# Compositions of polynomials with themselves and prime numbers

Suppose that we start with some polynomial $P_1(n)=\sum_{i=0}^ka_in^i$ and define $P_2(n)$ as $P_2(n)=P_1(P_1(n)$) and, more generally, $P_m(n)=P_{m-1}(P_1(n))$.

I do not know how much is known of can the sequence $m \to P_m(n)$ have an infinite number of primes for some choices of $P_1$ and $n$ and it seems to me as quite a hard a question, but I would like to know at least this:

Does there exist some polynomial $P_1$ and some natural number $n$ so that the sequence $m \to P_m(n)$ consists of prime numbers only?

• My guess is that if it is true, then nobody know how to prove it. – José Carlos Santos Dec 15 '17 at 9:05
• @JoséCarlosSantos Expect the unexpected. – user480281 Dec 15 '17 at 9:09
• If you drop the infinite number, it becomes mostly trivial: take a polynomial $P_1$ that merely does a permutation among a few primes, and start your sequence at one of them. – Professor Vector Dec 15 '17 at 9:26
• What about $P(n)=n$ ? :) If the iteration produces infinite many distinct values, it would be more than surprising if ALL these values could be prime numbers. And I agree Jose's comment. A proof that a specific polynomial does the job (if existent) would be almost certain out of reach. Note that it is even unknown whether $n^2+1$ contains infinite many primes. – Peter Dec 15 '17 at 10:01
• @AntoinePalAdeen In some sense, almost every question about prime numbers is harder than Goldbach+ Riemann hypothesis + abc + twin primes. The only tool we have to make some realistic conjectures on such sequences is the random model for the primes. See for example Apostol's book listing the very few topics which lead to interesting theorems about primes. – reuns Dec 15 '17 at 17:20