bump on this one

Hint: a-b|P(a)-P(b)

Firstly if $P(0)=0 (P(x)=xQ(x))$, then the sequence is constant because all terms are zero and hence the condition is satisfied. Also, it is obvious that all constant polynomials work. Now suppose that $P(x)$ is not constant and $P(0)$ is not equal to $0$. Let $P(x)=\sum_{i=0}^{d}a_ix^i$, where $d=degP>0$ and $a_0=P(0) \not = 0$. Firstly, it is trivial that $P(0) | P^i(0)$ for all $i>0$. Now define the sequence $x_n$ such that $x_0=0$ and $x_{n+1}=P(x_n)$. Then it is obvious that $x_n=P^n(0)$ for all $n$. Let $p$ be a positive integer such that $p | P^i(0)$ for some $i>1$, then for all positive integers $k$ we have that: $P^{ki}(0) \equiv P^{(k-1)i}(0) \dots \equiv P^{2i}(0) \equiv P^i(0) \equiv 0 (mod p)$. Now taking $n=p$ in the condition, we get that $x_i$ is eventualy constant $mod p$, hence it must be eventually zero $mod p$, as we have proved that $p$ divides infinitely many numbers of the sequence. Hence there exist $N$ such that $p | x_i$ for all $i>N$. Hence we have that $p$ divides $x_i$ and $x_{i+1} = P(x_i)=\sum_{j=0}^{d} a_jx_i^j$ for large $i$. Hence $p$ must divide $a_0=P(0)$ (as $p$ divides $a_jx_i^j$ for all $j>0$). Hence we have shown that if a number divides $P^i(0)$ for some $i$, then $p$ must also divide $P(0)$. So, we have that $P^i(0) | P(0) => |P^i(0)| = |P(0)| $ for all $i$. If $P(P(0))=P(0)$, then we have that $P(x)=(x-P(0))Q(x)+P(0)=(x-c)Q(x)+c$ for some $Q(x) \in Z[x]$, where $c=P(0)$. For $x=0$ we get that $Q(0)=0 => Q(x)=xR(x)=>P(x)=x(x-c)R(x)+c$ for some $R(x) \in Z[x]$. Then we have that $P(c)=c => P(P(0))=P(0)$ and hence inductively $P^i(0)=P^{i+1}(0)$ for all $i>0$, hence the condition is obviously satisfied. Now suppose that $P(P(0))=-P(0)$. In this case we have 2 subcases: 1st subcase: $P(P(P(0)))=P(0)=-P(P(0))$. In this case it is trivial by induction that $P^{2i+1}(0)=P^{2i-1}(0)=...P(0)$ and $P^{2i}(0)=P^{2i-2}(0)...=P(P(0))=-P(0)$. However, taking $n$ large enough in the condition we have that for large enough $i$: $x_{2i+1}-x_{2i}=2P(0)$ is divisible by $n$, which is a contradiction. 2nd subcase: $P(P(P(0)))=P(P(0))=-P(0)$ Using that $P(P(0))=-P(0)$ we get that there exist $Q(x) \in Z[x]$ such that $P(x) = (x-P(0))Q(x)-P(0)$ For $x=P(P(0))$ we get that: $P(P(P(0)))=(P(P(0))-P(0))Q(P(P(0)))-P(0)=0 => Q(P(P(0)))=0=>Q(-P(0))=0 => Q(x)=(x+P(0))R(x) => P(x)=(x-P(0))(x+P(0))R(x)-P(0) => P(x) =(x-c)(x+c)R(x)-c$. For $x=0$ we get that: $P(0)=-c^2R(0)-c => cR(0)=-2=>R(0)=-2/c=>R(x)=xS(x)-2/c=>P(x)=(x-c)(x+c)(xS(x)-2/c)-c$ for $c=1,-1,2$ or $-2$ (because $P$ has integer coefficients). It is easy to show inductively that $P^{i+1}(0)=P^i(0)$ for all $i>1$ for all polynomials of these form, hence they indeed satisfy the condition. In conclusion the required polynomials are all of the following form: $P(x)=c$, $P(x)=xQ(x)$, $P(x)=x(x-c)R(x)+c$ and $P(x)=(x-c)(x+c)(xS(x)-2/c)-c$ for some $c \in \{1,-1,2,-2\} $, where $Q(x),R(x),S(x)$ are polynomials with integer coefficients.

Construct, Polynomial P(0) with integer coefficient 0 such that P(0)=0. Then P(P(0))=P(0)=0 Therefore the sequence becomes 0,0,0,0,... Therefore,it is proved and we are done

Banglarmanush wrote: Construct, Polynomial P(0) with integer coefficient 0 such that P(0)=0. Then P(P(0))=P(0)=0 Therefore the sequence becomes 0,0,0,0,... Therefore,it is proved and we are done it tells you to find all of the polynomials, not point out one of them?

Banglarmanush wrote: Then,the only possible polynomial is P(x)=0 Not sounding rude but all of ur posts are like this... most don't make sense I recommend u t read other posts.... and learn methods and ways of proof

Solution from Twitch Solves ISL: The answer is: Any polynomial of the form $P(x) = Q(x)x(x-c)+c$, where $Q \in {\mathbb Z}[x]$ and $c \in {\mathbb Z}$. Any polynomial of the form $P(x) = Q(x)(x+c)(x-c)-c$, where $Q \in {\mathbb Z}[x]$ has $Q(0) = -\frac{2}{c}$ and $c \in \{\pm1,\pm2\}$. The main claim of the problem is that the sequence can be described as follows. Claim: The sequence is either \begin{align*} 0,\; & c,\; c,\; c,\; \dots \\ 0,\; & c,\; -c,\; -c,\; \dots \end{align*}for some integer $c$. Proof. If $P(0)=0$ we're done, so assume $c \coloneqq P(0) \neq 0$. Then every term must in the sequence is a multiple of $c$, by working modulo $|c|$. On the other hand, I contend every term in the sequence is $\pm c$ now. Indeed, if some term $M$ in the sequence is a different multiple of $c$, then take modulo $|M|$ to find the sequence is not eventually constant modulo $M$. (In the case $M = 0$, take modulo any number larger than $c$.) The claim then follows from this. $\blacksquare$ The answer then follows from the two cases of the claim. In the first case where the sequence is eventually constant, the only constraint is that we have $P(0)=P(c)=c$ for some $c$, which is equivalent to the bullet above. In the second case, we first read the constraint $P(c)=P(-c)=-c$ to get $P(x) = Q(x)(x+c)(x-c)-c$, and then plug in $x=0$ to conclude.

Just want to comment that one can invoke Kobayashi's Theorem to strengthen the statement to that sequence being constant modulo infinitely many primes p [which is not needed in the original version]