# theorem on sums of two squares by Fermat

Suppose that an odd prime number $p$ can be written as the sum

$$p={a}^{2}+{b}^{2}$$ |

where $a$ and $b$ are integers. Then they have to be coprime^{}.
We will show that $p$ is of the form $4n+1$.

Since $p\nmid b$, the congruence^{}

$$b{b}_{1}\equiv \mathrm{\hspace{0.33em}1}\phantom{\rule{veryverythickmathspace}{0ex}}(modp)$$ |

has a solution ${b}_{1}$, whence

$$0\equiv p{b}_{1}^{2}={(a{b}_{1})}^{2}+{(b{b}_{1})}^{2}\equiv {(a{b}_{1})}^{2}+1\phantom{\rule{veryverythickmathspace}{0ex}}(modp),$$ |

and thus

$${(a{b}_{1})}^{2}\equiv -1\phantom{\rule{veryverythickmathspace}{0ex}}(modp).$$ |

Consequently, the Legendre symbol^{} $\left(\frac{-1}{p}\right)$ is $+1$, i.e.

$${(-1)}^{\frac{p-1}{2}}=\mathrm{\hspace{0.33em}1}.$$ |

Therefore, we must have

$p=\mathrm{\hspace{0.33em}4}n+1$ | (1) |

where $n$ is a positive integer.

Euler has first proved the following theorem presented by
Fermat and containing also the converse^{} of the above claim.

Theorem
(Thue’s lemma (http://planetmath.org/ThuesLemma)). An odd prime $p$ is
uniquely expressible as sum of two squares of integers if and
only if it satisfies (1) with an integer value of $n$.

The theorem implies easily the

Corollary. If all odd prime factors of a positive
integer are congruent to 1 modulo 4 then the integer is a sum
of two squares. (Cf. the proof of the parent article and the article
“prime factors^{} of Pythagorean hypotenuses (http://planetmath.org/primefactorsofpythagoreanhypotenuses)”.)

Title | theorem on sums of two squares by Fermat |
---|---|

Canonical name | TheoremOnSumsOfTwoSquaresByFermat |

Date of creation | 2014-10-25 17:44:02 |

Last modified on | 2014-10-25 17:44:02 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 13 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 11A05 |

Classification | msc 11A41 |

Classification | msc 11A67 |

Classification | msc 11E25 |