Proving that if $\gcd(m, n) = 1$, and if $d | mn$, then there exist unique numbers $a$ and $b$ such that $a|m$, $b|n$, and $d = ab$. What do I know?
If d | mn, there exist an integer k such that dk = mn.
I also know that because gcd(m, n) = 1 there exist some integers x and y such that mx + ny = 1.
I am having trouble to prove the statement because I don't even know how to start. Am I missing a key insight?
 A: *

*List item

We claim that $d=\gcd(d,m) \cdot \gcd(d,n)$.
Since $d \mid mn$ and $\gcd(m,n)=1$,
$$ d = \gcd(d,mn) = \gcd(d,m) \cdot \gcd(d,n). $$

*

*List item

We next claim that $d=ab$ with $a \mid m$ and $b \mid n$ implies $a=\gcd(d,m)$ and $b=\gcd(d,n)$.
Note that $a$ is a common divisor of $d$ and $m$. Therefore, $a \mid \gcd(d,m)$. If $a \ne \gcd(d,m)$, then there exists a prime $p$ such that $pa \mid \gcd(d,m)$. But now from $p \mid pa$ and $pa \mid m$, we have $p \mid m$. Moreover, $pa \mid d$ and $a \mid d$ implies $p \mid \frac{d}{a}=b$. This implies $p \mid n$, and leads to the contradiction that $m$ and $n$ share $p$ as a common factor. Thus, $a=\gcd(d,m)$.
Analogously, we have $b=\gcd(d,n)$. $\blacksquare$
A: Since $(m,n)=1$, $\exists x,y:$
$$
1=mx+ny\tag1
$$
Multiply $(1)$ by $d\frac{(m,d)(n,d)}{(m,d)(n,d)}$:
$$
d=(m,d)(n,d)\left(\frac{d}{(n,d)}\frac{m}{(m,d)}x+\frac{d}{(m,d)}\frac{n}{(n,d)}y\right)\tag2
$$
$(2)$ says that $(m,n)=1\implies(m,d)(n,d)\mid d$.
Since $d\mid mn$, we have
$$
d=(mn,d)|(m,d)(n,d)\tag3
$$
$(3)$ is true because
$$
\overbrace{(mx_1+dy_1)}^{(m,d)}\overbrace{(nx_2+dy_2)}^{(n,d)}=\overbrace{mn(x_1x_2)+d(nx_2y_1+dy_1y_2+mx_1y_2)}^\text{divisible by $(mn,d)$}\tag4
$$
$(3)$ says that $d\mid mn\implies d\mid(m,d)(n,d)$.
Therefore, $(2)$ and $(3)$ say that $(m,n)=1$ and $d\mid mn$ imply that
$$
d=\overset{a\\\raise{3pt}{\downarrow}}{(m,d)}\overset{b\\\raise{3pt}{\downarrow}}{(n,d)}\tag5
$$
Since $a$ is the greatest factor of $d$ that divides $m$ and $b$ is the greatest factor of $d$ that divides $n$, neither can be greater. Their product is $d$, so neither can be less. Thus, they are unique.
A: We have that $dk_0 = mn$
Let $a,b \in \mathbb{N}$ such that $a,b\gt1$ and $a|m $ and $b |n$ (Assuming $m,n\gt1$)
So, $m = ak_1$ and $n = bk_2$. Thus, $mn = abk_1k_2$ giving $dk_0 = abk_1k_2$.
Since $k_1|m$ and $k_2|n$, we know that $k_1k_2|mn$.
Now we can let $k_0 = k_1k_2$, giving $d = ab$.
Finally, since $gcd(m,n)=1$, we know that $a$ and $b$ must be unique. ($b\ne a)$
