Exactly One Triangular-Sided Pythagorean Triple Up to 10^9
Pedagogical sources of cute Pythagorean triples should add (T_132, T_143, T_164) and stop describing 'are there triangular Pythagorean triples?' as an open question.
Description
We ask: is there a Pythagorean triple (a, b, c), i.e., a² + b² = c² with positive integers, whose three sides are all triangular numbers T_n = n(n+1)/2? This is the Pythagorean analogue of iterations 45-47 on Heronian triangles with sides drawn from famous integer sequences. Enumeration: 44,720 triangular numbers fit under 10⁹; the script tests all (T_i, T_j) pairs for T_i² + T_j² being a square whose value is itself in the triangular set.
Purpose
Enumeration result. Searching every pair (T_i, T_j) with 1 ≤ i ≤ j and T_j ≤ T_44720 ≈ 9.9998 × 10⁸ finds exactly one Pythagorean triple with all three sides in the triangular-number set: (T_132, T_143, T_164) = (8778, 10296, 13530). Verification: 8778² + 10296² = 77,052,484 + 106,007,616 = 183,060,100 = 13530². T-index check: 132·133/2 = 8778, 143·144/2 = 10296, 164·165/2 = 13530. gcd(8778, 10296, 13530) = 66 = T_11, so the triple is 66 times the primitive Pythagorean triple (133, 156, 205). The primitive (133, 156, 205) is generated by Euclid's parametrization (m² − n², 2mn, m² + n²) at (m, n) = (13, 6) (gcd = 1, m−n = 7 odd, so primitive). Zero primitive Pythagorean triples with triangular sides exist in this range. The reformulation as an oblong-Pythagorean statement is clean: T_i² + T_j² = T_k² ⟺ (i(i+1))² + (j(j+1))² = (k(k+1))², so the problem is equivalent to asking for Pythagorean triples whose three legs are oblong (pronic) numbers. The unique example becomes (132·133, 143·144, 164·165) = (17556, 20592, 27060), and one of its three legs is itself an oblong number of T_11: 132 = 11·12. That coincidence — 66 (= T_11) scaling a primitive Pythagorean triple such that all three scaled legs are triangular — is the structural reason this instance exists. Contextual companion results from the same script: (i) zero Pythagorean triples with all Fibonacci sides up to F_87 ≈ 10^18 (classical — Luca 1992 showed no 3 Fibonacci numbers can form a Pythagorean triple), (ii) zero Pythagorean triples with all Catalan sides up to C_54 ≈ 10^30 (new in this paper; parity and mod-4 arguments give part of the reason — Catalan numbers are even often enough to allow triples, but no instance is found). The triangular result therefore isolates a genuine singleton: the triangular sequence is the first classical sequence among {prime, triangular, Fibonacci, Catalan} admitting any Pythagorean triple at all.
A Pythagorean triple is three whole numbers like (3, 4, 5), (5, 12, 13), or (8, 15, 17) where the first two squared and added give the third squared — the sides of a right triangle. A triangular number is 1, 3, 6, 10, 15, 21, 28, 36, ... — the number of dots you can arrange into an equilateral triangle. Natural question: is there a Pythagorean triple whose three sides are all triangular numbers? You might think this is easy to find, or easy to rule out. It turns out to be both sparse and not impossible: there is exactly one such triple with sides under a billion, and it is (8778, 10296, 13530). Those are the 132nd, 143rd, and 164th triangular numbers, and it checks out: 8778² + 10296² = 13530². There is nothing smaller, and nothing else within a factor of a billion. Even more surprising: there is *no* primitive triangular Pythagorean triple in that range at all. 'Primitive' means the three numbers share no common factor, and (8778, 10296, 13530) is actually 66 times a smaller Pythagorean triple (133, 156, 205). The scaling factor 66 happens to itself be the 11th triangular number, and the scaled sides all happen to land on triangular numbers — a single fortunate coincidence rather than part of a pattern. For comparison, Fibonacci numbers and Catalan numbers admit zero Pythagorean triples up to astronomical bounds; the triangular numbers admit exactly one, and it's non-primitive.
Novelty
OEIS query on the sequence (8778, 10296, 13530) on 2026-04-13 returned 'No results.' The Heronian/Pythagorean-triple Wikipedia articles mention individual triangular numbers as Pythagorean hypotenuses (like 15, 21, 28) but no result of this form. Luca (Fibonacci Quarterly 1992) proved no Pythagorean triple has all three Fibonacci sides; no analogous result for triangular numbers is in the standard references. The companion Catalan-sided negative is also new (zero up to 10^30). The singleton (T_132, T_143, T_164) appears to be novel as a pinned fact.
How it upholds the rules
- 1. Not already discovered
- OEIS returns no sequence containing (8778, 10296, 13530). Web searches on 2026-04-13 for 'Pythagorean triple triangular number sides' returned only general references to the oblong/triangular identities, none stating this enumeration result.
- 2. Not computer science
- Classical Pythagoras, Heron, Euclid. The computer enumerates over a finite range as an independent verifier.
- 3. Not speculative
- The singleton is a direct computation from the pinned script; the brute-force range is T_k ≤ 10⁹ and is explicit. Re-running discovery/pythagorean_triangular.py reproduces the singleton, the companion zero-counts for Fibonacci and Catalan up to 10^30, and the Euclid parametrization (m, n) = (13, 6).
Verification
(1) Brute force over 44,720 triangular numbers up to T_44720 ≈ 9.9998 × 10⁸ finds one triangular-sided Pythagorean triple: (8778, 10296, 13530), as reported by the script. (2) Manual check: 8778² + 10296² = 183,060,100 = 13,530². The triangular indices 132, 143, 164 are computed from the inverse formula x = (−1 + √(1 + 8·T))/2, which gives exact integers. (3) Non-primitivity: gcd(8778, 10296, 13530) = 66, and 66·133 = 8778, 66·156 = 10296, 66·205 = 13530. (4) (133, 156, 205) is a primitive Pythagorean triple (gcd = 1) with Euclid parameters (m, n) = (13, 6): 13² − 6² = 133, 2·13·6 = 156, 13² + 6² = 205. (5) Companion check: zero Fibonacci-sided Pythagorean triples up to F_87 ≈ 10^18 (matches the classical Luca theorem), zero Catalan-sided Pythagorean triples up to C_54 ≈ 10^30.
Sequences
(8778, 10296, 13530) = (T_132, T_143, T_164) = 66 × (133, 156, 205)
(m, n) = (13, 6) → (m² − n², 2mn, m² + n²) = (133, 156, 205); scaling factor 66 = T_11
Fibonacci Pythagorean triples (up to F_87 ≈ 10^18): 0 (Luca 1992) · Catalan Pythagorean triples (up to C_54 ≈ 10^30): 0 · Triangular Pythagorean triples (up to T_44720 ≈ 10⁹): 1 · Primitive triangular Pythagorean triples (up to T_44720): 0
Next steps
- Extend the search to T_k ≤ 10^12 or higher via a smarter parametric iteration over Euclid (m, n) pairs with a per-primitive scaling search — this would determine whether the singleton is truly unique or just the smallest.
- Prove or disprove: the singleton (T_132, T_143, T_164) is the only triangular-sided Pythagorean triple for all triangular numbers (no bound).
- Investigate the primitive Pythagorean triples (m, n) = (13, 6), (m, n) = ?, etc., whose specific scalings land on all-triangular sides, and find the structural condition k·(m²−n²), 2kmn, k(m²+n²) all triangular in closed form.
- Combine with iterations 44-47 as a unified paper on 'Restricted Side-Sequences in Pythagorean and Heronian Triangles.'
Artifacts
- Triangular / Fibonacci / Catalan Pythagorean enumeration: discovery/pythagorean_triangular.py