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My exploration of this recent question

$\qquad$Proving $P$ for $N(n+N-1)$$\qquad$

provoked the following (speculative)

Conjecture:

If $a_1,...,a_m$ are positive rational numbers whose sum is an integer, and the sum of whose reciprocals is $1$, then $a_1,...,a_m$ must be integers.

Partial results:

  • $\;$For $m=1$, the conjecture holds, and the proof is trivial.
  • $\;$For $m=2$, the conjecture also holds, and the proof is easy.
  • $\;$For $m=3$, while I don't have a proof, limited testing suggests that the conjecture holds.

Remarks:

  • If the conjecture is true, a proof for the general case (arbitrary $m$) might be very hard.
  • Or perhaps there's an easy counterexample.

To test the waters, can anyone resolve the conjecture for the case $m=3\,?$

quasi
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1 Answers1

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$${10\over 3} + {5\over 3} + 10 = 15$$

$${3\over 10} + {3\over 5} + {1\over10} = 1$$

cr001
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  • Nice! I don't know how I missed that! – quasi Sep 20 '17 at 05:48
  • This happens all the time, I sometimes overlook easy counter examples too. – cr001 Sep 20 '17 at 05:49
  • What if I weaken the conjecture to at least one of $a_1,...,a_m$ is an integer? Is it still easy to break? – quasi Sep 20 '17 at 05:52
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    Yes, not as easy to find but ${11\over 2}, {11\over 3}, {11\over 6}$ would be one group of numbers that are all non-integers. – cr001 Sep 20 '17 at 06:00
  • I fixed the bug in my program. With no integers, we have $$\frac{11}{2}+\frac{11}{3}+\frac{11}{6}=11$$ $$\frac{2}{11}+\frac{3}{11}+\frac{6}{11}=1$$ – quasi Sep 20 '17 at 06:01
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    I see we got the same counterexample. I guess the conjecture is dead. I was initially fooled by a program which had a bug. – quasi Sep 20 '17 at 06:03
  • So we found the same group of numbers.Essentially if we have any reciprocals adding up to one we can just multiply by their sum to get the group. However the number of such groups might not be limited to those generated using this method. – cr001 Sep 20 '17 at 06:04