I was just trying to find all the integer triplets $(a,b,c)$ such that $ab+bc+ca=abc$. Actually this question popped up in my mind when I was just mathaholic.

By hit and trial I know that unordered triplet $(4,4,2)$ and $(0,0,0)$ satisfy the condition. But I couldn't find more.

I shall be thankful if you can give me some hint to prove that there exist infinite many such triplets (If really exist) otherwise if there are finite solutions then an algorithm to generate them all.


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    $\begingroup$ There can only be finitely many solutions (and not even that many), because the multiplication in the right-hand side quickly becomes too large for the addition on the left-hand side to keep up. $\endgroup$ – Arthur Dec 14 '16 at 17:22
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    $\begingroup$ Try dividing both sides by $abc$ $\endgroup$ – Simply Beautiful Art Dec 14 '16 at 17:22

If you divide both sides by $abc$, you end up with


Assuming they are positive integers greater than $0$ and $a\le b\le c$

Also assume $a=1$. Clearly, this is not possible.

Assume $a=2$. Some solutions involve $(b,c)=(4,4) $ and $(b,c)=(3,6)$.

Assume that $a=3$. The only solution here is $(b,c)=(3,3)$.

For $a>3$, there are no solutions.

If one includes negative integers, see that $(a,b,c)=(-n,1,n)$ is an infinite amount of such solutions.

  • $\begingroup$ Why for a>3, there are no solutions?? $\endgroup$ – I am Back Dec 14 '16 at 17:32
  • $\begingroup$ Notice that we have $3<a\le b\le c$. Also, use monotonacity of $1/x$. I used this to deduce there are no other positive solutions for other values of $a$. $\endgroup$ – Simply Beautiful Art Dec 14 '16 at 17:34
  • $\begingroup$ $(1,-n,n)$,$(-n,n,1)$,$(n,-n,1),(n,1,-n),(-n,1,n)$ will also be the solutions. $\endgroup$ – Error 404 Dec 14 '16 at 17:37
  • $\begingroup$ @VikrantDesai obviously one can rearrange the order of $a,b,c$, which is why I said $a\le b\le c$. $\endgroup$ – Simply Beautiful Art Dec 14 '16 at 17:43
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    $\begingroup$ $a\leq b \leq c$ should make $(-n, 1,n)$ the infinite set of solutions instead of $(1, n, -n)$, should it not? $\endgroup$ – Arthur Dec 14 '16 at 18:40

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