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The possible hypotenuses of the Pythagorean triangles form the infinite sequence

$5,\,10,\,13,\,15,\,17,\,20,\,25,\,26,\,29,\,30,\,34,\,35,\,37,\,39,\,40,\,41,% \,45,\,\ldots$ |

the mark of which is A009003 in the corpus of the integer sequences of OEIS. This sequence has the subsequence A002144

$5,\,13,\,17,\,29,\,37,\,41,\,53,\,61,\,73,\,89,\,97,\,101,\,109,\,113,\,137,\,\ldots$ |

of the odd Pythagorean primes.

Generally, the hypotenuse $c$ of a Pythagorean triangle (Pythagorean triple) may be characterised by being the contraharmonic mean

$c\;=\;\frac{u^{2}\!+\!v^{2}}{u\!+\!v}$ |

of some two different integers $u$ and $v$ (as has been shown in the parent entry), but also by the

Theorem. A positive integer $c$ is the length of the
hypotenuse of a Pythagorean triangle if and only if at least
one of the prime factors of $c$ is of the form $4n\!+\!1$.

Lemma 1. All prime factors of the hypotenuse $c$ in a primitive Pythagorean triple are of the form $4n\!+\!1$.

This can be proved here by making the antithesis that there exists a prime $4n\!-\!1$ dividing $c$. Then also

$4n\!-\!1\mid c^{2}\;=\;a^{2}\!+\!b^{2}\;=\;(a\!+\!ib)(a\!-\!ib)$ |

where $a$ and $b$ are the catheti in the triple. But $4n\!-\!1$
is prime also in the ring $\mathbb{Z}[i]$ of the Gaussian
integers, whence it must divide at least one of the factors
$a\!+\!ib$ and $a\!-\!ib$. Apparently, that would imply that
$4n\!-\!1$ divides both $a$ and $b$. This means that the
triple $(a,b,c)$ were not primitive, whence
the antithesis is wrong and the lemma true. $\Box$

Also the converse is true in the following form:

Lemma 2. If all prime factors of a positive integer $c$ are of the form $4n\!+\!1$, then $c$ is the hypotenuse in a Pythagorean triple. (Especially, any prime $4n\!+\!1$ is found as the hypotenuse in a primitive Pythagorean triple.)

Proof. For proving this, one can start from Fermat’s
theorem, by which the prime numbers of such form are sums of
two squares (see the
Theorem on sums of two squares by Fermat).
Since the sums of two squares form a set closed under
multiplication^{}, now also the product^{} $c$ is a sum of two
squares, and similarly is $c^{2}$, i.e. $c$ is the hypotenuse
in a Pythagorean triple. $\Box$

Proof of the Theorem. Suppose that $c$ is the hypotenuse of a Pythagorean triple $(a,b,c)$; dividing the triple members by their greatest common factor we get a primitive triple $(a^{{\prime}}\!,b^{{\prime}}\!,c^{{\prime}})$ where $c^{{\prime}}\mid\;c$. By Lemma 1, the prime factors of $c^{{\prime}}$, being also prime factors of $c$, are of the form $4n\!+\!1$.

On the contrary, let’s suppose that a prime factor $p$ of
$c=pd$ is of the form $4n\!+\!1$. Then Lemma 2 guarantees
a Pythagorean triple $(r,s,p)$, whence also $(rd,sd,c)$ is
Pythagorean and $c$ thus a hypotenuse. $\Box$

## Mathematics Subject Classification

11D09*no label found*51M05

*no label found*11E25

*no label found*

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