# exclusion of integer root

The equation

 $p(x)\;:=\;a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots+a_{0}\;=\;0$

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

 $2\mid 0=p(x_{0})\quad\textrm{and}\quad 2\nmid p(0)=a_{0},$

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

 $2\mid a_{n}\!+\!a_{n-1}\!+\!\ldots\!+\!a_{1}\!+\!a_{0}\;=\;p(1).$

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

Title exclusion of integer root ExclusionOfIntegerRoot 2013-03-22 19:08:21 2013-03-22 19:08:21 pahio (2872) pahio (2872) 5 pahio (2872) Theorem msc 12D10 msc 12D05 DivisibilityInRings