Problem from BdMO 2021

Let $\text{gcd}(x,y)=h$ , and let $x=x'h$ , $y=y'h$ (note that $x'$ and $y'$ share no common factors). Then $\text{lcm}(x,y)=x'y'h$ . The quantity we want to find is now $x'y'$ and the equation becomes $\begin{aligned} 2h\left(x'+y'\right)&=x'y'h+h \\ x'y'-2\left(x'+y'\right)+1 &=0 \\ x'y'-2\left(x'+y'\right)+4 &=3 \\ \left(x'-2\right)\left(y'-2\right) &=3 \end{aligned}$

The only solution to this in positive integers is $(x',y')=(3,5)$ (or $(5,3)$ ), and it's easy to verify these satisfy the original equation; hence the answer we need is $\boxed{15}$ .

It is clear that no solution for $x = y$

$LHS = 2 (x + y ) = 2 (x + x ) = 4x$

$RHS = x + x = 2x$

WLOG, assume $x < y$

$LHS = 2 (x+y) = 2x + 2y < 4y \implies \text{lcm}(x,y) =y, 2y, 3y$

There is no solution for $\text{lcm}(x,y) = y , 2y$

$2(x+y) = \text{lcm}(x,y) + \text{gcd}(x,y) \implies 2x + 2y - \text{lcm}(x,y) = \text{gcd}(x,y) \implies \text{gcd}(x,y) > x \implies \text{contradiction}$

For $\text{lcm}(x,y) = 3y$

$2(x+y) = 3y + \text{gcd}(x,y) \implies \text{gcd}(x,y) = 2x-y = a$

For some integer $k_1 < k_2$ , let $\text{gcd}(x,y) = 2x-y = a$

$x = k_1 a , y = k_2 a$

$2x-y=a \implies 2k_1 - k_2 = 1$

$\implies k_2 = 2k_1 - 1$

Since $k_1$ and $k_2=2k_1 - 1$ are relatively prime,

$\text{lcm}(x,y) = k_1 k_2 a =3y= 3 k_2 a \implies k_1 = 3, k_2 = 5 \implies \text{lcm}(x,y) = 15a$

$\therefore \cfrac{\text{lcm}(x,y)}{\text{gcd}(x,y)} = \cfrac{15a}{a} = \boxed{15}$