Polynomial and prime factors
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?
$\endgroup$ 31 Answer
$\begingroup$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 $p_i$.
$\endgroup$ 3More in general
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