Let $degP(x)=n $ and the coefficient of $P(x)$ is $a>0$ +) If $n=1$, we are done +) If $n>2$ Lemma 1 : $P(x)$ $>0$, increase and $\rightarrow \infty$ $\forall x \geq M$ for some $M>0 $ Proof: This is familiar Lemma 2: For any $P(x) \in Z\left [ x \right ]$, if $P(x)$ is $k$-th power for infinitely $x$, then exist $H(x): P(x)=H(x)^k$ Proof: Let m be deg of $P(x)$. There are two steps: +) Prove that exisr $Q(x) \in Q\left [ x \right ]: \underset{x \to +\infty}{lim }(Q(x)-\sqrt[k]{P(x)})=0$ by Extreme Principle +) Use equality $M(Q(a_i)-b_i)=M\left [Q(a_i)- \sqrt[k]{P(a_i)} \right ]+M\left [ \sqrt[k]{P(a_i)}-b_i \right ]$ Claim 1: $a_n \rightarrow \infty$ Proof: Choose $c_0 \in N^*: 2^{c_0}> M$. We know that exist $a_j: a_j=t^{c_0}\geq 2^{c_0}>M$ for some $t \in N_{>1}^{*}$ So from Lemma $(a_n)$ $>0$, increase and $\rightarrow \infty$ from $j$ Claim 2: For any $b$, exist infinitely $t: a_t$ is a power of nb $\forall n \in N^*$ Proof: Trivial cuz we know that exist $2nb$-th, $3nb$-th,.... in the sequence We will prove the statement: " For all $n \in N^*$, let $P (x)$ be a non constant integer polynomial and positive integer $n$. The sequence $a_0, a_1, ...$ is defined by $a_0 = n$ and $a_k = P (a_{k-1})$ for $k \ge 1$. If for any positive integer $b$, the sequence contains a $b$-th power of some positive integer greater than $1$. " by induction on $n$ Proof: Very trivial from all Lemmas and Claims above

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