# representants of quadratic residues

Theorem. Let $p$ be a positive odd prime number. Then the integers

${1}^{2},{\mathrm{\hspace{0.17em}2}}^{2},\mathrm{\dots},{\left({\displaystyle \frac{p-1}{2}}\right)}^{2}$ | (1) |

constitute a representant system of incongruent quadratic residues^{} modulo $p$. Accordingly, there are $\frac{p-1}{2}$ quadratic residues and equally many nonresidues modulo $p$.

*Proof.* Firstly, the numbers (1), being squares, are quadratic residues modulo $p$. Secondly, they are incongruent, because a congruence^{} ${a}^{2}\equiv {b}^{2}\phantom{\rule{veryverythickmathspace}{0ex}}(modp)$ would imply

$$p\mid a+b\mathit{\hspace{1em}}\text{or}\mathit{\hspace{1em}}p\mid a-b,$$ |

which is impossible when $a$ and $b$ are different integers among $1,\mathrm{\hspace{0.17em}2},\mathrm{\dots},\frac{p-1}{2}$. Third, if $c$ is any quadratic residue modulo $p$, and therefore the congruence ${x}^{2}\equiv c\phantom{\rule{veryverythickmathspace}{0ex}}(modp)$ has a solution $x$, then $x$ is congruent with one of the numbers

$$\pm 1,\pm 2,\mathrm{\dots},\pm \frac{p-1}{2}$$ |

which form a reduced residue system^{} modulo $p$ (see absolutely least remainders). Then ${x}^{2}$ and $c$ are congruent with one of the numbers (1).

Title | representants of quadratic residues |
---|---|

Canonical name | RepresentantsOfQuadraticResidues |

Date of creation | 2013-03-22 19:00:35 |

Last modified on | 2013-03-22 19:00:35 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 7 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 11A15 |

Synonym | representant system of quadratic residues |

Related topic | GaussianSum |

Related topic | DifferenceOfSquares |

Related topic | DivisibilityByPrimeNumber |