Sorry

My life was so busy for the past two days I didn't even  have time to post things on here. Luckily, Thanksgiving break has started so I have plenty of time to learn things. Currently, I'm reading a number theory book that I randomly picked out at my library, it's by Nagell. It turns out that this book has a lot of good exercises and I've decided to post some of the interesting ones on here. The following ones come from the chapter on congruences.

53. Let $f(x)$ be an integral polynomial in $x$, and let $n,m$ be natural numbers. Show that there exists an integer $x_0$ such that each of the numbers \[ f(x_0),f(x_0+1),...,f(x_0+n-1)\] has at least $m$ different prime divisors.

Lemma: For an integer polynomial non-constant $f(x)$, show that there is an infinite number of primes $p$ that can divide $f(x)$.

Proof: Assume the contrary that there's only a finite number of primes denoted by $p_1,p_2,...,p_k$ that can divide $f(x)$. Let $e_i$ denote the number satisfying $p^{e_i-1}||f(0)$, then $p^{e_i-1}||f(mp^{e_i})$ for $m\in \mathbb {Z}$. Consider the number function $g(n)=n\prod_{i=1}^{k}p^{e_i}$ where $n\in \mathbb {Z}$, thus $\prod_{i=1}^{k}p^{e_i-1}||f(g(n))$. Since there clearly exists $n$ such that $f(g(n))>\prod_{i=1}^{k}p^{e_i-1}$, this must mean that there exists a prime not one of $p_1,p_2,...,p_k$ dividing $f(g(n))$, which contradicts our original assumption.

Main Proof:  By the Lemma above, one can select $n$ groups of $m$ primes that can divide $f(x)$: $S_1,S_2,...,S_n$ such that $S_i\cap S_j=\emptyset$ for $1\le i<j\le n$. By Chinese remainder theorem, there exist $r_i$ such that $\prod_{p\in S_i}p|f(r_i)$. In addition by Bezout's Lemma, one can find $n$ consecutive integers $x_0,x_0+1,...,x_0+n-1$ such that $x_0+i-1\equiv r_i \pmod {\prod_{p\in S_i}}$ for $i=1,2,...,n$. Hence $\prod_{p\in S_i}|f(x_0+i-1)$ which means each $f(x_0+i-1)$ is divisible by $m$. $\Box$.