Ordering the Primitive Pythagorean Triples by Pellian Sequences

When you combine Euclid's formula for generating triples with the recursion in Pell sequences, you get an ordering of the primitive Pythagorean triples. This is because that recursion, where you double your current term and add that to the prior term to get the next term, preserves the leg difference for triples. It is similar to Fibonacci recursion, but it requires that doubling before you add the terms.

https://rationalargumentator.com/issue232/Pellian.pdf

ORDERING THE PRIMITIVE PYTHAGOREAN TRIPLES

BY GENERALIZED PELLIAN SEQUENCES

While it is well-known that the Pell numbers: 0, 1, 2, 5, 12, 29, 70, . . . generate

the Pythagorean triples with consecutive legs, it seems to be a great secret that the rest of

the primitive triples can be generated, ordered and largely sorted by leg difference using

similar sequences. But before getting to them, let’s review. Recall that the primitive

Pythagorean triples are the sets {a, b, c} that each represent the lengths of the sides of a

a right triangle that are all whole numbers with gcd(a,b,c) = 1. Letting b be the even

leg, each of these triples can be represented uniquely by a pair of positive integers u and

v, where v > u, gcd(u,v) = 1, u and v are of opposite parity and {a, b, c} = {v^2 – u^2,

2uv, v^2 + u^2}. Now, taking the Pell numbers in consecutive pairs starting at 1, 2 and

substituting them for u and v so that u = 1, v = 2 ; then u = 2, v = 5 ; then u = 5, v =

12 , and so on, gives us the triples {3, 4, 5}; {21, 20, 29}; {119, 120, 169}, . . . Note that

the leg differences are 1 ( d = | b – a | = 1).

For primitive triples, the next greater leg difference happens to be 7. There are

actually two sequences, having the same recursion relation, P_n + 2 = 2P_n + 1 + P_n , as

the Pell numbers, that generate all of the primitive triples with d = 7 without

redundancy. And they are: 2, 3, 8, 19, 46, . . . and 1, 4, 9, 22, 53, . . . Substituting their

consecutive pairs of terms for u and v yields the desired triples:

{5, 12, 13}, {55, 48, 73}, {297, 304, 425}, {1755, 1748, 2477}, . . .

{15, 8, 17}, {65, 72, 97}, {403, 396, 565}, {2325, 2332, 3293}, . . .

Again, note that the leg differences are 7, and observe that we can represent the initial

two terms of the sequences as n, n + m and n – m, 3n – 2m, where n = 2 and m = 1.

In this case, d = | b – a | = 2n^2 – m^2 = 2(2)^2 – 1^2 = 7.

After d = 7, the next greater possible difference corresponds to when n = 3 and

m = 1. Thus, d = 2n^2 – m^2 = 2(3)^2 – 1^2 = 17. And the twin sequences that generate

the primitive triples with this leg difference are

n, n + m, . . . = 3, 4, 11, 26, 63, . . .

n – m, 3n – 2m, . . . = 2, 7, 16, 39, 94, . . .

But sometimes there are multiple permissible n and m values that give the same

difference. For example, n = 8, m = 3 and n = 10, m = 9 both correspond to

d = 2(8)^2 – 3^2 = 2(10)^2 – 9^2 = 119. Each pair of n and m values initiates twin

sequences. So it requires four infinite sequences to generate the primitive triples with

d = 119.

If we focus on the leg difference from a distance, everything falls into place.

Fixing the leg differences of primitive triples results in Pell-type equations of the form

2n^2 – m^2 = d. And the solutions to such equations are recursive sequences (generalized

Pellian sequences). It is easy to show that our recursion relation preserves leg difference

(i.e., all consecutive pairs of terms satisfy the Pell-type equation if the initial terms do).

PROOF: Observe that if P_n and P_n+1 are any two consecutive terms of any of our

sequences, including the Pell numbers themselves, the leg difference of the triple they

yield would be d = | (P_n+1)^2 – (P_n)^2 – 2P_n P_n+1 | .

By the recursion, P_n +2 = 2P_n+1 + P_n , and the leg difference of the triple

generated by P_n+1 and P_n+2 would be d = | (P_n+2)^2 – (P_n+1)^2 – 2P_n+1 P_n+2 |

= | (2P_n+1 + P_n)^2 – (P_n+1)^2 – 2P_n+1(2P_n+1 + P_n) |

= | 4(P_n+1)^2 + 4P_nP_n+1 + (P_n)^2 – (P_n+1)^2 – 4(P_n+1)^2 – 2P_nP_n+1 |

= | (P_n)^2 – (P_n+1)^2 + 2P_n P_n+1 |

= | (P_n+1)^2 – (P_n)^2 – 2P_n P_n+1 | .

QED

So, let me sum up the overall result and prove that it holds true.

THEOREM: All of the primitive Pythagorean triples can be generated, ordered and

largely sorted without redundancy by substitution into {v^2 – u^2, 2uv, v^2 + u^2 } of

consecutive pairs of terms of the Pell numbers and similar sequences formed by the same

recursion relation, P_n + 2 = 2P_n+1 + P_n, and initial values n, n + m and

n – m, 3n – 2m, where n and m are positive integers, n > m, gcd(m,n) = 1,

and m is odd. (The leg difference of triples generated from both sequences is

d = 2n^2 – m^2 . And the Pell numbers can be considered the special case of a singleton

sequence: n, n + m, . . . , where n = m = 1.)

Since I’ve shown that the recursion preserves leg differences and it is easy to

verify that the gcd(n,n + m) = gcd(n – m ,3n – 2m) = 1 and that the recursion

preserves this relative primality as well as the opposite parities, it remains for me to prove

that the generation of the primitive triples is exhaustive and without redundancy.

PROOF: Let {a, b, c} be an arbitrary primitive Pythagorean triple. There exists a pair of

positive integers, u and v, such that v > u, gcd(u,v) = 1, u and v are of opposite parity,

and {a, b, c} = {v^2 – u^2, 2uv, v^2 + u^2}.

Case 1: If v < 2u, then n = u, m = v – u . u, v are the initial terms n, n + m .

Case 2: If v = 2u, then u = 1, v = 2, the second and third Pell numbers.

Case 3: If v > 3u, then n = v – 2u, m = v – 3u . u, v are the initial terms n – m,

3n – 2m.

Case 4: If 2u < v < 3u, then u, v are advanced terms in one of the sequences and can

be determined by backtracking, via the reverse recursion relation P_n–2 = P_n – 2P_n–1

for as long as the terms remain positive and strictly decreasing, until one of the following

occurs.

Case 4a: If P_n–2 < 0, then P_n < 2P_n–1 and u, v are advanced terms of the sequence

n, n + m, . . . where n = P_n–1 , m = P_n – P_n–1 .

Case 4b: If P_n–2 = 0, then P_n = 2P_n–2 and u, v are advanced terms of the Pell

numbers.

Case 4c: If P_n–2 > P_n–1, then P_n > 3P_n–1 and u, v are advanced terms of the

sequence n – m , 3n – 2m, . . . where n = P_n – 2P_n–1 , m = P_n – 3P_n–1 .

This covers all possibilities and proves that u and v must be in some specific position in

one of the sequences.

QED

Keith Raskin

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