exclusion of integer root

Theorem.  The equation

$$p(x):=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{\dots }+a_0=\mathrm{\hspace{0.33em}0}$$

with integer coefficients $a_i$ has no integer roots (http://planetmath.org/Equation), if $p(0)$ and $p(1)$ are odd.

Proof.  Make the antithesis, that there is an integer $x_0$ such that  $p(x_0)=0$.  This $x_0$ cannot be even, because else all terms of $p(x_0)$ except $a_0$ were even and thus the whole sum could not have the even value 0.  Consequently, $x_0$ and also its powers (http://planetmath.org/GeneralAssociativity) have to be odd.  Since

$$2\mid 0=p(x_0)\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}2\nmid p(0)=a_0,$$

there must be among the coefficients $a_n,a_{n-1},\mathrm{\dots },a_1$ an odd amount of odd numbers.  This means that

$$2\mid a_n+a_{n-1}+\mathrm{\dots }+a_1+a_0=p(1).$$

This however contradicts the assumption on the parity of $p(1)$, whence the antithesis is wrong and the theorem .