I honestly have no idea how to approach this question. I have done easier questions eg. show $gcd(a,b)lcm(a,b)=ab$ by letting $a=Alcm(a,b)$ and $b=Blcm(a,b)$ and then using the fact that $A$ and $B$ are coprime. However this approach only works in two variables. Any help with the question in the title would be greatly appreciated, as well as the thought process behind the answer so that I can approach similar problems.
Prove $lcm(a,gcd(b,c))=gcd(lcm(a,b),lcm(a,c))$
algebra-precalculuselementary-number-theory
Related Solutions
According to this post (the 2nd and 3rd equations), we should have
$$F(x,y,z)=\frac{\text{GCD}(x,y)\text{GCD}(x,z)\text{GCD}(y,z)}{\text{GCD}(x,y,z)^2}$$
$$=\frac{\text{LCM}(x,y)\text{LCM}(x,z)\text{LCM}(y,z)}{\text{LCM}(x,y,z)^2};$$
this corresponds to the fact that
$$\text{mid}(x,y,z)=\min(x,y)+\min(x,z)+\min(y,z)-2\min(x,y,z)$$
$$=\max(x,y)+\max(x,z)+\max(y,z)-2\max(x,y,z).$$
But $\text{mid}$ is a purely order-theoretic function; it shouldn't depend on addition or subtraction. Indeed, this answer describes $\text{mid}$ in terms of $\max$ and $\min$. Here's a more symmetric expression:
$$\text{mid}(x,y,z)=\min(\max(x,y),\max(x,z),\max(y,z))$$
$$=\max(\min(x,y),\min(x,z),\min(y,z)).$$
Applied to the exponents in prime factorization, this gives a formula for $F$:
$$F(x,y,z)=\text{GCD}\Big(\text{LCM}(x,y),\text{LCM}(x,z),\text{LCM}(y,z)\Big)$$
$$=\text{LCM}\Big(\text{GCD}(x,y),\text{GCD}(x,z),\text{GCD}(y,z)\Big).$$
These dual expressions are sandwiched between the "meet" and "join" of $(x,y,z)$ in any lattice, and are equal in any distributive lattice.
Using the facts that $\text{LCM}(x,0)=0$ and $\text{GCD}(x,0)=x$, we get
$$F(x,y,0)=\text{LCM}(x,y)$$
$$F(x,0,0)=0$$
$$F(0,0,0)=0.$$
Generalizing to $n$ variables (and arbitrary lattices), we can define functions
$$\begin{align}F_1(x_1,\cdots,x_n)&=\text{GCD}\Big(\text{LCM}(x_1),\text{LCM}(x_2),\text{LCM}(x_3),\cdots,\text{LCM}(x_n)\Big) \\ F_2(x_1,\cdots,x_n)&=\text{GCD}\Big(\text{LCM}(x_1,x_2),\text{LCM}(x_1,x_3),\text{LCM}(x_2,x_3),\cdots,\text{LCM}(x_{n-1},x_n)\Big) \\ F_3(x_1,\cdots,x_n)&=\text{GCD}\Big(\text{LCM}(x_1,x_2,x_3),\cdots,\text{LCM}(x_{n-2},x_{n-1},x_n)\Big) \\ &\;\;\vdots \\ F_n(x_1,\cdots,x_n)&=\text{GCD}\Big(\text{LCM}(x_1,x_2,x_3,\cdots,x_n)\Big)\end{align}$$
and
$$\begin{align}G_1(x_1,\cdots,x_n)&=\text{LCM}\Big(\text{GCD}(x_1),\text{GCD}(x_2),\text{GCD}(x_3),\cdots,\text{GCD}(x_n)\Big) \\ G_2(x_1,\cdots,x_n)&=\text{LCM}\Big(\text{GCD}(x_1,x_2),\text{GCD}(x_1,x_3),\text{GCD}(x_2,x_3),\cdots,\text{GCD}(x_{n-1},x_n)\Big) \\ G_3(x_1,\cdots,x_n)&=\text{LCM}\Big(\text{GCD}(x_1,x_2,x_3),\cdots,\text{GCD}(x_{n-2},x_{n-1},x_n)\Big) \\ &\;\;\vdots \\ G_n(x_1,\cdots,x_n)&=\text{LCM}\Big(\text{GCD}(x_1,x_2,x_3,\cdots,x_n)\Big).\end{align}$$
It follows easily that
$$\text{GCD}=F_1\mid F_2\mid F_3\mid\cdots\mid F_n=\text{LCM}$$
and
$$\text{GCD}=G_n\mid G_{n-1}\mid\cdots\mid G_2\mid G_1=\text{LCM}.$$
Furthermore, $F_k=G_{n+1-k}$ in any distributive lattice. And for the special case of a total order, $(F_1,F_2,F_3,\cdots,F_n)$ is a permutation of $(x_1,x_2,x_3,\cdots,x_n)$.
$\textbf{Hint:}$ Consider one prime at a time.Say we have a prime $p$.
Its highest exponent dividing $m,n,k$ are respectively $a_1,a_2,a_3$ Now,if we can deduce that $a_1=a_2$ for every prime we consider,we would be done.
$\textbf{Solution:}$consider one side: $lcm(m,m+k)$. If $a_1 \ge a_3$ then, $m+k$ has highest exponent $a_3$ and m has $a_1$.then,$lcm(m,m+k)$ has $a_1$ since its the greater one.
Again,if $a_1 \le a_3$ then,$m+k$ has highest exponent $a_1$ ,so is $m$. Hence, $lcm(m,m+k)$ has $a_1$ as its highest divisor. same is true for the other side of the equation.
Since,both sides are equal $a_1=a_2$ for every prime we consider.
Best Answer
For any prime $p$, let $p_a$ be its exponent in the prime factorisation of $a$, and similarily define $p_b, p_c$. Thus, for instance, $a = 2^{2_a}\cdot 3^{3_a}\cdot 5^{5_a}\cdots$, where most of these factors are $1$.
Now consider the exponent of $p$ in the prime factorisations of the left and right hand sides. On the left-hand side it's $\max(p_a, \min(p_b, p_c))$ and on the right-hand side it's $\min(\max(p_a, p_b), \max(p_a, p_c))$. We see that if $p_a<p_b, p_c$, then we get $\min(p_b, p_c)$ on both sides, and otherwise we get $p_a$ on both sides.
In either case they're both equal, and since $p$ was arbitrary this means that the prime factorisation of $\operatorname{lcm}(a, \gcd(b, c))$ is the same as the prime factorisation of $\gcd(\operatorname{lcm}(a, b), \operatorname{lcm}(a, c))$, so the two numbers must be equal.