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 $d = ab$.

Question

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?

Answer 1

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. ($b\ne a)$

Answer 2

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.