Prove identity involving $\operatorname{lcm}(x, y, z)$

Question

I received this problem to prove that: $$ \operatorname{lcm}(x, y, z) = \frac{xyz}{\operatorname{gcd}(x, y, z)\cdot\operatorname{gcd}(\operatorname{lcm}(x, y), \operatorname{lcm}(x, z), \operatorname{lcm}(y, z))} $$ Knowing that: $$\operatorname{lcm}(x, y) = \frac{xy}{\operatorname{gcd}(x, y)}$$

I started saying that: \begin{align*} \operatorname{lcm}(x, y, z) &= \operatorname{lcm}(\operatorname{lcm}(x, y), z)\\ &= \frac{z\cdot\operatorname{lcm}(x, y)}{\operatorname{gcd}(z, \operatorname{lcm}(x, y))}\\ &=\operatorname{lcm}(x, y, z)\\ &= \frac{z\cdot\operatorname{lcm}(x, y)}{\operatorname{gcd}(z, \operatorname{lcm}(x, y))}\\ &= \frac{xyz}{\operatorname{gcd}(x, y)\cdot\operatorname{gcd}(z,\operatorname{lcm}(x, y))} \end{align*}

And I got stuck, how do I prove that it equals:

$$ \frac{xyz}{\operatorname{gcd}(x, y, z)\cdot\operatorname{gcd}(\operatorname{lcm}(x, y), \operatorname{lcm}(x, z), \operatorname{lcm}(y, z))} $$

Answer 1

Lemma: For any nonnegative integers $a_i, b_i, c_i$, we have $$\mathrm{min}\{\mathrm{max}\{a_i, b_i\},\mathrm{max\{b_i, c_i}\},\mathrm{max\{c_i, a_i}\}\} + \mathrm{min}\{a_i, b_i, c_i\} +\mathrm{max}\{a_i, b_i, c_i\} = a_i + b_i + c_i.$$

Proof: Without loss of generality, we can impose an arbitrary ordering, say $a_i \leq b_i \leq c_i$. Then $$\mathrm{min}\{\mathrm{max}\{a_i, b_i\},\mathrm{max\{b_i, c_i}\},\mathrm{max\{c_i, a_i}\}\} + \mathrm{min}\{a_i, b_i, c_i\} +\mathrm{max}\{a_i, b_i, c_i\} = \mathrm{min\{b_i, c_i, c_i\} + a_i + c_i = b_i + a_i + c_i = a_i + b_i + c_i}. \square$$

Now, we have $$\max\{a_i, b_i, c_i\} = a_i + b_i + c_i - \mathrm{min}\{\mathrm{max}\{a_i, b_i\},\mathrm{max\{b_i, c_i}\},\mathrm{max\{c_i, a_i}\}\} - \mathrm{min}\{a_i, b_i, c_i\}$$ so that \begin{align*} \mathrm{lcm}(x, y, z) = \prod_{p}p_i^{\max\{a_i, b_i, c_i\}} &= \frac{\prod_p p_i^{a_i + b_i + c_i}}{\prod_p p_i^{\mathrm{min}\{\mathrm{max}\{a_i, b_i\},\mathrm{max\{b_i, c_i}\},\mathrm{max\{c_i, a_i}\}\}} \prod_{p} p_i^{\mathrm{min}\{a_i, b_i, c_i\}}} \\ &= \frac{xyz}{\operatorname{gcd}(x, y, z)\cdot\operatorname{gcd}(\operatorname{lcm}(x, y), \operatorname{lcm}(x, z), \operatorname{lcm}(y, z))} \end{align*} as desired. $\blacksquare$

Answer 2

It's enough to do it one prime at a time.

Let $p$ be a prime that divides one or more of $x,y$ or $z$, and let $k_x$ be the highest power so that $p^{k_x}|x$ and $k_y$ the highest power so that $p^{k_y}|y$ and so on (note: some of these may be equal to $0$).

By symmetry we can assume $k_x \le k_y \le k_z$. If that seems too glib, just relabel that $w = $ the $x,y$ or $z$ where $p$ divides by the lowest power, $u=$ the $x,y$ or $z$ where $p$ divides by the middle power, and $v = x,y$ or $z$ where $P$ divides by the highest.

The highest power of $p$ that divides $lcm(x,y,z)=lcm(w,u,v)$ is $p^{\max(k_x,k_y,k_z)} = p^{k_v}$.

The highest power of $p$ that divide the RHS of the equation id $p^{(k_x+ k_y + k_z)- [\min(k_x,k_y,k_z) + \min(\max(k_x,k_y),\max(k_x,k_z),\max(k_y,k_z)]}=$

$p^{(k_w+ k_u + k_v)- [\min(k_w,k_u,k_v) + \min(\max(k_w,k_u),\max(k_w,k_v),\max(k_u,k_v)]}=$

$p^{(k_w+ k_u + k_v)- [k_w + \min(k_u,k_v,k_v)]}=$

$p^{(k_w+ k_u + k_v)- [k_w + k_u]}=$

$p^{k_v}=$

$p^{\max(k_x,k_y,k_z)}$.

So for every $p$ that divides any of $x,y,z$ the power of $p$ that divides both sides is exactly the same!

So do that for every prime that divides $x,y,$ or $z$ and the result follows.