[seqfan] Pentathagorean triples: two easy proofs.
Brad Klee
bradklee at gmail.com
Fri Mar 20 06:36:48 CET 2020
https://oeis.org/draft/A330657
Theorem. Pentathagorean triples satisfy a triangle inequality n+k>m.
By definition, n*(3*n - 1) + k*(3*k - 1) - m*(3*m - 1) = 0. Under change
of coordinates, {n -> (x + 1)/6, k -> (y + 1)/6, m -> (z + 1)/6}, this
constraint becomes: x^2 + y^2 -(z^2+1) = 0. After reverse transform,
the triangle inequality, x+y >= sqrt[z^2+1] > z, implies n+k > m.
Theorem. [n,m-1,m] is a pentathagorean triple if and only if n=3*z+1.
By definition after substitution k=m-1, then m = 1/6*(4 - n + 3*n^2).
Split cases by n=3*z, n=3*z+1, n=3*z+2, and then respectively:
m = 1/2*(1 - z + 9*z^2) + 1/6
m = 1/2*(2 + 5*z + 9*z^2)
m = 1/2*(4 + 11*z + 9*z^2) + 1/3
Clearly the first and last case can never have m an integer, while
the second case always has m an integer.
Ex. n = 4, z=1, m = 8. [n,k,m] = [4,7,8] satisfies the definition that
n*(3*n - 1) + k*(3*k - 1) - m*(3*m - 1) = 0.
Cheers,
--Brad
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