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Is it possible to find 3 distinct positive integers a, b, c such that the result abc/(ab + bc + ca) is also an integer?

Karri Chandrasekhar
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  • If you can find two distinct positive integers whose harmonic mean is an integer, you can find an answer to this question. Do you see why? – David K Sep 11 '22 at 00:42
  • @DavidK As 3 = (1+1) + 1, and since for 2 specific positive integers, it is possible to find HM - that is an integer, can it be generalized like this? Or principle of mathematical induction? Not sure, though, what your hint suggests! – Karri Chandrasekhar Sep 11 '22 at 07:13
  • I only had in mind that from two you can get three. The same method does not give you $n+1$ distinct numbers from $n$ distinct numbers if $n>2,$ because the harmonic mean of the $n$ numbers might already be in the set of $n$ numbers. – David K Sep 11 '22 at 15:18

1 Answers1

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Yes! Note that the distinct integers $2, 3, 6$ satisfy $2\cdot 3\cdot 6 = 36$ and $$2\cdot 3 + 2\cdot 6 + 3\cdot 6 = 6 + 12 + 18 = 36$$ so the quotient $$\frac{2\cdot 3\cdot 6}{2\cdot 3 + 2\cdot 6 + 3\cdot 6}$$ is equal to $1$.

Nicolas Agote
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