prime factors of Pythagorean hypotenuses


The possible hypotenusesMathworldPlanetmath of the Pythagorean trianglesMathworldPlanetmath (http://planetmath.org/PythagoreanTriangle) form the infiniteMathworldPlanetmath sequenceMathworldPlanetmath

5, 10, 13, 15, 17, 20, 25, 26, 29, 30, 34, 35, 37, 39, 40, 41, 45,

the mark of which is http://oeis.org/search?q=a009003&languagePlanetmathPlanetmath=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,

of the odd Pythagorean primes.

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

c=u2+v2u+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 factorsMathworldPlanetmath 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-1c2=a2+b2=(a+ib)(a-ib)

where a and b are the catheti in the triple.  But 4n-1 is prime also in the ring [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. 

Also the converseMathworldPlanetmath 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 numbersMathworldPlanetmath 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 multiplicationPlanetmathPlanetmath, now also the product c is a sum of two squares, and similarly is c2, i.e. c is the hypotenuse in a Pythagorean triple. 

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,b,c) where  cc.  By Lemma 1, the prime factors of c, 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. 

Title prime factors of Pythagorean hypotenuses
Canonical name PrimeFactorsOfPythagoreanHypotenuses
Date of creation 2014-01-31 10:48:20
Last modified on 2014-01-31 10:48:20
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 12
Author pahio (2872)
Entry type Theorem
Classification msc 11D09
Classification msc 51M05
Classification msc 11E25