1. ## GCD Multiplicative Property

i'm reading Introduction to Analytic Number Theory, Tom M. Apostol and i have a doubt about an exercise in chapter 1. Follow the exercise:

Prove the following multiplicative property of the gcd:

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

i already tried using Bezout identity and $(a,b)=\frac{a\cdot b}{[a,b]}$

2. Let
$(a,b)=m$,
$(h,k)=n$,
$a=ma_{1}$,
$b=mb_{1}$,
$h=nh_{1}$,
$k=nk_{1}$,
then $(a_{1},b_{1}), (h_{1},k_{1})=1$

Then the question becomes:
$(a_{1}h_{1},b_{1}k_{1})=(a_{1},k_{1})(b_{1},h_{1})$

$(\frac{a_{1}}{(a_{1},k_{1})}\times\frac{h_{1}}{(b_ {1},h_{1})},\frac{k_{1}}{(a_{1},k_{1})}\times\frac {b_{1}}{(b_{1},h_{1})})=1$, which is obviously true.

EDIT: Then I recommend waiting for the more experienced forum members to answer.

3. I think that your way is not formal. A prove different

4. Originally Posted by streethot
i'm reading Introduction to Analytic Number Theory, Tom M. Apostol and i have a doubt about an exercise in chapter 1. Follow the exercise:

Prove the following multiplicative property of the gcd:

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

i already tried using Bezout identity and $(a,b)=\frac{a\cdot b}{[a,b]}$
it's obvious that if $\gcd(x,y)=\gcd(x,z)=\gcd(t,y)=\gcd(t,z)=1,$ then $\gcd(xt, yz)=1. \ \ \ \ \ \ \ (1)$

now let $a_1=\frac{a}{\gcd(a,b)}, \ b_1=\frac{b}{\gcd(a,b)}, \ k_1=\frac{k}{\gcd(k,h)}, \ h_1=\frac{h}{\gcd(k,h)},$ and:

$x=\frac{a_1}{\gcd(a_1,k_1)}, \ y=\frac{b_1}{\gcd(b_1,h_1)}, \ z=\frac{k_1}{\gcd(a_1,k_1)}, \ t=\frac{h_1}{\gcd(b_1,h_1)}.$ it is obvious that $x,y,z,t$ satisfy

the conditions in $(1)$ and thus we have $\gcd(xt,yz)=1,$ which is exactly your identity!