Ternary generalizations of $\,\gcd(n,m)\,{\rm lcm}(n,m) = nm$

Question

$\gcd(n,m)\,{\rm lcm}(n,m) = nm.\,$ 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.

Answer 1

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.

Answer 2

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$(A,B)=A+B$