TITLE: How to show $(x \wedge m)(x \wedge n) = x$ when $m \wedge n = 1$ and $x \mid mn$
QUESTION [5 upvotes]: I have been trying to evaluate this GCD with different manners:

For instance: $(x \wedge m)(x \wedge n) = \prod_{p \in \mathbb{P}} p^{\min\{ v_p(x), v_p(m) \}} \times \prod_{p \in \mathbb{P}} p^{\min \{ v_p(x), v_p(n) \}}$

I am not sure how to show that $\min\{ v_p(x), v_p(m) \} + \min\{v_p(x), v_p(n) \} = v_p(x)$.

Also, I have been trying to rewrite $x$ as a divisor of $mn$ and extract some common factors without success.
Euclid algorithm didn't work.

What is the evident way to show this without lengthy proof using decomposition in prime numbers of $x, m, n$?
EDIT2: The proof posted earlier was false.
Notations : $\wedge$ is the GCD. $\mathbb{P}$ is the set of prime numbers.

REPLY [2 votes]: Since $m\land n=1$, for each $p$ you have either $v_p(m)=0$ or $v_p(n)=0$. Since $x\mid mn$, $v_p(x)\le v_p(mn)=v_p(m)+v_p(n)$.
If $v_p(m)=0$, then
$$
\min\{ v_p(x), v_p(m) \} + \min\{v_p(x), v_p(n) \} = 
0+\min\{v_p(x), v_p(n) \}=
\min\{v_p(x), v_p(mn) \}=v_p(x)
$$
Similarly if $v_p(n)=0$.

Alternative proof. Since $m\land n=1$, also $(x\land m)\land(x\land n)=1$,
$$
(x\land m)(x\land n)=(x\land m)\lor(x\land n)
$$
By distributivity,
$$
(x\land m)\lor(x\land n)=x\land(m\lor n)=x\land (mn)=x
$$