Prove $(ah,bk)=(a,b)(h,k)\left(\frac{a}{(a,b)},\frac{k}{(h,k)}\right)\left(\frac{b}{(a,b)},\frac{h}{(h,k)}\right)$

Question

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.

Answer 1

Homogeneous problems like the above question are generally simplified (as explained here) by reducing to $\rm\color{#0a0}{coprime}$ case by $\rm\color{#c00}{canceling\ gcds}$ throughout (using gcd & lcm distributive laws).

$\rm\color{#c00}{Divide}$ it by $\,(a,b)(h,k),\ $ let $\,A = \frac{a}{(a,b)},\, B = \frac{b}{(a,b)},\, H = \frac{h}{(h,k)},\, K = \frac{k}{(h,k)}$

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 $\ \,1 = \color{#0a0}{(a_1,k_1) = (b_1,h_1)}\,\smash[b]{{= \color{#0af}{(a_1,b_1) = (h_1,k_1)}}}$

true by Euclid's Lemma or coprimes to $\,n\,$ are closed under products, since $\rm\color{#0a0}{greens} = 1$ by here, and $\rm\color{#0af}{blues} = 1$ being respective factors of coprimes $\,A,B;\,$ and $\,H,K$ resp.

Remark $ $ To discover the formula (vs. verify it as above), we apply the gcd distributive law to factor out $\,(a,b)(h,k)\,$ from $\,(ah,bk),\,$ leaving $\,(AH,BK),\,$ then from this we factor out $(A,K)(B,H),\,$ leaving $(a_1h_1,b_1k_1)$ which $=1$ as proved above. This proves that $\,(ah,bk) = (a,b)(h,k)(A,K)(B,H)=\,$ sought formula.