Is it possible that there is a polynomial $f$ with integer coefficients and an integer $a$ such that the sequence $f(a), f(f(a)), \ldots$ consists only of primes and tends to infinity? Are there any specific polynomials $f$ that are conjectured to have this property?
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1$\begingroup$ @Dietrich: Why must any such polynomial be constant and what does that have to do with Goldbach? $\endgroup$SJR– SJR2013-12-12 15:44:29 +00:00Commented Dec 12, 2013 at 15:44
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1$\begingroup$ Goldbach says that a polynomial producing prime values for almost all $n$ is constant (see en.wikipedia.org/wiki/Formula_for_primes). But we need more here. Still I guess that the polynomial must be constant. $\endgroup$Dietrich Burde– Dietrich Burde2013-12-12 15:48:23 +00:00Commented Dec 12, 2013 at 15:48
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$\begingroup$ One way of looking at it that might help with an impossibility proof: $f()$ must be irreducible (trivially), but for every prime $p_i=f^{\circ i}(a)$, $0$ must be in the preperiod of the iteration map of $f() \bmod p_i$ but not part of any period - that is, there can be no $k$ with $f^{\circ k}(0)= 0\bmod p_i$. This can happen for constant polynomials (and by the CRT we can construct a polynomial corresponding to an arbitrary map/function modulo any given prime), but it should be possible to show that for any given polynomial it can't happen once the modulus gets sufficiently large. $\endgroup$Steven Stadnicki– Steven Stadnicki2013-12-12 18:36:09 +00:00Commented Dec 12, 2013 at 18:36
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$\begingroup$ @Steven: Yes, it seems like a strange requirement that $f$ must never return to a multiple of any of its iterates. But I have no intuition about how the modulus being large makes this "no return" policy unlikely. $\endgroup$SJR– SJR2013-12-13 04:56:15 +00:00Commented Dec 13, 2013 at 4:56
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$\begingroup$ @SJR Essentially, because it makes the polynomial 'less arbitrary'. Keep in mind that for any fixed prime $p$ and any arbitrary function $f():\mathbb{F}_p\to\mathbb{F}_p$, we can find a polynomial $P$ whose values on $\mathbb{F}_p$ match the given $f$. But for a fixed polynomial, once you choose sufficiently large primes it seems plausible that the behavior 'settles down'. $\endgroup$Steven Stadnicki– Steven Stadnicki2013-12-13 05:09:58 +00:00Commented Dec 13, 2013 at 5:09
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It is not known whether there exists a univariate polynomial with integer coefficients of degree at least $2$ that assumes an infinite number of values that are prime. See the Bunyakovsky conjecture in this context - http://en.wikipedia.org/wiki/Bunyakovsky_conjecture. The polynomial $f$ with $f^n(a)$ prime would produce infinitely many primes as values. I would believe that $f$ must be constant, but I don't know how difficult it is to prove this. Some of these questions are very hard, as mentioned before.
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1$\begingroup$ Good point! But it might be known that no such polynomial as described in the question can exist. There might be a simple proof of this. Indeed, this is what I suspect. $\endgroup$SJR– SJR2013-12-12 16:02:20 +00:00Commented Dec 12, 2013 at 16:02