Iranian Mathematical Olympiad

Suppose that the image of integer numbers under the polynomial $P(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$ contains an infinite geometric progression with common ratio $a\in \mathbb{Z}-\{0\}$. For each $b\in \mathbb{Z}$ we have $$P(ax+b)=a_n{a}^n{x}^n+(n{a}_n{a}^{n-1}b+a_{n-1}{a}^{n-1})x^{n-1}+\dots$$ $$a^{n}P(x)=a_n{a}^n{x}^n+a_{n-1}{a}^n{x}^{n-1}+\cdots +a_0{a}^n.$$ We can find some $b_1,b_2\in \mathbb{Z}$ and $N\in \mathbb{N}$ such that for every $x\ge N$, $$P(ax+b_2)<a^{n}P(x)<P(ax+b_1),$$ and for every $x\le -N$, $$P(ax+b_2)<a^{n}P(x)<P(ax+b_1),\text{or }P(ax+b_1)<a^{n}P(x)<P(ax+b_2).$$ For each $P(x)$ in the geometric progression, $aP(x),a^{2}P(x),\dots ,a^{n}P(x),\dots$ are all in $P(\mathbb{Z})$, hence there exists some $y\in \mathbb{Z}$ such that $a^{n}P(x)=P(y)$. If $|x|$ is sufficiently large (there are infinitely many values of $P(x)$ in the geometric progression), using the above inequalities, $ax+b_1<y<ax+b_2$. Therefore, $y-ax$ is a constant value between $b_1$ and $b_2$. Hence there exists some constant number $c$ such that the equation $a^{n}P(x)=P(ax+c)$ has infinitely many solutions and consequently $P(ax+c)$ and $a^{n}P(x)$ are two equal polynomials. Now, our goal is to find all polynomials $P(x)\in \mathbb{Z}[x]$ satisfying the equation $a^{n}P(x)=P(ax+c)$. Let $Q(x)=ax+c$. If $\alpha$ is a root of $P(x)$, setting $x=\alpha$ in the equation implies that $P(Q(\alpha ))=0$ and hence $Q(\alpha ),Q^2(\alpha ),\dots$ are all roots of $P$. Since $P(x)$ has a finite number of roots, there are some natural numbers $m_1>m_2$ such that $Q^{m_1}(\alpha )=Q^{m_2}(\alpha )$. This implies $Q^{m_1-m_2}(\alpha )=\alpha$, since $Q$ is injective. Note that if $\beta =\frac{c}{1-a}$, then $Q(\beta )=\beta$ and hence $Q^{m_1-m_2}(\beta )=\beta$. On the other hand, $Q^{m_1-m_2}(x)-x$ is a linear polynomial. Therefore, it has at most one root, hence $\alpha =\frac{c}{1-a}$ is the only root of $P(x)$. Consequently, $P(x)$ has the form $r(x-\frac{p}{q}{)}^n=\frac{r}{q^n}(qx-p{)}^n$, where $p,q$ and $r$ are three integers such that $(p,q)=1$. $P(x)\in \mathbb{Z}[x]$, hence $$\frac{r}{q^n}{p}^n\in \mathbb{Z}\Rightarrow q^n\mid r{p}^n\stackrel{(q,p)=1}{\Rightarrow }{q}^n\mid r\Rightarrow \exists s\in \mathbb{Z};r=q^{n}s.$$ Therefore, we have $P(x)=s(qx-p{)}^n$, for some $p,q,s\in \mathbb{Z}$ that $(p,q)=1$. We claim that each polynomial of this form satisfies the problem's condition. It is enough to show that the polynomial $qx-p$ satisfies the property. This is equivalent to showing the existence of a geometric progression whose elements are all congruent to $-p$ modulo $q$. Obviously, $\{-p(q+1{)}^m{\}}_{m\ge 0}$ is an example of such progression and this completes our proof.