$|f(x)|$ has 2020 divisors for infinately many $x\in\Bbb N$ 
Let $f(X)\in\Bbb Z[X]$ be a polynomial of degree $n \ge 1$. Prove that there exist infinately many $x\in\Bbb N$ such that $|f(x)|$ has more than $2020$ distinc prime divisors.

Firstly I want to announce that this is how I remember the problem, so the problem statement might be false, but I thought about it and it seems legit.
My approach was to firstly prove that the mentioned polynomial is inifnately often a composite number.
So let $a\in\Bbb N$ and $f(a) = p$ for some prime $p$. Now let's consider a sequence $a_i = a + kp$ and let $i \in {0,1,2,...}$. It is easy to see $a_i\equiv a\pmod p$ and hence $f(a_i)\equiv f(a)\pmod p$, but we know that a polynimial of degree $n$ can achive a certain value at most $n$ times. So in at most $3n$ cases $f(a) = 0,-p,p$ and as we have inifnately many choices of $i$ we will eventually get a value that is multiple of $p$ but not equal to $0,-p,p$. Now my idea was to prove the problem statement using the above idea but for numbers with $2$ distinct prime divisors and extend this idea for numbers with any number of distinct prime divisors ($2020$ included). But I got stucked. I smilarily proved that there are only finite numbers of cases in which the function $f(x)$ is $0,-pq,pq$ for some primes $p,q$ but how do I know if I'm not getting the multiples of the form $p^xq^y$ for some nonnegative $x,y$? How should I proceed? Or maybe I should change strategy? Any help appreciated.
 A: By Schur's theorem, the set of primes that divide some $f(n)\neq0$ (for $n\in\mathbb N$) is infinite.
So pick $2021$ primes $p_1,p_2,\dots,p_{2021}$ and positive integers $a_1,\dots,a_{2021}$ such that $p_i\mid f(a_i)$. Then by the Chinese Remainder Theorem pick $y$ solving the system
$$y\equiv a_i\pmod{p_i},$$ for each $1\leq i\leq2021$. Then we have infinitely many such $y$, and
$$f(y)\equiv f(a_i)\equiv0\pmod{p_i},$$
so each such $f(y)$ has at least $2021$ prime divisors.
A: You're definitely on the right track. See if you can prove the following statement by induction on $n$.

For all $n$, there exists some integer $x$ for which $|f(x)|$ has at least $n$ divisors.

You've essentially gone through one level of the inductive step (from "there exists a value that isn't $\pm 1$" to "there exist infinitely many composite values") -- the rest of the approach should generalize similarly. Start with a value of $x$ for which $|f(x)|$ has at least $n$ divisors, and try to find some other $x'$ for which $|f(x')|$ has more divisors than $|f(x)|$.
