elementary number theory gcd lcm

Question

if $m,n,k$ are any three positive integers prove that $$(m,n)(m,k)(n,k)[m,n,k]^2=[m,n][m,k][n,k](m,n,k)^2$$ where $(a,b)$ denotes gcd of $a$ and $b$, $[a,b]$ denotes lcm of $a$ and $b$.

I tried with the formula $(a,b)[a,b]=ab$

Answer 1

Let $p$ be a prime number and $\nu_p$ be the $p$-adic valuation at $p$. Wlog we can assume $\nu_p(m)\leq\nu_p(n)\leq\nu_p(k)$. Then the applying $\nu_p$ to both members of the required equation gives, respectively, \begin{align} &\nu_p(m)+\nu_p(m)+\nu_p(n)+2\nu_p(k)& &\nu_p(n)+\nu_p(k)+\nu_p(k)+2\nu_p(m) \end{align} which are equals. Then the assertion follows by unique factorization theorem.

Answer 2

Definition Let $p$ be prime. Write $m_p$ for the greatest exponent such that $$p^{m_p} \mid m.$$

Basically, we are writing $m$ in the form $$m = \prod_{p \text{ prime}} p^{m_p}.$$

It is easy to see that:

Proposition: Multiplication corresponds to addition on the exponent level (for example, $(mn)_p = m_p + n_p$), GCD corresponds to minimum (for example, $(m, n)_p = \min(m_p, n_p)$), and LCM corresponds to maximum (for example, $[m, n]_p = \max(m_p, n_p)$).

Example The equality $(a, b)[a, b] = ab$ follows from $\min(a_p, b_p) + \max(a_p, b_p) = a_p + b_p$.

Solution to the problem asked On the LHS you have $$LHS_p = \min(m_p, n_p) + \min(m_p, k_p) + \min (n_p, k_p) + 2\max(m_p, n_p, k_p),$$ and on the RHS $$RHS_p = \max(m_p, n_p) + \max(m_p, k_p) + \max(n_p, k_p) + 2 \min(m_p, n_p, k_p).$$

As each expression is symmetric under swapping $m_p, n_p, k_p$, we can assume $m_p \le n_p \le k_p$. Then $$ LHS_p = 2m_p + n_p + 2k_p = RHS_p,$$ what finishes the proof.

Edit: It seems that Fabio Lucchini's solution was posted 10 minutes earlier than mine, and the idea is the same. I will however keep this post as here the calculation is more verbose.