TITLE: Prove that $m^{2013}-m^{20}+m^{13}-2013$ has at least $N$ prime divisors
QUESTION [13 upvotes]: for positive integer $N>1$,There always exists $m$ such that
$$m^{2013}-m^{20}+m^{13}-2013$$
has at least $N$ prime divisors
Thank you all, this is good problem, but I don't know how to solve it.

REPLY [3 votes]: I guess it works for any non-constant polynomial with integer coefficients.
Let $f(x)=\sum_{i=0}^na_ix^i$, $n\ge 1$, with $a_n\ne 0$ and all $a_i$ integers. We show that $\{p \textrm{ prime: there is an integer } m \textrm{ such that } p \textrm{ divides } f(m)\}$ is an infinite set. 
If $a_0=0$ the assert is clear. Let's assume $a_0\ne 0$.
Suppose the set is finite and let $p_1,...,p_k$ be all of them. Let $g(x)=f(a_0x)/a_0$. Then $g(x)$ is a non-constant polynomial with integer coefficients with constant term 1. Let $m_c=c\cdot p_1\cdots p_k$ where $c$ is any integer. Since $g(m_c)$ has no prime factor other than $p_1$, ..., $p_k$ while $g(m_c)$ is congruent to 1 mod every $p_i$, it follows that $g(m_c)=1$ for every $c$. But $g(x)=1$ has at most $n$ roots, a contradiction.
To answer the original question, use Chinese Remainder Theorem as Hagen von Eitzen did above.