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My teacher gave me the following lemma to prove :

Let $P(x)$ be a polynomial with integer coefficients. If some integer $t$ satisfies $$P(P((...P(t)...))) =t$$ for some number of iterations, then prove that $t$ should either satisfy $P(t)=t$ or $P(P(t)) = t$.

$(t \in \mathbb{Z})$

I have tried some usual approaches but failed. My teacher told me that this lemma is related to some IMO problem of the year 2005 or 2006. So I expect some clever and ingenious proof. I have tried my best but it does not yield.

Can anyone help me to prove this conundrum ?

Any help will be gratefully acknowledged.

Thanks in Advance ! :-)

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    Applying $P$ to both sides of the hypothesis yields $P(t)=t$ right? – 2017-02-17
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    @stud_iisc No, the hypothesis is that $P^n(t)=t$ for some $n$. Applying $P$ gives you $P(t)=P^{n+1}(t)$, you don't know if $P^{n+1}(t)=t$. – 2017-02-17
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    @charMD Okay, then I suppose the hypothesis must be written appropriately like you said. Thanks. – 2017-02-17
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    By the phrase "for some t" is t a single real number or is it restricted to a domain such as the integers? – 2017-02-17
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    Because of Sarkowski order (see https://en.wikipedia.org/wiki/Sharkovskii's_theorem), if a function has an orbit of period $\ge 3$, then it has also an orbit of period $4$. Then to answer your problem, it is enough to prove that $P$ does not have orbits of order $4$, i.e. if $P^4(t)=t$, then there is $i \in [\![1,3]\!]$ such that $P^i(t)=t$. – 2017-02-17
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    **Please do not delete your questions**, such as [this one](http://math.stackexchange.com/q/2148717/). – 2017-02-17
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    The claim is false as currently stated. Counter-example : $P(x)=2x^2+3x-1$ and $t$ is any root of $Q(x)=8x^3 + 16x^2 + 2x - 1$. Note that $P(P(x))=(-2x^2 - 4x - 1)+(x+1)Q(x)$ and $P(-2x^2 - 4x - 1)=x+(x+2)Q(x)$. – 2017-02-23
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    @EwanDelanoy But does $P^n(t)=t$? – 2017-03-10
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    I think there was a statement similar to this, proved in Silverman's Arithmetic of Dyn Sys - I will read that trying to find it again, but try looking there? – 2017-03-10
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    @mdave yes, $P^n(t)=t$ with $n=3$, check it. By the way, I think the correct version of your question is with "t integer" rather than "t real" (and in this form the question has already been answered here on MSE although I can't find it back). It is a simple divisibility argument – 2017-03-10

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Lemma $1$: if $s$ and $t$ are two distinct integer values then $s-t$ divides $P(s)-P(t)$.

Proof: suppose $P(x)=a_nx^n+\dots+a_0$, then $P(s)-P(t)=a_n(s^n-t^n)+a_{n-1}(s^{n-1}-t^{n-1})+\dots a_1(s-t)$. Clearly every summand is a multiple of $s-t$ because of the classical factorization $s^k-t^k=(s-t)(s^{k-1}+s^{k-2}t+\dots+t^k)$.

Suppose that we have elements $s_1,s_2,\dots,s_n,s_{n+1}=s_1$ such that $P(s_i)=s_{i+1}$ for all $1\leq i \leq n$ and the first $n$ elements are all distinct. We then have that $s_1-s_2|s_2-s_3\dots | s_1-s_n$. This implies that the magnitudes of all the distances are the same, and in order for the first $n$ elements to be different we clearly need $n=2$ or $n=1$.