prime factors of Pythagorean hypotenuses
of the odd Pythagorean primes.
of some two different integers and (as has been shown in the parent entry), but also by the
Lemma 1. All prime factors of the hypotenuse in a primitive Pythagorean triple are of the form .
This can be proved here by making the antithesis that there exists a prime dividing . Then also
where and are the catheti in the triple. But
is prime also in the ring of the Gaussian
integers, whence it must divide at least one of the factors
and . Apparently, that would imply that
divides both and . This means that the
triple were not primitive, whence
the antithesis is wrong and the lemma true.
Also the converse is true in the following form:
Lemma 2. If all prime factors of a positive integer are of the form , then is the hypotenuse in a Pythagorean triple. (Especially, any prime 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 is a sum of two
squares, and similarly is , i.e. is the hypotenuse
in a Pythagorean triple.
Proof of the Theorem. Suppose that is the hypotenuse of a Pythagorean triple ; dividing the triple members by their greatest common factor we get a primitive triple where . By Lemma 1, the prime factors of , being also prime factors of , are of the form .
On the contrary, let’s suppose that a prime factor of
is of the form . Then Lemma 2 guarantees
a Pythagorean triple , whence also is
Pythagorean and thus a hypotenuse.
|Title||prime factors of Pythagorean hypotenuses|
|Date of creation||2014-01-31 10:48:20|
|Last modified on||2014-01-31 10:48:20|
|Last modified by||pahio (2872)|