Prove $(ah,bk)=(a,b)(h,k)\left(\frac{a}{(a,b)},\frac{k}{(h,k)}\right)\left(\frac{b}{(a,b)},\frac{h}{(h,k)}\right)$ I had an idea and was wondering if it works. I seem to have gotten away quite cheaply. The many multiplications on the right side made me consider the prime divisors: 
$p\mid (ah,bk)\Rightarrow p\mid ah \land p\mid bk \Rightarrow \left(p\mid a\lor p\mid h\right)\land\left(p\mid b\lor p\mid k\right)\Rightarrow \left(p\mid a\land p\mid b\right)\lor\left(p\mid a\land p\mid k\right)\lor\left(p\mid h\land p\mid b\right)\lor\left(p\mid h\land p\mid k\right)$. 
Which matches exactly the right side! I am almost tempted to replace the implications with equivalences, use the definition of the gcd and call it a day! What do you think?
Reference
Apostol, Introduction to Analytic Number Theory, exercise 1.24.
 A: $\rm\color{#c00}{Divide}$ it by $\,(a,b)(h,k),\ $ let $\,A = \frac{a}{(a,b)},\, B = \frac{b}{(a,b)},\, H = \frac{h}{(h,k)},\, K = \frac{h}{(h,k)}$
reduces it to $\, (AH,BK) = (A,K) (B,H)\,\ $ if $\,\ (A,B)\! =\! 1\! =\! (H,K),\, $
by gcd Distributive Law.
$\rm\color{#c00}{Divide}$ prior by $\,(A,K)(B,H),\,$ let $\,a_1 = \frac{A}{(A,K)},\, b_1 = \frac{B}{(B,H)},\, h_1 = \frac{H}{(B,H)},\, k_1 = \frac{K}{(A,K)}$
reduces it  to $\,(a_1h_1,b_1k_1)= 1\,\ $ if $\ \,1 = (a_1,k_1) = (b_1,h_1)\,\smash[b]{\underbrace{= (a_1,b_1) = (h_1,k_1)}}$
whose proof is clear by Euclid's Lemma or FTA $ $ [We used that $\,a_1,\,b_1$ are coprime since they are respective factors of the coprimes $\,A,B.\,$ Similarly for $\,h_1,k_1\,$]. Recall FTA = Fundamental Theorem of Arithmetic, i.e. the existence and uniqueness of prime factorizations.
Remark $ $ Homogeneous problems like the above are generally simplified (as explained here) by reducing to the coprime case by $\rm\color{#c00}{cancelling\ gcds}$ throughout (using gcd & lcm distributive laws).
