The Hypotenuse of a Primitive Pythagorean Triple

B. Covello

March 25, 2013

Hypotenuse of PPT with form 4n+1

Definition 1. A primitive Pythagorean triple (PPT) is a triple of numbers (a, b, c) having no commonfactors and satisfying:

a2 + b2 = c2

Recall that we will get every primitive Pythagorean triple (a, b, c) with a odd and b even by using the formulas:

a = st, b =s2 − t2

2, c =

s2 + t2

2

Where s > t ≥ 1 are chosen to be any odd integers with no common factors.

Theorem 1. n is the hypotenuse of a PPT iff all of its prime factors are of the form 4k + 1.

Proof. If n has all prime factors of the form 4k+1, then Lemma 1 and Lemma 2 imply the n is the hypotenuseof a PPT. Hardy and Wright [1, p.13] state:

“If a and b have no common factor, then any odd prime divisor of a2 + b2 is of the form 4n+ 1”

Thus, if n is the hypotenuse of a PPT, then it may be expressed as the sum of co-primes, one of the co-primesbeing even, and the other odd.

Lemma 1. If m and n are hypotenuses of PPTs and gcd(m,n) = 1, then mn is also a hypotenuse of a PPT.

Proof. Let m = a2 + b2, n = c2 + d2 with gcd(a, b) = gcd(c, d) = 1. WLOG let a and c be even and b and dbe odd.

mn = (a2 + b2)(c2 + d2)

By Fibonacci’s sum of square identity...

(a2 + b2)(c2 + d2) = (ac− bd)2 + (bc+ ad)2

Note that these pairs are of opposite parity. Thus, we must show that these pairs have gcd=1.Suppose there exist some p such that p|(ac− bd) and p|(bc+ ad). So p|(c(ac− bd)) and p|(d(bc+ ad)). Thisimplies that p|(c(ac− bd) + d(bc+ ad))

c(ac− bd) + d(bc+ ad) = ac2 − cbd+ ad2 + cbd = a(c2 + d2)

p|(a(c2 + d2))

It is in this manner that one can show p|(b(c2 + d2)), p|(c(a2 + b2)), p|(d(a2 + 22)). Since gcd(a, b) =gcd(c, d) = 1, p|(m,n). Since gcd(m, 1) = 1, this implies that p = 1. Thus the pairs (ac− bd) and (bc+ ad)must be co-prime. Since these pairs are co-prime and have opposite parity, mn = (ac− bd)2 + (bc+ ad)2 isthe hypotenuse of another PPT.

Lemma 2. If p is a prime with the form 4j + 1, then pk is the hypotenuse of a PPT when k ≥ 1.

Proof. By induction on k.By Fermat’s Theorem “The Sum of Two Squares”, a prime number x in the form 4j + 1 may be written asthe sum of two squares as follows:

x = 4j + 1 = e2 + f2

Base Case k=1Since x is odd, e and f are of opposite parity. We must now show that gcd(e, f) = 1.Suppose not. Then ∃z such that gcd(e, f) = z. Then z2|x =⇒ z = 1.

1

So 4j + 1 must be the hypotenuse in the PPT (e2 − f2, 2ef, x). So x is the hypotenuse of some PPT withgcd(e, f) and e, f have opposite parity. Thus the result is true for k=1.Inductive StepAssume true for k=n so that xn = g2 + i2 where g and i satisfy the conditions for PPT and have oppositeparity. We must prove this is true for xn+1.

xnx1 = xn+1 = (e2 + f2)(g2 + i2)

By Fibonacci’s sum of squares identity... = (eg − fi)2 + (ei+ fg)2

Following the same reasoning in lemma 1, ∃q such that q|e(g2 + i2), f(g2 + i2), g(e2 + f2), i(e2 + f2), whereq is a common prime factor within the pairs. Thus q must be a factor of e and f or q must be a factor ofg2 + i2. Since gcd(e, f) = 1, and q does not divide e or f , q|(g2 + i2) =⇒ q|(xk) =⇒ q|x. Since x is prime,q = x. Additionally, since gcd(e, f) = 1 and gcd(g, i) = 1, xn+1 must be the hypotenuse of a PPT. Thus, wehave proven lemma 2 by induction.

Fibonacci’s Identity and the Relation to Gaussian Integers

Definition 2. The Gaussian integers are given by the ring such that:

Z[i] = {a+ bi : a, b ∈ Z}

For α = a+ bi, the conjugate of α is α = a− bi, and the norm is defined as:

N(α) = |α|2 = αα = a2 + b2

Note that c = |α| denotes the length of the vector α in the complex plane. Thus N(α) = c2. Hence thePythagorean theorem may be noted:

αα = ab + b2 = c2 = N(α)

Theorem 2. Fibonacci’s Identity

(a2 + b2)(c2 + d2) = (ac− bd)2 + (ad+ bc)2

= (ac+ bd)2 + (ad− bc)2

Proof. We begin with two multiplying together Gaussian integers:

|(a+ bi)||c+ di| = |(a+ bi)(c+ di)||(a+ bi)||(c+ di)| = |(ac− bd) + i(ad+ bc)|

(a+ bi)|2|(c+ di)|2 = |(ac− bd) + i(ad+ bc)|2

(a2 + b2)(c2 + d2) = (ac− bd)2 + (ad+ bc)2

The proof of Fibonacci’s Identity therefore, lies in the multiplication of two Gaussian integers.

Conjecture 1. In lemma 1 we found that multiplying consecutive terms of c having the form 4n + 1 givesthe next c in the pythagorean triple (a, b, c). For example: 5 ∗ 1 = 5; 13 ∗ 5 ∗ 1 = 65; 17 ∗ 13 ∗ 5 ∗ 1 = 1105,where 5, 65, 1105 represent a given c. For c = 5 there is 1 triple (3, 4, 5), for c = 65, there are two triples(33, 56, 65), (16, 63, 65). For c = 1105 there are three triples, (47, 1104, 1105), (817, 744, 1105), (1073, 264, 1105).

I conjecture that the number of distinct primitive pythagorean triples for any given c of the form 4n+ 1 arefound in multiples of 2n.

1 References

[1] Hardy, G H “An Introduction to the Theory of Number”, Oxford University Press, 1960

2