# Find pairwise coprime $a$ $b$ $c$ with certain property [duplicate]

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?

• @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. Commented Dec 28, 2023 at 20:47
• I assumed that $m=1/a+1/b+1/c$ was a positive integer - sorry, I have to apologise. Commented Dec 28, 2023 at 22:46
• @DietrichBurde : no worries ... have a nice day! I dismissed other ME post as solution for this one. Commented Dec 28, 2023 at 22:53
• 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 $\,a,b,c\ge 1$ Commented Dec 28, 2023 at 23:11
• 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. Commented Dec 29, 2023 at 2:41

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 $$a$$.
• @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\ \$ Commented Dec 28, 2023 at 23:33