Q:

Number Theory: Proving polynomials with parity

Prove that if P(x) is a polynomial with integer coefficients such that P(0) and P(1) are both odd, then it has no integer roots.

A:

Well, $a\equiv b\mod 2$ implies $P(a)\equiv P(b)\mod 2$, and so if $P(0)$ and $P(1)$ are both odd, so is $P(n)$ for all other integers $n$.

