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# exclusion of integer root

Theorem. 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, 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 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 right.

Related:

DivisibilityInRings

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Theorem

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## Mathematics Subject Classification

12D10*no label found*12D05

*no label found*

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