Subfield: Number Theory


Question: Consider a polynomial $P(x)=\left(x+d_{1}\right)\left(x+d_{2}\right) \cdot \ldots \cdot\left(x+d_{9}\right)$, where $d_{1}, d_{2}, \ldots, d_{9}$ are nine distinct integers. Prove that there exists an integer $N$ such that for all integers $x \geq N$ the number $P(x)$ is divisible by a prime number greater than 20 .

Solution:
Note that the statement of the problem is invariant under translations of $x$; hence without loss of generality we may suppose that the numbers $d_{1}, d_{2}, \ldots, d_{9}$ are positive.

The key observation is that there are only eight primes below 20, while $P(x)$ involves more than eight factors.

We shall prove that $N=d^{8}$ satisfies the desired property, where $d=\max \left\{d_{1}, d_{2}, \ldots, d_{9}\right\}$. Suppose for the sake of contradiction that there is some integer $x \geq N$ such that $P(x)$ is composed of primes below 20 only. Then for every index $i \in\{1,2, \ldots, 9\}$ the number $x+d_{i}$ can be expressed as product of powers of the first 8 primes.

Since $x+d_{i}>x \geq d^{8}$ there is some prime power $f_{i}>d$ that divides $x+d_{i}$. Invoking the pigeonhole principle we see that there are two distinct indices $i$ and $j$ such that $f_{i}$ and $f_{j}$ are powers of the same prime number. For reasons of symmetry, we may suppose that $f_{i} \leq f_{j}$. Now both of the numbers $x+d_{i}$ and $x+d_{j}$ are divisible by $f_{i}$ and hence so is their difference $d_{i}-d_{j}$. But as

$$
0<\left|d_{i}-d_{j}\right| \leq \max \left(d_{i}, d_{j}\right) \leq d<f_{i}
$$

this is impossible. Thereby the problem is solved.
Observe that for each index $i \in\{1,2, \ldots, 9\}$ the product

$$
D_{i}=\prod_{1 \leq j \leq 9, j \neq i}\left|d_{i}-d_{j}\right|
$$

is positive. We claim that $N=\max \left\{D_{1}-d_{1}, D_{2}-d_{2}, \ldots, D_{9}-d_{9}\right\}+1$ satisfies the statement of the problem. Suppose there exists an integer $x \geq N$ such that all primes dividing $P(x)$ are smaller than 20. For each index $i$ we reduce the fraction $\left(x+d_{i}\right) / D_{i}$ to lowest terms. Since $x+d_{i}>D_{i}$ the numerator of the fraction we thereby get cannot be 1 , and hence it has to be divisible by some prime number $p_{i}<20$.

By the pigeonhole principle, there are a prime number $p$ and two distinct indices $i$ and $j$ such that $p_{i}=p_{j}=p$. Let $p^{\alpha_{i}}$ and $p^{\alpha_{j}}$ be the greatest powers of $p$ dividing $x+d_{i}$ and $x+d_{j}$, respectively. Due to symmetry we may suppose $\alpha_{i} \leq \alpha_{j}$. But now $p^{\alpha_{i}}$ divides $d_{i}-d_{j}$ and hence also $D_{i}$, which means that all occurrences of $p$ in the numerator of the fraction $\left(x+d_{i}\right) / D_{i}$ cancel out, contrary to the choice of $p=p_{i}$. This contradiction proves our claim.
Given a nonzero integer $N$ as well as a prime number $p$ we write $v_{p}(N)$ for the exponent with which $p$ occurs in the prime factorization of $|N|$.

Evidently, if the statement of the problem were not true, then there would exist an infinite sequence $\left(x_{n}\right)$ of positive integers tending to infinity such that for each $n \in \mathbb{Z}_{+}$the integer $P\left(x_{n}\right)$ is not divisible by any prime number $>20$. Observe that the numbers $-d_{1},-d_{2}, \ldots,-d_{9}$ do not appear in this sequence.

Now clearly there exists a prime $p_{1}<20$ for which the sequence $v_{p_{1}}\left(x_{n}+d_{1}\right)$ is not bounded; thinning out the sequence $\left(x_{n}\right)$ if necessary we may even suppose that

$$
v_{p_{1}}\left(x_{n}+d_{1}\right) \longrightarrow \infty
$$

Repeating this argument eight more times we may similarly choose primes $p_{2}, \ldots, p_{9}<20$ and suppose that our sequence $\left(x_{n}\right)$ has been thinned out to such an extent that $v_{p_{i}}\left(x_{n}+d_{i}\right) \longrightarrow \infty$ holds for $i=2, \ldots, 9$ as well. In view of the pigeonhole principle, there are distinct indices $i$ and $j$ as well as a prime $p<20$ such that $p_{i}=p_{j}=p$. Setting $k=v_{p}\left(d_{i}-d_{j}\right)$ there now has to be some $n$ for which both $v_{p}\left(x_{n}+d_{i}\right)$ and $v_{p}\left(x_{n}+d_{j}\right)$ are greater than $k$. But now the numbers $x_{n}+d_{i}$ and $x_{n}+d_{j}$ are divisible by $p^{k+1}$ whilst their difference $d_{i}-d_{j}$ is not - a contradiction.