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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?

Bill Dubuque
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  • @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. – FirstName LastName Dec 28 '23 at 20:47
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    I assumed that $m=1/a+1/b+1/c$ was a positive integer - sorry, I have to apologise. – Dietrich Burde Dec 28 '23 at 22:46
  • @DietrichBurde : no worries ... have a nice day! I dismissed other ME post as solution for this one. – FirstName LastName Dec 28 '23 at 22:53
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    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 $,a,b,c\ge 1$ – Bill Dubuque Dec 28 '23 at 23:11
  • @BillDubuque : with permission ... your argument applies dupe multiple times and I gladly grant forum etiquette permission to proceed towards dupe, though, of course, not an exact dupe but a (multiple times) consequence. However Merosity's argument does not even need dupe. I would not like his reputation to be harmed. You can lower my reputation instead. because: My question may have been poor and may have triggered cherry picking by experts way above me but below you. Nevertheless I am thankful to at least have received proof without real need to use more general knowledge. ME has helped me. – FirstName LastName Dec 29 '23 at 02:26
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    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. – Bill Dubuque Dec 29 '23 at 02:41
  • @BillDubuque : I never doubted you would not also be capable to avoid dupe :-) – FirstName LastName Dec 29 '23 at 02:43

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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 $a$.

Dietrich Burde
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Merosity
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  • Thanks a lot, I can stop searching ... I should have been able to find this contradiction myself :-( – FirstName LastName Dec 28 '23 at 19:50
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    Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Dec 28 '23 at 23:12
  • @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? – FirstName LastName Dec 28 '23 at 23:28
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    @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\ \ $ – Bill Dubuque Dec 28 '23 at 23:33
  • @BillDubuque : OK :-) Right: closed under product ... perhaps it depends on how one defines immediate consequences. Your proof in OP comment is nice and simple, but requires at least some steps to finish. Have a nice day! – FirstName LastName Dec 28 '23 at 23:41
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    @First If you have questions about that then you should post them there, not here. – Bill Dubuque Dec 28 '23 at 23:49