Find all pairs of positive integers that add up to $667$ and their $\frac{\text{lcm}}{\text{gcd}} =120$ 
Let $x,y$ be positive integers, $x<y$, and $x+y=667$. Given that $\dfrac{\text{lcm}(x,y)}{\text{gcd}(x,y)}=120,$ find all such pairs $(x,y)$.

The only way I can think of solving this is trying all possibilities where one number is odd and the other even, and testing them all. Using this, I found one solution, $(115,552)$, but I'm wondering if there is a more efficient way to do this problem.
 A: If a number divides $x$ and $y$ then the number divides $x+y$. Therefore $\gcd(x,y)$ is a divisor of $667$.
Since $667=23\times 29$ there are only $4$ cases.
If $\gcd(x,y)=1$ we get $\text{lcm}(x,y)=120$, clearly impossible since one of the numbers is larger than $120$.
If $\gcd(x,y)=23$ then $\text{lcm}(x,y)=23\times 120$, dividing everything by $23$ we must find coprime $x'$ and $y'$ that add to $29$ and have product equal to $120$. Solving the quadratic equation leads to $(5,24)$ and $(24,5)$
If $\gcd(x,y)=29$ then we must find coprime $x'$ and $y'$ that add to $23$ and have product $120$. Solving the quadratic leads to $(8,15)$ and $(15,8)$
If $\gcd(x,y)=23\times 29$ then we must find $x',y'$ that add to $1$ and have product $120$, clearly impossible.

So the only solutions are $(5\times 23,24\times 23),(24\times 23, 5\times 23),(8\times 29, 15\times 29),(15\times 29,8\times 29)$
A: Let $\dfrac xX=\dfrac yY=(x,y)=d$(say) $\implies(X,Y)=1$
$120=\dfrac{XYd}d=XY$
As $120=1\cdot120,2\cdot60,3\cdot40,4\cdot30,5\cdot24,6\cdot20,8\cdot15,10\cdot12$
The possible set of values of $(X,Y)$ such that $X<Y;(X,Y)=1$ are $\{(1,120);(3,40);(5,24);(8,15)\}$
Now use the fact that $\displaystyle667=x+y=d(X+Y)\implies\dfrac{23\cdot29}{X+Y}=d$ which is a positive integer to find the possible set of values of $(X,Y)$ to be $\{(5,24);(8,15)\}$
