Prime divisors of values of a polynomial on an infinite set

Question

This may be a well known problem:

Let $f$ be a polynomial with integer coefficients. Is it possible for the set of prime divisors of values of $f$ on an infinite set of integers, to be finite?

I guess such a set exists only for polynomials $f$ of the form $f(x) = c(ax+b)^n, a,b,c\in \mathbb{Z}, c\neq 0$. But I can't give a proof.

I'm specially interested in the case $f(x) = x^2+1$.

Thanks!

Answer 1

The answer is no for irreducible $f(x) \in \mathbb{Z}[x]$ of degree at least $3$. In this case, let $F(X,Y)$ be the homogenization of $f$, so that $f(x) = F(x,1)$. Let $\{p_1, \cdots, p_k\}$ be a fixed set of distinct rational primes. Then it was shown by Mahler and many subsequent authors that the equation $$\displaystyle F(x,y) = p_1^{z_1} \cdots p_k^{z_k}$$ where $z_i$ range over all integers and $x,y$ are integers, has at most finitely many solutions, with the record $$\displaystyle (5 \times 10^6 \deg f)^k$$ being obtained by Evertse; see

J.-H. Evertse, "The number of solutions of the Thue–Mahler equation" J. Reine Angew. Math. , 482 (1997) pp. 121–149.

Therefore, this also gives an upper bound for the number of solutions to the equation $f(x) = n$, where $n$ is a number all of whose prime divisors come from some finite set.

Answer 2

Another way would be to appeal to Faltings theorem. Indeed, if $f(n)$ has finitely many prime factors $p_1,\dots p_k$ for infinitely many $n,$ then the equation, say $$f(x)=Dy^3$$ would have infinitely many solutions for some fixed $D,$ which is a product of $\prod_{i=1}^kp_i^{\alpha_i}$ for $0\le \alpha_i\le 2.$

EDIT: since OP is interested in the case of quadratics, one can use the same reasoning as above to reduce everything to the equation

$$x^2+D=Ay^n$$

for fixed $A.$ This is Ramanujan-Nagel type equation. See https://en.wikipedia.org/wiki/Ramanujan–Nagell_equation for the description of solution.

Answer 3

Assume that it holds for, say, $x^2+1$ with given primes $p_1,\dots,p_m$. Consider factorization of $x^2+1=(x+i)(x-i)$ in Gaussian integers. For inifinitely many $x$ we have $x+i=A y^7$ for the same $A$. That is, $2i=A y^7-\bar{A} {\bar y}^7$, factor RHS in 7 multiples $A^{1/7}y-\bar{A}^{1/7}\bar{y}$. All but 1 are about $|y|$ by absolute value, hence another is about $1/|y|^6$. It means that algebraic number $(A/\bar{A})^{1/7}$ is approximated by $\bar{y}/y$ with accuracy $|y|^{-7}$. Taken real part we get approximation with the same accuracy by rational with denominator $|y|^2$. This may happen only finite number of times by Roth's theorem.

Answer 4

For $f(x)=x^2+1$, one can reach the desired conclusion by applying an elementary method of Stormer. "Elementary" means that nothing as deep as Siegel's theorem is used, just some theory of Pellian equations. Stormer's method is discussed on Wikipedia:

https://en.wikipedia.org/wiki/St%C3%B8rmer%27s_theorem

(However, there the method is explained in terms of a slightly different problem: Finding all $n$ with $n, n+1$ supported on a given finite set of primes.)

A nice reference is the paper of Lehmer cited there. You might also want to look at the following:

Najman, Filip(CT-ZAGR)
Smooth values of some quadratic polynomials.
Glas. Mat. Ser. III 45(65) (2010), no. 2, 347–355.