TITLE: Prove that $f(x)\in \mathbb{Z}[x]$ such that $f(0)$ and $f(1)$ are odd has no integer roots
QUESTION [21 upvotes]: Suppose $f(x) \in \mathbb{Z}[x]$ is such that $f(0)$ and $f(1)$ are odd. How do I show that $f(x)$ has no integer roots?

REPLY [4 votes]: A slight (but cute, in my opinion) variation of the the other methods is this one: 
If $f\in \Bbb Z[x]$ has an integer root $m$, then by Ruffini's rule $f(x)=(x-m)(b_n x^n+\cdots + b_0)=(x-m)g(x)$ for some $b_0,\cdots, b_m\in \Bbb Z$.
Then, $$\begin{cases} f(0)=-m\cdot g(0)\\ f(1)=(1-m)\cdot g(1)\end{cases}$$
But at least one between $-m$ and $(1-m)$ must be even.