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I am looking for pairwise coprime natural numbers $a$ $b$ $c$ for which $n = \dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}$ is also a natural number.

I can find examples (note $a$ $b$ $c$ must differ) for which one of their three gcds is $1$ and the two other are prime numbers.

Maybe such triplets do not exist?

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  • $\begingroup$ @DietrichBurde : I apologize to not see how the solutions for $= 1$ imply answer for any $= 1/n$. But then I did not manage to prove contradiction either so I am perhaps not good person to judge about duplication. Feel free to elaborate (perhaps in separate answer) or to to duplicate. I do like the answer though and I appreciate and up-voted all your comments. $\endgroup$ Commented Dec 28, 2023 at 20:47
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    $\begingroup$ I assumed that $m=1/a+1/b+1/c$ was a positive integer - sorry, I have to apologise. $\endgroup$ Commented Dec 28, 2023 at 22:46
  • $\begingroup$ @DietrichBurde : no worries ... have a nice day! I dismissed other ME post as solution for this one. $\endgroup$ Commented Dec 28, 2023 at 22:53
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    $\begingroup$ Clearing denoms $\Rightarrow d\!=\!ab\!+\!bc\!+\!ca\mid abc.\,$ But $(d,a)\!=\!(bc,a)\!=\!1\,$ by $\,(b,a)\!=\!1\!=\!(c,a)\,$ and dupe. Similarly $(d,b)\!=\!1\!=\!(d,c)\,$ so $\,d=(d,abc)\!=\!1$ by dupe, contra $\,a,b,c\ge 1$ $\endgroup$ Commented Dec 28, 2023 at 23:11
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    $\begingroup$ The proof in M's answer is essentially a contradiction-form of my argument, but using a prime-divisor form of Euclid's Lemma vs. the gcd form that I used. To rewrite my proof that way: $\,1\neq(d,abc)\underset{\rm wlog}\Rightarrow 1\neq(d,a)=(bc,a),\,$ contra $\,(b,a)\!=\!1\!=\!(c,a).\,$ Using gcds is more general. $\endgroup$ Commented Dec 29, 2023 at 2:41

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Multiply through by $abc$ in the numerator and denominator, this shows that we must have $ab+ac+bc$ divides $abc$. Suppose $p$ divides $ab+ac+bc$, then $p$ must divide $abc$. By Euclid's lemma and without loss of generality we may assume $p$ divides $a$. But if $p$ divides $a$ and $ab+ac+bc$, then it must also divide $bc$, which is a contradiction since they're relatively prime to $a$.

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  • $\begingroup$ Thanks a lot, I can stop searching ... I should have been able to find this contradiction myself :-( $\endgroup$ Commented Dec 28, 2023 at 19:50
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    $\begingroup$ Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. $\endgroup$ Commented Dec 28, 2023 at 23:12
  • $\begingroup$ @BillDubuque : Thanks for spotting duplicate. Does question map one-on-one to #3555122 which was : 'if x is coprime with each pi then x is coprime with p1...pn' or is some deduction and raesoning still required and is OP some sort of corolarry? $\endgroup$ Commented Dec 28, 2023 at 23:28
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    $\begingroup$ @First The proof is an immediate consequence of coprimes are closed under product - as I explained in the closing comment, i.e. $\,a,b,c\,$ coprime to $d\Rightarrow abc\,$ coprime to $\,d\ \ $ $\endgroup$ Commented Dec 28, 2023 at 23:33
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    $\begingroup$ @First If you have questions about that then you should post them there, not here. $\endgroup$ Commented Dec 28, 2023 at 23:49

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