Brian Covello's paper on primitive pythagorean triples. Information below is taken from wikipedia.org A Pythagorean triple consists of three positive integers a, b, and c, such that a2 + b2 = c2. Such a triple is commonly written (a, b, c), and a well-known example is (3, 4, 5). If (a, b, c) is a Pythagorean triple, then so is (ka, kb, kc) for any positive integer k. A primitive Pythagorean triple is one in which a, b and c are coprime. A right triangle whose sides form a Pythagorean triple is called a Pythagorean triangle. The name is derived from the Pythagorean theorem, stating that every right triangle has side lengths satisfying the formula a2 + b2 = c2; thus, Pythagorean triples describe the three integer side lengths of a right triangle. However, right triangles with non-integer sides do not form Pythagorean triples. For instance, the triangle with sides a = b = 1 and c = √2 is right, but (1, 1, √2) is not a Pythagorean triple because √2 is not an integer. Moreover, 1 and √2 do not have an integer common multiple because √2 is irrational. (c − a)(c − b) / 2 is always a perfect square. This is particularly useful in checking if a given triple of numbers is a Pythagorean triple, but it is only a necessary condition, not a sufficient one. The triple {6, 12, 18} passes the test that (c − a)(c − b)/2 is a perfect square, but it is not a Pythagorean triple. When a triple of numbers a, b and c forms a primitive Pythagorean triple, then (c minus the even leg) and one-half of (c minus the odd leg) are both perfect squares; however this is not a sufficient condition, as the triple {1, 8, 9} is a counterexample since 12 + 82 ≠ 92. At most one of a, b, c is a square. (See Infinite descent#Non-solvability of r2 + s4 = t4 for a proof.) The area (K = ab/2) is an even congruent number. The area of a Pythagorean triangle cannot be the square[7]:p. 17 or twice the square[7]:p. 21 of a natural number. Exactly one of a, b is odd; c is odd.[8] Exactly one of a, b is divisible by 3.[9] Exactly one of a, b is divisible by 4.[9] Exactly one of a, b, c is divisible by 5.[9] The largest number that always divides abc is 60.[10] All prime factors of c are primes of the form 4n + 1. Every integer greater than 2 that is not congruent to 2 mod 4 (in other words, every integer greater than 2 which is not of the form 4n + 2) is part of a primitive Pythagorean triple. Every integer greater than 2 is part of a primitive or non-primitive Pythagorean triple. For example, the integers 6, 10, 14, and 18 are not part of primitive triples, but are part of the non-primitive triples 6, 8, 10; 14, 48, 50 and 18, 80, 82. There exist infinitely many Pythagorean triples in which the hypotenuse and the longer of the two legs differ by exactly one (such triples are necessarily primitive).

1. 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 common
factors 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 hypotenuse
of 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-primes
being 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 d
be 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)). This
implies 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
is
the 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 as
the sum of two squares as follows:
x = 4j + 1 = e2
+ f2
Base Case k=1
Since 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

2. So 4j + 1 must be the hypotenuse in the PPT (e2
− f2
, 2ef, x). So x is the hypotenuse of some PPT with
gcd(e, f) and e, f have opposite parity. Thus the result is true for k=1.
Inductive Step
Assume true for k=n so that xn
= g2
+ i2
where g and i satisfy the conditions for PPT and have opposite
parity. We must prove this is true for xn+1
.
xn
x1
= 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
), where
q is a common prime factor within the pairs. Thus q must be a factor of e and f or q must be a factor of
g2
+ 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, we
have 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 the
Pythagorean 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 gives
the 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 are
found in multiples of 2n
.
1 References
[1] Hardy, G H “An Introduction to the Theory of Number”, Oxford University Press, 1960
2