n*(n+1)/2 is not a homogeneous form, so multiplying by a constant is not allowed. There are duplicates though, for instance both 6,20,21 and 20,6,21 appear On Thu, Feb 27, 2020 at 6:23 PM Allan Wechsler <acwacw@gmail.com> wrote:
Are those just the "primitive" triples, so that one has to include k*[n,k,m] as well in order to get a complete list? Or does that form hit all triples? Also, does this procedure hit all triples just once, or are there occasions where different choices of n, Q, (and k?) give the same triple?
On Thu, Feb 27, 2020 at 4:54 PM Neil Sloane <njasloane@gmail.com> wrote:
I mentioned this on Feb 19 2020 and several people sent very helpful replies.
For squares, we have the theorem that all solutions to x^2+y^2=z^2 in integers are given by Pythagorean triples, t*[u^2-v^2, 2uv, u^2+v^2].
Expanding on the replies from Michael Collins and Rich, it looks like the following is the analog for triangular numbers T_n = n(n+1)/2. ALL solutions to T_n+T_k=T_m are given by the list of what one might call Trithagorean triples: these are the triples
[n,k,m] = [n, T/Q-(Q+1)/2,T/Q+(Q-1)/2] where n >= 2, T=n(n+1)/2, and Q is any odd divisor of T less than n,
plus these triples with the first two coordinates swapped.
I haven't found this in the literature, but it can hardly be new. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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