I am struggling with these two problems:
1) Let $p$ be a polynomial with integer coefficients. Show that for each sequence of $k$-times iterating the polynomial, $n,p(n),p(p(n))=p^{(2)} (n),\ldots,p^{(k)}(n)$, we have to have $k\leqslant 2$, if $p^{(k)}(n)=n$ and $p(n) \neq n,p^{(2)} (n)\neq n,\ldots,p^{(k-1)}(n)\neq n$.
After thinking for some time about this problem, I have the strong impression, that I somehow have to make of use the fact, that the integer roots of a polynomial with integer coefficients, because the remainder of $p^{(k)}$ is $p^{(k)} (r)$, where $r$ is the remainder of $p$.
2) One has to show if there exists a polynomial such that $p(\frac{1}{k})=\frac{1}{2k+1}$, for all $k\in \mathbb{N}$.
My hunch here is, that this would be impossible, since $p$ would have to "oscillate" to much...but I can't make this precise.
Any help would be much appreciated.
EDIT: Second problem solved. What I've got for the first one with user9325's help: After proving that the distance between the iterates has to be constant, one can distinguish between the following cases: If $p(n)-n=0$, the we are finished. If $\alpha:=p(n)-n \neq0$, then we have to have $p^{(2)}(n)=n+2\alpha,\ldots,p^{(k-1)}(n)=n+(k-1)\alpha$, since $n-p(n)=p(n)-p^{(2)}(n)=\ldots=p^{(k-2)}(n)-p^{(k-1)}(n)=p^{(k-1)}(n)-n$. But $n-p(n)=p^{(k-1)}(n)-n$ implies that $p^{(k-1)}(n)=n-\alpha$. Combining these to equations concerning $p^{(k-1)}(n)$ gives us that $k=0$. This should be a contradiction but I am not sure how to make that crystal clear...I have the feeling, that I can't really see through this problem yet...