Prove that $p(x)=x+1$. 
Let $p(x)$ be a polynomial with integer coefficients such that $p(n) > n$ for every positive integer $n$. Define a sequence by $x_1 = 1, x_{i+1}=p(x_i)$ for $i\ge 1$. Suppose for any positive integer m, there exists some $r$ so that $m | x_r$. Prove that $p(x)=x+1$.


Source: problem 10 from this problem set.

One should theoretically be able to show $p(x)-x-1$ has infinitely many zeroes, though this may be very hard. I know that $p(n)\equiv p(m)\mod a$ for any positive integer a and integers $m,n$ with $m\equiv n\mod a$. The hint says to prove that $(x_{n+1}-x_1) | x_i$ for some $1\leq i\leq n.$ Based on the proof's conclusion, we should be able to make $n$ arbitrary. Then we should be able to prove that $i=n$ is the unique index satisfying the above constraint. In the latter case, we must have $x_2-x_1 | x_1$. If we do prove this though, it could probably be useful to prove the desired claim by induction. The sequence $(x_n)$ contains all the positive integers as prime factors, which seems to relate to the notion of surjectivity ($p$ turns out to be a surjective map from positive integers to $\mathbb{Z}_{\ge 2}$).
 A: The AoPS thread Easy indicates a problem source of the "Iranian selection test for IMO $2004$". The thread includes several solution methods, with this answer adapting (e.g., by correcting some mistakes with its equivalent of \eqref{eq2A}) and expanding on the one used in Post #$9$.
Since $p(n) \gt n$ for all positive integers $n$, then for all $i \ge 1$ we have $(x_{i+1} = p(x_i)) \gt x_i$. Thus, the $x_i$ form a strictly increasing sequence so, with $x_1 = 1$, then $x_i \ge i$. For any $k \ge 3$, define
$$N = x_k - 1 \tag{1}\label{eq1A}$$
The problem states there exists an $r$ such that $N \mid x_r$, so let $m$ be the minimum such index value. Since $x_k - 1 \ge x_{k-1}$, then $m \ge k-1$. With $m \gt k-1$, then $m \neq k$ because $x_k-1 \gt 1$, so $\gcd(x_k-1,x_k)=1 \; \to \; N \nmid x_k$, which means $m \gt k$.
Define $p^{0}(x) = x$ and $p^{i+1}(x) = p(p^{i}(x)) \; \forall \; i \gt 0$ (e.g., $p^2(x) = p(p(x))$). Next, note for any integers $a$ and $b$ that $a - b \mid p(a) - p(b)$, so $p(a) - p(b) \mid p^{2}(a) - p^{2}(b) \; \to \; a - b \mid p^{2}(a) - p^{2}(b)$, thus possibly repeating this multiple times gives that $a - b \mid p^{j}(a) - p^{j}(b) \; \forall \; j \ge 0$. Using this, along with $x_s = p^{s-1}(1)$ for all $s \ge 1$, we have
$$x_m = p^{m-1}(1) = p^{m-k}(p^{k-1}(1)) = p^{m-k}(x_k) \equiv p^{m-k}(1) = x_{m-k+1} \pmod{N} \tag{2}\label{eq2A}$$
However, this contradicts the minimality of $m$. Thus, $m = k-1$ so $x_k - 1 \mid x_{k-1} \; \to \; x_{k-1} \ge x_k - 1$ but, since $x_k - 1 \ge x_{k-1}$, we have $x_k = x_{k-1} + 1$. However, then the polynomial
$$q(x) = p(x) - x - 1 \tag{3}\label{eq3A}$$
has infinitely many roots, i.e., at $x = x_{k-1}$ for all $k \ge 3$. Thus, $q(x)$ must be the $0$ polynomial, which means that
$$p(x) = x + 1 \tag{4}\label{eq4A}$$
A: Here's a reference to an extremely high-level theorem which easily beats this problem into submission. If someone can prove it more simply that would be much better.

This can be proved by a special case of Chebotarev's Density Theorem. The statement of the theorem is complicated, but the special case of it can be restated to say: if $p(x)$ is a degree $n$ polynomial and $p_k$ is the $k$th prime, then
$$\lim_{N\to\infty} \frac{|\{ p_k\leq N|p_k\text{ does not divide }p(x)\text{ for any }x\in\mathbb{Z}\}|}{\pi(N)}=\frac{n-1}{n}$$
(Someone please correct me if I am incorrectly stating this case). Basically, for any non-linear polynomial there are infinite primes $p_{k_i}$ such that
$$p(x)\not\equiv 0\ (\text{mod }p_{k_i})\text{ for all }i\in\mathbb{N}$$
Thus, it is only possible that $p(n)$ from your original question is linear. The rest of the proof is simple.
