Define an H-Triple (a, b, c), where a > b > c > 1, as one where a, b and c = $\frac{ab+1}{a+b}$ are all integers. (There are 'lots' of these!)
Conjecture; if (a, b, c) and (d, e, f) are distinct H-Triples, then abc$\neq$def.
That is, the product of the terms of an H-Triple is unique.
Can anybody help with this? Computers suggest any counterexample will not be small.
The reason the problem arises? Consider the family of elliptic curves $xy\frac{xy+1}{x+y}=k$ and the integer points that lie on them. If (a,b,c) is an H-Triple and k = abc then (a,b) is an integer point on this elliptic curve. The uniqueness conjecture would narrow down the possibility that there might be any others.
Here is an idea to search for a counterexample. We have the following universal $H$-triples, namely $$ (a, b ,c)=(2k+1,2k-1,k),\; (r^2+r-1,r+1,r) $$ for all $k,r\ge 1$. Equality of their products means $$ 4k^3-k=r^4+2r^3-r $$ This Diophantine equation has the "trivial" solution $(k,r)=(2,2)$. In this case the triples coincide and are $(a, b,c)=(5,3,2)$ with product $abc=30$. With a computer one could try to find a non-trivial solution in positive integers, if it exists. This would yield two different $H$-triples with equal product. Of course one may use also different universal triples, like $(a,b,c)=(2l^2+4l+1,2l+3,2l+1)$ and others.