Polynomial $P$ satisfies $P(n)>n$ for all positive integers $n$. Every positive integer $m$ is a factor of some number of the form $P(1),P(P(1)),P(P(P(1))),\ldots $. Prove that $P(x)=x+1$.
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Denote the iterates by $x_0 = 1, x_{n+1} = P(x_n)$. Assume that the coefficients of $P$ are integral. If at some point $P(x_n) > 2x_n$, then I claim that $m = P(x_n)-x_n$ does not divide any iterate. First, $x_n < m$, so $x_0,\ldots,x_n$ cannot be divisible by $m$. Second, we prove by induction that for $k \geq n$, $x_k \equiv x_n \pmod{m}$: $x_{k+1} = P(x_k) \equiv P(x_n) \equiv x_n \pmod{m}$. Since $x_n < m$, we see that $m$ doesn't divide any of the iterates. We conclude that always $P(x_n) \leq 2x_n$. Thus $P(x) = ax+b$ with $a \leq 2$. On the one hand $P(1) > 1$, and on the other hand $P(1) \leq 2$. Thus $P(1) = 2$, and therefore either $a = 1$ or $a = 2$. If $a = 2$ then $P(x) = 2x$, and we generate only powers of $2$. Thus $a = 1$ and $P(x) = x + 1$. |
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