prime factors of Pythagorean hypotenuses

The possible hypotenuses of thePythagorean triangles (http://planetmath.org/PythagoreanTriangle) form theinfinite sequence

$$5,\mathrm{\hspace{0.17em}10},\mathrm{\hspace{0.17em}13},\mathrm{\hspace{0.17em}15},\mathrm{\hspace{0.17em}17},\mathrm{\hspace{0.17em}20},\mathrm{\hspace{0.17em}25},\mathrm{\hspace{0.17em}26},\mathrm{\hspace{0.17em}29},\mathrm{\hspace{0.17em}30},\mathrm{\hspace{0.17em}34},\mathrm{\hspace{0.17em}35},\mathrm{\hspace{0.17em}37},\mathrm{\hspace{0.17em}39},\mathrm{\hspace{0.17em}40},\mathrm{\hspace{0.17em}41},\mathrm{\hspace{0.17em}45},\mathrm{\dots }$$

the mark of which ishttp://oeis.org/search?q=a009003&language=english&go=SearchA009003 in the corpus of the integersequences of http://oeis.org/OEIS.  This sequence has the subsequence A002144

$$5,\mathrm{\hspace{0.17em}13},\mathrm{\hspace{0.17em}17},\mathrm{\hspace{0.17em}29},\mathrm{\hspace{0.17em}37},\mathrm{\hspace{0.17em}41},\mathrm{\hspace{0.17em}53},\mathrm{\hspace{0.17em}61},\mathrm{\hspace{0.17em}73},\mathrm{\hspace{0.17em}89},\mathrm{\hspace{0.17em}97},\mathrm{\hspace{0.17em}101},\mathrm{\hspace{0.17em}109},\mathrm{\hspace{0.17em}113},\mathrm{\hspace{0.17em}137},\mathrm{\dots }$$

of the odd Pythagorean primes.

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

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

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

Theorem.  A positive integer $c$ is the length of thehypotenuse of a Pythagorean triangle if and only if at leastone of the prime factors of $c$ is of the form $4n+1$.

Lemma 1.  All prime factors of the hypotenuse $c$ ina primitive Pythagorean triple are of the form $4n+1$.

This can be proved here by making the antithesis that thereexists 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 Gaussianintegers, 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 thetriple $(a,b,c)$ were not primitive, whencethe antithesis is wrong and the lemma true.  $\mathrm{\square }$

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 ina Pythagorean triple.  (Especially, any prime $4n+1$ isfound as the hypotenuse in a primitive Pythagorean triple.)

Proof.  For proving this, one can start from Fermat’stheorem, by which the prime numbers of such form are sums oftwo squares (see thehttp://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 undermultiplication, now also the product $c$ is a sum of twosquares, and similarly is $c^2$, i.e. $c$ is the hypotenusein a Pythagorean triple.  $\mathrm{\square }$

Proof of the Theorem.  Suppose that $c$ is thehypotenuse of a Pythagorean triple $(a,b,c)$; dividing thetriple members by their greatest common factor we get aprimitive triple $(a^{\prime },b^{\prime },c^{\prime })$ where  $c^{\prime }\mid c$.  ByLemma 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 guaranteesa Pythagorean triple $(r,s,p)$, whence also $(rd,sd,c)$ isPythagorean and $c$ thus a hypotenuse.  $\mathrm{\square }$