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$\gcd(n,m)\,{\rm lcm}(n,m) = nm.\,$ Can this theorem work with 3 integers? And how to prove it? I tried doing this with 2 integers n,m , but I can't figure out how to do it with 3.

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    $abc=a(bc)$. What are DBD and MBK? – 2017-01-30
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    Oh sorry, Im not native english speaker so I miswrote them, DBD is GCD and MBK is LCM – 2017-01-30

2 Answers 2

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From here:

We have that

Theorem: $\rm\ \ lcm(a,b,c)\, =\, \dfrac{abc}{(bc,ca,ab)}$

Proof: $\!\begin{align}\qquad\qquad\rm\ a,b,c&\mid\rm\ k\\ \iff\quad\rm abc&\mid \rm\,\ kbc,kca,kab\\ \iff\quad\rm abc&\mid \rm (kbc,kca,kab)\, =\, k(bc,ca,ab)\\ \iff\rm \ \dfrac{abc}{(bc,ca,ab)} &\:\Bigg| \rm\,\ k\end{align}$

where $(bc, ca, ab) $ means the gcd of $ab, bc, ca $. Hope it helps.

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    As a courtesy, when you quote verbatim [an answer](http://math.stackexchange.com/a/319690/242) posted by another user, you should cite that answer. – 2017-01-30
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    @BillDubuque Yes, yes, very sorry. In fact I was then going to cite that answer, but due to unavoidable circumstances, totally forgot about it. Will never do it again. Very sorry once again. – 2017-01-30
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    No problem, I presumed it was an oversight. – 2017-01-30
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Let the highest power of prime $p$ that divides $a,b,c$ be $A,B,C$ respectively.

So, the highest power of prime $p$ that divides the GCD will be min$(A,B,C)$

and the highest power of prime $p$ that divides the LCM will be max$(A,B,C)$

We need min$(A,B,C)+$max$(A,B,C)=A+B+C$ for any prime that divides at least one of $a,b,c$

WLOG min$(A,B,C)=A,$ and max$(A,B,C)=C\implies A+C=A+B+C\iff B=0$

So, $(a,b,c)$ must be $1$ for LCM$(a,b,c)\cdot$GCD$(a,b,c)=abc$

For the two integer case, trivially min$(A,B)+$max$(A,B)=A+B$