Let $f\in \mathbb Z[X]$ be a polynomial of positive degree.How to show that there are infinitely many prime numbers $p$ such that the polynomial $f$ has a zero in $\mathbb Z/p \mathbb Z$ ?
I have no idea of how to start even.
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2By $\Bbb{Z}_p$ do you mean the $p$-adic integers or the field of $p$ elements? – Servaes Feb 12 '19 at 18:31
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@Servaes:See the edit – Damon Feb 12 '19 at 18:33
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That edit wasn't too helpful :) – darij grinberg Feb 12 '19 at 18:38
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@darijgrinberg:I mean a field with p elements – Damon Feb 12 '19 at 18:39
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Ah, I fixed it. – darij grinberg Feb 12 '19 at 18:40
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It would help your cause to provide some context. As per our guide for new askers. Also, did you search the site? This may well be a dupicate of an older question! – Jyrki Lahtonen Feb 13 '19 at 08:00
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To all future voters on this thread. The question is sub-par. But Carl Schildkraut's nice answer is worth preserving. I think merging this to the chosen dupe target (or to another answer linked to my pick, if judged better) would be a possibility. Carl's answer will defend its place among those older answers quite well if you ask me. – Jyrki Lahtonen Feb 13 '19 at 08:06
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A merger may raise some notational compatibility issues. Unless I missed something here that would be limited to denoting the polynomial with $g$ instead of $f$. Ping me, if you choose to merge, and want help wih such edits. – Jyrki Lahtonen Feb 13 '19 at 08:08
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Hint: Assume there are finitely many primes that divide any values of $f$. Then, for all $n$,
$$f(n)=\pm p_1^{e_1}\cdots p_k^{e_k}$$
for some nonnegative integers $e_1,\cdots,e_k$.
If you look at the range $[-N,N]$ for some large $N$, what proportion of numbers in that range are values of $f$ (asymptotically)? What proportion of those numbers can be written as $\pm p_1^{e_1}\cdots p_k^{e_k}$?
Carl Schildkraut
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Carl Schildkraut:$\frac{\tau (f(n))}{N}$?,Where $\tau (f(n))$ means number of positive divisors of $f(n).$ – Damon Feb 12 '19 at 19:10
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@Damon not quite. I mean what proportion of integers in the range $[-N,N]$ can be written as $f(n)$ for some integer $n$. E.g. for $f(x)=x^2+1$ and $N=30$ only the values ${1,5,10,17,26}$ can be reached, which is $$\frac{5}{2\cdot 30+1}=\frac{5}{61}$$ of the number of integers in the interval itself. – Carl Schildkraut Feb 12 '19 at 21:34
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1@Damon Try to bound the range of those exponents $e_i$ using the logarithm of $N$. Then do the same for the values of $f$, this time using a relevant root of $N$. Then remember what you know about the asymptotics of $\log N$ vs. $N^\alpha,\alpha>0$, as $N\to\infty$. – Jyrki Lahtonen Feb 13 '19 at 07:47