TITLE: Is there a polynomial such that $F(p)$ is always divisible by a prime greater than $p$?
QUESTION [24 upvotes]: Is there an integer-valued polynomial $F$ such that for all prime $p$, $F(p)$ is divisible by a prime greater than $p$?  For example, $n^2+1$ doesn't work, since $7^2+1 = 2 \cdot 5^2$.  I can see that without loss of generality it can be assumed that $F(0) \ne 0$.  Also, it is enough to find a polynomial where the property is true for sufficiently large prime $p$, since we could multiply that polynomial by some prime in the sufficiently large range and fix all the smaller cases. 
I think it is possible that there are no such polynomials, is there any good hint for proving this?
I can't find any solutions to $\text{gpf}(p^4+1) \le p$ for prime $p \le 10000$, where $\text{gpf}$ is the greatest prime factor, but there are plenty for $\text{gpf}(p^3+1) \le p$, for example $\text{gpf}(2971^3+1) = 743 \lt 2971$.  So I guess $F(p) = p^4+1$ might be an example.  I also checked higher powers for small $p$ and couldn't find solutions there either, so $k \ge 4 \rightarrow \text{gpf}(p^k+1) \gt p$ is plausible.

REPLY [4 votes]: Probably not, but this might be very hard to prove.  For each $\epsilon > 0$ the asymptotic fraction of integers $x<N$ that are multiples of primes $<x^\epsilon$ (a.k.a. "$x^\epsilon$-smooth") is positive, though it decays rapidly with $\epsilon$.  (See for instance this Wikipedia page.)  So it reasonable to expect that for any polynomial $F$ a positive fraction of primes $p$ will have $F(p)$ a product of primes less than $p$ (choose any positive $\epsilon < 1 \,/ \deg F$).
[Added a minute later: I see that Dan Brumleve said much the same in a comment on the original question, referring instead to the Wikipedia page on the Dickman function.]