Divisors of a product. Is there a proof that if $d \mid mn$, where $m$ and $n$ are coprime, then $d=d_1d_2$ where $d_1 \mid m$ and $d_2 \mid n$, where the $d_i$ are comprime? I was working on Project Euler and came across this fact but the link to the proof had expired.

$\color{#c00}{cd = mn}\,\Rightarrow\, (d,m)(d,n) = ((d,m)d,(d,\color{#c00}m)\color{#c00}n) = (d,\color{#0a0}{m,n},\color{#c00}c)\color{#c00}d = d,\,$ by $\color{#0a0}{(m,n)=1}$

Hence $\ d = (d,m)(d,n)\ $ where $\ (d,m)\mid m,\,\ (d,n)\mid n\ \ \ $ **QED**

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**Or** $\,\ d\mid mn\iff d\mid dn,mn \color{#c0f}\iff d\mid (dn,mn)\overset{\color{#0a0}{\rm D\,L}} = (d,m)n \iff d/(d,m)\mid n$

Therefore $\ d\, =\, (d,m) \dfrac{d}{(d,m)} \ $ where $\ (d,m)\mid m\ $ and $\ \dfrac{d}{(d,m)}\mid n\ \ \ $ **QED**

Above we used $\, d\mid j,k\color{#c0f}\iff d\mid (j,k),\,$ the universal property of the gcd, and we also used $\,\rm\color{#0a0}{DL} =$ the gcd Distributive Law, both here and in the first proof.