I think this has been asked before, but I couldn't find it on math.SE. I googled it too, but I wasn't lucky enough to find it there either. So, here's the problem:

Demonstrate that for any $a,b,c \in \mathbb{N}$: $$\displaystyle [a,b,c] = \frac{abc}{(ab,bc,ca)}$$

It's not very hard to verify this equality once we know the prime factorization theorem. We can set:

$$a = p_1^{\alpha_1} p_2^{\alpha_2}\cdots p_n^{\alpha_n}$$ $$b = p_1^{\beta_1} p_2^{\beta_2} \cdots p_n^{\beta_n}$$ $$c = p_1^{\gamma_1} p_2^{\gamma_2}\cdots p_n^{\gamma_n}$$

Where $0 \leq \alpha_i,\beta_i,\gamma_i $ and $p_i$ is a prime number that appears in the prime factorization of at least one of the three.

Therefore, the equality is true if and only if this equality holds for all $1 \leq i \leq n$:

$$\min\{\alpha_i+\beta_i,\beta_i+\gamma_i,\gamma_i+\alpha_i\} + \max\{\alpha_i,\beta_i,\gamma_i\} = \alpha_i+\beta_i+\gamma_i$$

It's very easy to verify this equality. Therefore, the theorem holds if we are allowed to use prime factorization theorem.

How can we demonstrate that the equality is true without using the prime factorization theorem?

  • $\begingroup$ Maybe this identity is true in a GCD domain? $\endgroup$ – clark May 17 '14 at 14:08
  • 2
    $\begingroup$ @math.n00b No, a GCD domain is any domain where any two elements not both $\,0\,$ have a gcd. A Bezout Domain is a GCD domain where a Bezout identity exists for every gcd, which need not be true generally, e.g. $\,\Bbb Q[x,y]\,$ is UFD so GCD, and $\,\gcd(x,y) = 1\,$ but $\,x f + y\, g \ne 1,\,$ else $\, 0 = 1\,$ by eval @ $\,x,y = 0,0.\ $ The proof in my answer works in any gcd domain. $\endgroup$ – Bill Dubuque May 17 '14 at 16:14

Notice that we can wirte $[a,b,c]=[[a,b],c]$ and let say $k=[a,b]=\dfrac {ab}{(a,b)}$

$[k,c]=\dfrac {kc}{(k,c)}=\dfrac {abc}{(a,b)(k,c)}$ and notice that $r(x,y)=(rx,ry)$

$$=\dfrac {abc}{((a,b)k,(a,b)c)}$$ $$=\dfrac {abc}{(ab,(a,b)c)} $$ $$=\dfrac {abc}{(ab,(ac,bc))}$$ $$=\dfrac {abc}{(ab,ac,bc)}$$

Note: $(a,b),[a,b]$ stand for $\gcd(a,b)$ and $\text{lcm}[a,b]$ respectively.

  • $\begingroup$ This can be proved symmetrically, without using the gcd * lcm identity, by using only the universal properties of gcd and lcm, and the distributive law for gcd - see my answer. $\endgroup$ – Bill Dubuque May 17 '14 at 16:20

A proof is mechanical using universal properties $\rm\,\color{#c00}{(UP)\ of\ lcm,gcd}\,$ and $\,\rm\color{#0a0}{duality}$

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.

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

  • $\begingroup$ You are following me ha :), nice answer. $\endgroup$ – mesel May 17 '14 at 17:36
  • $\begingroup$ @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate. $\endgroup$ – Bill Dubuque May 18 '14 at 16:21
  • $\begingroup$ This is a very good answer. I like how elementary and simple it is and it's a pity that I failed to reason like this on my own. :| Is it possible to generalize the concept of least common divisor to other rings? $\endgroup$ – math.n00b May 19 '14 at 18:18
  • $\begingroup$ @math.n00b Yes, see the universal definitions of gcd and lcm, which I have implicitly used in the above proof, viz. $\,a,b,c\mid n\iff {\rm lcm}(a,b,c)\mid n,\,$ and $\ abc\mid i,j,k\iff abc\mid (i,j,k).\ \ $ $\endgroup$ – Bill Dubuque May 19 '14 at 18:23
  • $\begingroup$ @math.noob I've expanded the answer to highlight the innate gcd,lcm duality. $\endgroup$ – Bill Dubuque May 19 '14 at 19:04

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