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Let $f(x)\in \mathbb{Z}[x]$ be a non constant polynomial with integer coefficients. Show that as $a$ varies over the integers, the set of divisors of $f(a)$ includes infinitely many primes.

To be frank, I have no idea where to start...

Trivial case is when constant term of $f(x)$ is zero.

In case of $f(x)=x(a_nx^n+\cdots+a_1)$ we have $p$ divides $f(p)$ for all primes $p$...

Other than this i have no idea...

Please give only hints..

user26857
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  • try supposing that the number of prime divisors were finite and try to extract a contradiction – MT_ Mar 04 '16 at 01:26
  • Hint: estimate the number of values of $f(x)$ that lie in the range ${1,2,\dots,N}$. Then, given any finite set $S$ of primes, estimate the number of integers in the range ${1,2,\dots,N}$ all of whose prime factors come from $S$. – Greg Martin Mar 04 '16 at 01:27
  • @Soke : I have done that as well,, Could not succeed.. –  Mar 04 '16 at 01:29
  • @GregMartin : Estimation? I have no idea how to do that.. Can you suggest some reference where this has been discussed... I would read that and come back.. –  Mar 04 '16 at 01:31
  • @Greg This sounds like a fun approach. So if $S={p_1, …, p_m}$ is a finite set of primes, the number of integers in the range ${1, 2, …, N}$ all of whose prime factors come from $S$ would be at most $(\log_{2}N+1)^{m}$. But perhaps this is too weak of a bound. More interestingly, how do you bound the number of values of $f(x)$ that lie in the range ${1, 2, …, N}$? Intuitively, this should be a big proportion when $N$ is large with respect to the degree of $f$, but I am not sure how to make it precise. – Prism Mar 04 '16 at 02:02
  • Yes, the number of values of $f(x)$ in the range ${1,\dots,N}$ (assuming the leading coefficient $a_d$ is positive) is asymptotic to $(N/a_d)^{1/d}$: when $x$ is large, the difference between $f(x)$ and $a_d x^d$ is negligible. – Greg Martin Mar 04 '16 at 05:09
  • @Greg Thanks, got it! That is a pretty slick proof. – Prism Mar 04 '16 at 05:53

2 Answers2

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We will show that for any $H$, however huge, there is a prime $p\gt H$ that divides some $f(a)$.

As you observed, we can assume that $f(x)$ has non-zero constant term $a_0$. Consider $f(a_0x)=a_0(1+xg(x))$.

The equations $1+xg(x)=1$ and $1+xg(x)=-1$ have only finitely many solutions. Let $N\ge H$ be chosen so that $N!$ is greater than any of these solutions. Then $1+N!g(N!)$ cannot be $1$ or $-1$, so is divisible by some prime $p$.

If $p\le N$, then $p$ divides $N!$ so $p$ cannot divide $1+N!g(N!)$. Thus $p\gt N\ge H$. Since $p$ divides $1+N!g(N!)$ it follows that $f(a_0N!)\equiv 0\pmod{p}$.

André Nicolas
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  • Can you give some motivation how did you come up with some thing like this?? I can read this and understand but the i will forget soon.. –  Mar 04 '16 at 01:56
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    I have a collection of problems I assigned to students, or solved in class, when teaching an elementary number theory course. This collection, and the proofs, got refined over the years. As you can see, this solution is a very close relative of Euclid's proof of the infinitude of primes. – André Nicolas Mar 04 '16 at 02:03
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Let $f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0\in \mathbb Z[x]$.

If $a_0=0$ it is evident that $p$ divides $f(p)$ for arbitrary primes so we make $a_0\ne 0$. Assume that there is only a finite number of prime divisors $p_1,p_2,\dots,p_N$ for $ f (k);\space \forall\, k\in \mathbb Z $, and make the product $P=p_1p_2\cdots p_N$. We have $$f(a_0P)=a_0\left(a_na_0^{n-1}P^n+a_{n-1}a_0^{n-2}P^{n-1}+\dots+ \space a_2a_0P^2+a_1P+1\right).$$ It is clear that no prime divisor of the factor $$ a_na_0^{n-1}P^n+a_{n-1}a_0^{n-2}P^{n-1}+\dots+ \space a_2a_0P^2+a_1P+1$$ can be one of the $p_1,p_2, \dots,p_N$. This is a contradiction.

user26857
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