# 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 .

Title | exclusion of integer root |
---|---|

Canonical name | ExclusionOfIntegerRoot |

Date of creation | 2013-03-22 19:08:21 |

Last modified on | 2013-03-22 19:08:21 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 5 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 12D10 |

Classification | msc 12D05 |

Related topic | DivisibilityInRings |