These two exercises I encountered recently seem to develop some type of connection between GCD and LCM I can't quite figure out.
Exercise 1:
Find all the numbers $x$ and $y$ such that:
$a) \ GCD(x,y)=15, \ LCM(x,y)=150$ $b) \ GCD(x,y)=120 \ LCM(x,y)=1320$
$c) \ GCD(x,y)=100 \ LCM(x,y)=990$
Exercise 2:
Find all the numbers $m,n$ such that $GCD(m,n)=pq , \ LCM(m,n)=p^2qs$
where $p,q,s$ are prime
The first thing that is known to me is that $GCD(x,y) \cdot LCM(x,y)= x \cdot y$
Also $LCM(x,y)$ is at most $x \cdot y$ while $GCD(x,y)$ is at most $\max \{x,y\}$. Last thing is that $GCD(x,y)|LCM(x,y)$.
Using all this I tried to solve the first exercise:
$a)$ First two obvious pairs are $x=15, y=150$ and $y=15, x=150$. Now neither of the numbers can be bigger than $150$ or smaller than $15$. So we are looking for numbers in the range $15-150$ that satisfy $x \cdot y = 15 \cdot 150$ Another such pair is $(x,y)=(30,75), \ (x,y)=(75,30)$.
Similarly for $b)$ we find that the only possible values are permutations of the set {$120,1320$} and in $c)$ since $100$ does not divide $990$ no such numbers exist.
Now exercise 2 is what made me think there is actually another connection I'm not quite aware of since now it's about arbitrary prime numbers and the previous method doesn't work anymore. My intuition is that it has something to do with $GCD$ or $LCM$ of the $GCD(x,y), \ LCM(x,y)$
Best Answer
$\begin{align}{\bf Hint}\ \ &\gcd(X,Y) = d,\ \ \ {\rm lcm}(X,Y) = m \ \ \ \text{yields by $\rm\color{#90f}{cancelling}$ $\,d\neq 0$}\\[.3em] \iff\ &\color{#0a0}{gcd(x,\,y)\ \ = 1},\qquad\ \ \, \color{#c00}{x\cdot y}\ =\, m/d,\ \ {\rm for}\ \ x = X/d,\,\ y = Y/d \end{align}$
$\begin{align}{\rm e.g.}\ \ \ &\gcd(X,Y) = 15,\ \, {\rm lcm}(X,Y) = 150\\[.3em] \iff\ &\gcd(x,\,y)\ \ =\ 1,\qquad\ \ \, x\cdot y\,\ =\,\ 10 \end{align}$
So it's equivalent to factoring $\,m/d = \color{#c00}{xy}\,$ into ${\rm \color{#0a0}{coprime}\ factors}\,\ \color{#0a0}{x,y}$.
This method quickly and simply solves all your problems.
$\rm\color{#90f}{Cancelling}$ to reduce to a coprime case is a common way to simplify homogeneous divisibility problems.