The Infinitude of Primitive Pythagorean Triples with Composite Numbers

What is the cardinality of primitive Pythagorean triples that consist of three composite numbers? Despite the numbers being co-prime, such triples do exist. This article delves into the existence and density of these unique triples, redefining the boundaries of traditional Pythagorean triples through mathematical proof.

Introduction to Primitive Pythagorean Triples

The first Pythagorean triple is the well-known (3, 4, 5), where a, b, and c are co-prime. When a, b, and c are co-prime, the Pythagorean triple is termed primitive. There are only 16 primitive Pythagorean triples with c 100:

(3, 4, 5) (5, 12, 13) (8, 15, 17) (7, 24, 25) (20, 21, 29) (12, 35, 37) (9, 40, 41) (28, 45, 53) (11, 60, 61) (16, 63, 65) (33, 56, 65) (48, 55, 73) (13, 84, 85) (36, 77, 85) (39, 80, 89) (65, 72, 97)

Among these, only two triples (33, 56, 65) and (36, 77, 85) have all three numbers composite. This article explores the existence of such triples and the density of such structures.

Existence of Composite Pythagorean Triples

Diophantus' formula for generating all Pythagorean triples is well-known:

[a p^2 - q^2]

[b 2pq]

[c p^2 q^2]

with (p eq q) and (p) and (q) having different parity. This formula reveals that for any natural numbers (p) and (q), the resulting triple can be composite. In fact, we see that:

[(p^2 - q^2)^2 (2pq)^2 (p^2 q^2)^2]

When we introduce a natural number (k 1) such that:

[a k(p^2 - q^2) kp^2 - kq^2]

[b 2kpq]

[c kp^2 q^2]

it becomes evident that there are infinite Pythagorean triples where (a, b,) and (c) are composite. This construction proves the existence of such triples.

Primitive Pythagorean Triples with All Composite Numbers

To prove that there are infinite primitive Pythagorean triples where all three numbers are composite, we start by considering the properties of (b 2pq). Since (b) is always even and greater than 2, (b) is always composite. Next, we analyze (a p^2 - q^2) and (c p^2 q^2).

If (p) and (q) are chosen such that (p q), we have:

(a p^2 - q^2 (p - q)(p q))

Since (p q), ((p - q) 1), and thus (a) cannot be prime. Therefore, (a) is always composite.

Now, let's set (q 2). Since there are infinitely many primes (p equiv 1 mod 5), we can choose such a prime. Then, (c p^2 4) is also composite, as (p^2 4 equiv 0 mod 5).

Finally, if (a c d 1), then:

(d 2) would imply (2) divides both odd numbers (a) and (c), which is impossible.

(d p) means (p p^2 - 4) and (p^2 - 4 p^2 - 4), which simplifies to (p 2), a contradiction since (p) is an odd prime.

Thus, we conclude that there are infinitely many primitive Pythagorean triples where all three numbers are composite.

Implications and Further Reading

The Brahmagupta–Fibonacci identity, which asserts that the ‘hypotenuse’ can be composite, further supports this conclusion. The identity states:

[left((a^2 b^2)(c^2 d^2)right) (ac - bd)^2 (ad bc)^2]

which proves that the product of two sums of squares is also a sum of squares, thus reinforcing the complexity and abundance of composite triples.

For a deeper dive into the subject, one can refer to the works of Diophantus and further explore the properties of Pythagorean triples using number theory principles.