Elegant proof: The two sets, S1 = { (n,k,m) in Z^3 : n*(n+1) + k*(k+1) - m*(m+1) = 0 } S2 = { (x,y,z) in (2Z+1)^3 : x^2 + y^2 - z^2 - 1 = 0 } are equivalent under change of coordinates (x,y,z) = 2(n,k,m)+1. From the condition of S2 and the triangle inequality, x + y >= sqrt(z^2+1) > z. This implies n + k > m - 1/2. Over Z^3, we may round to the next integer, thus n + k >= m, but equality can only be achieved when n=k=m=0. Choosing some particular value of x, we are looking for the intersection of an odd-integer lattice with a hyperbola. Generally, this is the same situation as in the case of Pythagorean triples. I don't know how to finish the proof just now, but think that an integer-curve theorist probably could. Also compare the following: http://mathworld.wolfram.com/PythagoreanTriple.html https://0x0.st/iKud.png ( blue: Pythagorean triples, red: triangle triples ) --Brad On Wed, Feb 19, 2020 at 12:52 PM Brad Klee <bradklee@gmail.com> wrote:
I did not prove, but would guess that . . .