Inverse chaos

Let $\gcd(x, y) = d$ for some positive integer $d.$ We can write $x = dx_1, y = dy_1,$ where $\gcd(x_1, y_1) = 1$ since otherwise the greatest common divisor of $x$ and $y$ would not be $d.$ We can also write

$\dfrac{1}{\text{lcm}(x, y)} = \dfrac{\gcd(x, y)}{xy} = \dfrac{d}{dx_1dy_1} = \dfrac{1}{dx_1y_1}.$

Our equation rewrites as

$\begin{aligned} \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{\gcd(x, y)} + \dfrac{1}{\text{lcm}(x, y)} &= \dfrac{1}{2} \\ \dfrac{1}{dx_1} + \dfrac{1}{dy_1} + \dfrac{1}{d} + \dfrac{1}{dx_1y_1} &= \dfrac{1}{2} \\ \dfrac{1}{d} \left (\dfrac{1}{x_1} + \dfrac{1}{y_1} + 1+ \dfrac{1}{x_1y_1} \right) &= \dfrac{1}{2} \\ \dfrac{1}{x_1} + \dfrac{1}{y_1} + \dfrac{1}{x_1y_1} &= \dfrac{d - 2}{2}. \end{aligned}$

Notice that this equation implies $d \geq 3,$ since the LHS of the equality is a positive value. Also, if we set $x_1, y_1 = 1,$ we get the maximum possible value of $\frac{1}{x_1} + \frac{1}{y_1} +\frac{1}{x_1y_1},$ which is 3. So, $\frac{d - 2}{2} \leq 3,$ or $d \leq 8.$

Continuing on, we have

$\begin{aligned} \dfrac{1}{x_1} + \dfrac{1}{y_1} + \dfrac{1}{x_1y_1} &= \dfrac{d - 2}{2} \\ 2(x_1 + y_1 + 1) &= (d - 2)x_1y_1 \\ (d - 2)x_1y_1 - 2x_1 - 2y_1 &= 2 \\ (d - 2)^2 x_1y_1 - 2(d - 2)x_1 - 2(d - 2)y_1 &= 2(d - 2) \\ (d - 2)x_1[(d - 2)y_1 - 2] - 2[(d - 2)y_1 - 2] &= 2(d - 2) + 4 \\ [(d - 2)x_1 - 2][(d - 2)y_1 - 2] &= 2d. \end{aligned}$

Now, we examine each possible value of $d$ :

$\begin{array}{c|c|c|c} d & \text{Corresponding equation} & \text{Solutions for } (x_1, y_1) & \text{Solutions for } (x, y) \\ \hline 3 & (x_1 - 2)(y_1 - 2) = 6 & (3, 8), (4, 5), (5, 4), (8, 3) & (9, 24), (12, 15), (15, 12), (24, 9) \\ 4 & (x_1 - 1)(y_1 - 1) = 2 & (2, 3), (3, 2) & (8, 12), (12, 8) \\ 5 & (3x_1 - 2)(3y_1 - 2) = 10 & (1, 4), (4, 1) & (5, 20), (20, 5) \\ 6 & (2x_1 - 1)(2y_1 - 1) = 3 & (1, 2), (2, 1) & (6, 12), (12, 6) \\ 7 & (5x_1 - 2)(5y_1 - 2) = 14 & \text{none} & \text{none} \\ 8 & (3x_1 - 1)(3y_1 - 1) = 4 & (1, 1) & (8, 8) \\ \end{array}$

In sum, there are $n = 11$ solutions to the original equation, which are:

$\begin{aligned} (x, y) &= (9, 24), (12, 15), (15, 12), (24, 9) \\ & \quad (8, 12), (12, 8), (5, 20), (20, 5) \\ & \quad (6, 12), (12, 6), (8, 8). \end{aligned}$

Thus, $\displaystyle \sum_{i = 1}^{11} (x_i^2 - y_i) = \boxed{1772}.$

I did more or less the same solution as Steven, just looking at bounding $x$ instead of $d.$

So write $x=dx_1,y=dy_1$ where $x_1,y_1$ are relatively prime. Then we have $\begin{aligned} \frac1{dx_1}+\frac1{dy_1}+\frac1{d}+\frac1{dx_1y_1} &= \frac12 \\ \frac2{x_1}+\frac2{y_1}+2+\frac2{x_1y_1} &= d \\ 2\left( 1+ \frac1{x_1} \right) \left( 1+\frac1{y_1} \right) &= d \end{aligned}$ Suppose for now that $x_1 \le y_1$ (i.e. $x \le y$ ).

Now suppose $x_1 \ge 3.$ Then the product on the left is at most $2 \cdot \frac43 \cdot \frac43 = \frac{32}9 < 4.$ So $d \le 3.$ But the left side is $> 2,$ so both sides must be exactly $3.$ In this case we get $\begin{aligned} \frac1{x_1}+\frac1{y_1}+\frac1{x_1y_1} &= \frac12 \\ 2y_1+2x_1+2 &= x_1y_1 \\ 6 &= (x_1-2)(y_1-2) \end{aligned}$ which leads to the solutions $(x_1,y_1) = (3,8),(4,5),$ which becomes $(x,y) = (9,24),(12,15).$

Now suppose $x_1=2.$ In this case we get $3 \left(1+\frac1{y_1} \right) = d,$ and the only way $d$ can be an integer is if $3/y_1$ is an integer, so $y_1 = 3.$ (Remember $x_1 \le y_1.$ ) This leads to the solution $(x,y) = (8,12).$

Finally, if $x_1=1,$ we get $4 \left( 1+\frac1{y_1} \right) = d,$ and this only works if $y_1 = 1,2,4,$ which leads to the solutions $(8,8), (6,12), (5,20).$

So there are six solutions in all with $x_1 \le y_1,$ which leads to eleven total solutions (not twelve because $(8,8)$ doesn't change when we switch $x$ and $y$ ). Doing the computation, we get an answer of $\fbox{1772}.$