Is there a polynomial such that $F(p)$ is always divisible by a prime greater than $p$?

Question

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.

Answer 1

$$10181^4 + 1 = 2 \cdot 17 \cdot 1657 \cdot 4657 \cdot 5113 \cdot 8009$$

I have many factoring variants. Many of the commands in the following involve string variables, these are used in the versions where the factorization will actually be printed out, but are irrelevant here.

part of file form.h  

#include <iostream>
#include <stdlib.h>
#include <fstream>
#include <sstream>
#include <list>
#include <set>
#include <math.h>
#include <iomanip>
#include <string>
#include <algorithm>
#include <iterator>
#include <gmp.h>
#include <gmpxx.h>
using namespace std;



int  mp_all_prime_factors_below_bound( mpz_class  i, mpz_class bound)
{
  int squarefac = 0;
  string fac;
  fac = " = ";
  mpz_class p = 2;
   mpz_class  temp = i;
  if (temp < 0 )
  {
    temp *= -1;
    fac += " -1 * ";
  }

  if ( 1 == temp) fac += " 1 ";
  if ( temp > 1)
  {
    int primefac = 0;
    while( temp > 1 && p <= bound && p * p <= temp)
    {
      if (temp % p == 0)
      {
        ++primefac;
        if (primefac > 1) fac += " cdot ";
       //  fac += stringify( p) ;
        temp /= p;
        int exponent = 1;
        while (temp % p == 0)
        {
          temp /= p;
          ++exponent;
        } // while p is fac
        if ( exponent > 1)
        {
          fac += "^" ;
          fac += stringify( exponent) ;
          if (p >2) ++squarefac ;
        }
      }  // if p is factor
      ++p;
   // cerr << " temp " << temp << endl;
    } // while p
    if (temp > 1 && primefac >= 1) fac += " ";
    if (temp > 1 && temp < bound * bound  ){ fac += " cdot "; fac +=   temp.get_str()   ;}

       if (temp > 1 && temp >= bound * bound  ){ fac += " cdot mbox{BIG} "; }
    //  if (squarefac) fac += "      WOW " ;
  } // temp > 1
 // cerr << " temp " << temp << endl;
  return ( temp  <= bound ) ;
} // mp_all_prime_factors_below_bound

Answer 2

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.]