# prime factors of Pythagorean hypotenuses

The possible hypotenuses of the Pythagorean triangles (http://planetmath.org/PythagoreanTriangle) 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 http://oeis.org/search?q=a009003&language=english&go=SearchA009003 in the corpus of the integer sequences of http://oeis.org/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

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 http://en.wikipedia.org/wiki/Proofs_of_Fermat's_theorem_on_sums_of_two_squaresTheorem 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.  $\Box$

Title prime factors of Pythagorean hypotenuses PrimeFactorsOfPythagoreanHypotenuses 2014-01-31 10:48:20 2014-01-31 10:48:20 pahio (2872) pahio (2872) 12 pahio (2872) Theorem msc 11D09 msc 51M05 msc 11E25