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So I've been thinking about this questions for ages but I haven't made any progress on it.

I need to prove that for every $f(X) \in \mathbb{Z}[X]$ with $f(0) = 1$, there exists a $n \in \mathbb{Z}$ such that $f(n)$ is divisible by at least 2019 distinct primes.

The only thing that I've seen is that this is easy to show when $f(X)$ has a root in $\mathbb{Z}$ but for the rest I haven't made any progress?

Does anyone know how I can solve this?

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  • $\begingroup$ Perhaps induction on the degree of the polynomial? You'd need to assume that the degree is at least one, of course. $\endgroup$ – Gary Moon Mar 20 at 13:42
  • $\begingroup$ Perhaps the answer there will help you: math.stackexchange.com/questions/601526/… $\endgroup$ – B.Swan Mar 20 at 14:37
  • $\begingroup$ I don't really see how that answer can help me? Like, I don't know which primes divide my function divides and they also don't need to be the same for every value? $\endgroup$ – Mee98 Mar 20 at 14:46
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Suppose you know $f(x)$ has a root $r_i$ modulo a prime $p_i$, for $m$ different such primes.

Let $P_m$ be the product of the $m$ primes.

Then since $f(0) = 1$, this tells you that $$ f(kP_m) \equiv 1\not\equiv 0 \pmod{ p_i} $$ for any integer $k$. So none of the existing $m$ primes divide this.

By choosing $k$ large enough, this is not zero (since a polynomial only have finitely many roots) and therefore it must have a prime factor $p_{m+1}$ not the same as the current $m$ you have found. Thus you get a new root $kP_m$ and prime $p_{m+1}$. i.e. $$ f(kP_m) \equiv 0 \pmod{p_{m+1}} $$

So you can find solution to $$ f(r_i) \equiv 0 \pmod{p_i} $$ for arbitrarily many primes $p_i$.

Finally, you can use Chinese Remainder Theorem to find a common root $r$ for all these primes. So that $f(r)$ will be divisible by each of the $p_i$.

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  • $\begingroup$ How exactly can you use the Chinese Remainder Theorem in the last step? $\endgroup$ – Mee98 Mar 20 at 16:29
  • $\begingroup$ @Mee98 The CRT system to solve is $$r \equiv r_i \pmod{p_i},$$ since if $r_i$ is a root mod $p_i$, then any $r \equiv r_i \pmod{p_i}$ is also a root. Alternatively, if we have $f(r) \equiv 0\pmod A, f(s) \equiv 0\pmod B$ then we can set $k \equiv (s-r)A^{-1}\pmod B$ to get $$f(r+kA)\equiv 0 \pmod{AB}$$, then repeat. (Adding one prime at a time to the product.) $\endgroup$ – Yong Hao Ng Mar 20 at 16:45
  • $\begingroup$ which works since $$f(r+ kA)\equiv f(r)\equiv 0 \pmod A$$ and $$f(r+kA) \equiv f(r + (s-r)A^{-1}A) \equiv f(r +(s-r))\equiv f(s) \equiv 0\pmod B$$So divisible by $A$ and $B$ means divisible by $AB$. $\endgroup$ – Yong Hao Ng Mar 20 at 16:50

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