7x)& =Q(x-7)Q(x-8)Q(x-3)\\ & +(x-1{)}^{n}Q(x-8)Q(x-3)+(x-2{)}^{n}Q(x-7)Q(x-3)\\ & +(x+3{)}^{n}Q(x-7)Q(x-8)+(x-1{)}^n(x-2{)}^{n}Q(x-3)\\ & +(x-2{)}^n(x+3{)}^{n}Q(x-7)+(x+3{)}^n(x-1{)}^{n}Q(x-8).\end{array}$$ 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 $n$. $\square$