Show that $abc=[ab,bc,ca]*(a,b,c)=(ab,bc,ca)*[a,b,c]$

Question

Let $a,b \in \mathbb N$. Show that

$$ abc=[ab,bc,ca](a,b,c)=(ab,bc,ca)[a,b,c] .$$

How would I prove this?

Answer 1

It is straightfoward using universal properties $\rm\,\color{#c00}{(UP)\ of\ lcm,gcd}.\,$ Below is the proof of one equality. The other follows immediately by duality (see the Remark).

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

$\begin{eqnarray}{\bf Proof}\quad\ \ \ {\rm lcm}(a,b,c)&\mid& n\\ \iff\: a,b,c&\mid& n,\ \ \text{by}\,\color{#c00}{\text{ UP lcm}}\\ \iff\quad abc&\mid& nbc,nca,nab,\ \ \ \,\text{by}\ \ \rm\color{#0a0}{duality},\ see\ below\\ \iff\quad abc&\mid&\!\! (nbc,nca,nab),\ \ \text{by}\,\color{#c00}{\text{ UP gcd}}\\ \iff\quad abc&\mid& n(bc,ca,ab) \\ \iff\ \ \dfrac{abc}{(bc,ca,ab)}&\Big|& n\end{eqnarray}$

Remark $\ $ The innate symmetry governing the proof is the involution $\ x\to x' = abc/x\,$ which highlights the duality $ $ lcm$' =$ gcd,$\ $ gcd$' =$ lcm, $ $ that arises from $\, \color{#0a0}{x\mid y\iff y'\mid x'},\,$ viz.

$\qquad\qquad \begin{eqnarray} {\rm lcm}(a,b,c) &=& {\rm lcm}(a,b,c)''\\ &=& {\gcd(a',b',c')'}\\ &=& \dfrac{abc}{\gcd(a',b',c')}\\ &=& \dfrac{abc}{\gcd(bc,ca,ab)}\end{eqnarray}$

Answer 2

Fix a prime p. Show that the prime power p^k that divides each side is the prime power that divides abc.

Hence, we can even conclude that the expression is equal to abc

Answer 3

let $n_a, n_b, n_c$ denote a prime power of $a,b,c$, so the equation is just $$\max(n_a+n_b, n_b+n_c,n_a+n_c)+\min(n_a,n_b,n_c) = \min(n_a+n_b,n_a+n_c,n_b+n_c)+max(n_a,n_b,n_c)=n_a+n_b+n_c$$ because when $n_a+n_b$ obtains maximum, $n_c$ obtains minimum, and vice versa.

Answer 4

If $\gcd(a,b,c) = d > 1$, then by writing: $a = pd$, $b = qd$, and $c = rd$, we can factor out $d^3$ in common. So we can assume $\gcd(a,b,c) = 1$. Then we need to prove: $abc = [ab,bc,ca]$. Let $[ab,bc,ca] = m$, then: $ab|m$, $bc|m$, and $ca|m$. Thus:

$abc|mc$, $abc|ma$, and $abc|mb$. Thus:

$abc|\gcd(ma,mb,mc) = m\cdot \gcd(a,b,c) = m\cdot 1 = m$. Thus: $abc|m$. We only need to prove: $m|abc$.

But observe that: $ab|abc$, $bc|abc$, and $ca|abc$. Thus: $[ab,bc,ca]|abc$. This means: $m|abc$.

Thus: $abc|m$, and $m|abc$. So: $m = abc$. Done.