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Find all triplets $(a,b,c)$ of positive integers so that $\gcd(a,b,c)=1$ and $$ \frac{2abc}{(a+b-c)(b+c-a)(c+a-b)} $$ is a positive integer.

What I've done: first I looked with Mathematica for solutions. For $a<150$, $b<1000$ and $c<1000$, I found $(1,1,1)$, $(7,117,121)$ and $(11,39,49)$ as the only valid solutions, which led me to believe these are the only ones. Unfortunately, I have no idea how to prove this assertion. What might be interesting is that these triplets would also be solutions if the problem had stated $abc$ instead of $2abc$, and that in that case the fraction would equal either $1$ or $13$. Now it can be $2$ or $26$. Also, there are more solutions if the fraction is allowed to be negative, but for the moment I'm not interested in those. Any help is appreciated. Thanks.

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I am afraid your conjecture is not correct. The situation is quite complex. I have been studying such cubic representation problems for some time.

The basic identity \begin{equation*} \frac{2abc}{(a+b-c)(b+c-a)(c+a-b)}=N \end{equation*} is equivalent to finding points of infinite order on the elliptic curve \begin{equation*} v^2=u^3+2(2N^2-2N-1)u^2+(4N+1)u \end{equation*}

For specifically positive solutions the situation is even more complicated, in that we also need a point where the u-coordinate is negative.

Such situations are quite rare, but other examples come from $N=74, 218$ which give the results in the comment. A novel solution is for $N=250$ with $(97, 10051, 10125)$ as a solution.

We can also use multiples of the points in the elliptic-curve formulation to get larger solutions. For example, $N=26$ has the solution

$a=4739\,2819\,4344\,87$

$b=2634\,1867\,7932\,41$

$c=7228\,3008\,5646\,67$

and we can keep going getting larger and larger solutions.

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  • $\begingroup$ Just curious: What kind of digit grouping is that - groups of four digits starting from the left as opposed to groups of three starting from the right? I know the latter with varying separator (period or comma or hyphen or thin space) but not the former in a math context $\endgroup$ – Hagen von Eitzen Mar 9 '16 at 9:05
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There are more solutions, e.g. $(27, 1805, 1813)$ , $(115, 5239, 5341)$.

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This problem is equivalent to looking for an integer-sided triangle such that \begin{equation*} \frac{R}{r}=N \end{equation*} where $R$ is the radius of the circumcircle and $r$ is the radius of the incircle, with $N \in \mathbb{Z}$.

Full details are in a paper I wrote in 2010 in Forum Geometricorum which can be retrieved from the journal web-ste.

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  • $\begingroup$ You got me there; that was indeed how I arrived on this problem. Since your excellent answer and article have shown solving this problem is virtually impossible to solve, a more interesting exercise might be to prove that some values of N, for example 3, will not give any solutions. $\endgroup$ – Mike Daas Mar 11 '16 at 6:14
  • $\begingroup$ @AllanMacLeod: Can you kindly look at this elliptic curve problem to see if you have a solution? $\endgroup$ – Tito Piezas III Jul 4 '16 at 2:59

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