Remember Pair Share? It carries over seamlessly to triples. Let T be the set of all triples (m,n,p) such that mnp = m+n+p. (Digressing: notice that these triples are not sets, since if two numbers are the same, we can't let them fall together; also, they are not ordered triples or 3-vectors, since we don't distinguish between (m,n,p) and (m,p,n), etc. I believe that the technical term for this intermediate kind of entity is "bag.")(0,0,0) is in there, of course, but so is (1,2,3). Cool!
Most pairs (m,n) allow one (m,n,p) in T, but not all. Can you find (m,n) that doesn't allow any (m,n,p)? Answer: (1,1) is a good choice (of very many). 1 x 1 x p = p; 1+1+p = p+2.
Can you generate all and only the integer triples in T? Are there finitely or infinitely many of them? Answer: relying on analogy with the case of pairs, you might be tempted to bite me--er, no, I mean, you might be tempted to answer "finitely many." But triples have a dimension that pairs lack: every (-n,0,n) qualifies: the answer is "infinitely many."
The question of whether the aggregate resultants (i.e., mnp = m+n+p) are finite or infinite in number is slightly harder.
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