Entire pursuit inspired by [url=https://twitter.com/jamestanton/status/1871827972866326921]this James Tanton tweet[/url].[br]Summary of [url=https://drive.google.com/file/d/1Xno2o8P_uTvdYIB2YIFNmVPYBq_50bWj/view?usp=sharing]my efforts documented in this pdf[/url].[br][br]It's well-known that there are infinitely-many "Pythagorean Triples" that satisfy A²+B²=C² for positive integers A, B, and C, and there exist formulas to generate all of them.[br]If we care to, we may define the "squaring function" S(N)=N² and express the Pythagorean Theorem as S(A)+S(B)=S(C).[br][br]Now consider the "triangular number function" T(N) = 1+2+3+…+N = N(N+1)/2.[br][br]The referenced tweet invites the challenge: How many "Triangular Triples" satisfy T(A)+T(B)=T(C) for positive integers A, B, and C, and does there exist a formula or algorithm to generate all of them?[br][br]I'm confident that I've found such an algorithm that finds all such Triangular Triples, and I've implemented it into the GeoGebra construction below. However, it falls short of a tidy closed-form formula that I'd consider to be ideal. Rather, at one point in my algorithm, a pair of linear Diophantine equations must be solved. It doesn't take much computing power to churn out the solutions to the Diophantine equations, but it still strikes me as a tad unsophisticated.[br]Also worth noting: If we consider {A, B, C} to be the same triple as {B, A, C}, then such duplicates do appear in this algorithm. For example, m=3 & n=2 & u=0 yields solution {4, 9, 10}, and m=4 & n=1 & u=0 yields solution {9, 4, 10}.[br][br]If using a keyboard, finer control of the sliders below may be achieved by first clicking on the slider and then using ←→ or ↓↑ arrow keys to decrement/increment the values.[br][br]While I'm content to consider my work on this challenge to be complete, if I ever get so inspired to tackle it again and if I can overcome this hurdle, it will certainly be the highlight of my life.