2024/045) Solve in positive integer $\frac{x}{y+7} + \frac{y}{x+7}= 1$

We are given  $\frac{x}{y+7} + \frac{y}{x+7}= 1$

or $x(x+7) + y(y+7) = (x+7)(y+7)$

or $x^2+7x + y^2 +7y = xy + 7x + 7y + 49$

or $x^2 + y^2 -xy = 49$

now $x^2- xy$ we need to form in form of squares by adding some teerm

let us multiply by 4 to above to get

    $4x^2 +  - 4xy + 4y^2 = 196$

or $4(x^2-4xy + y^2) +3y ^2 = 196$

or $(2x-y)^2 + 3y^2 = 196$

as it is sum of squares  we need to check a finite number of values(fron pair of values we consider only positive ones) 

$y = 3$ gives $2x-y =13$ giving $x = 8, y = 3$

$y= 5$ gives $2x-y = 11$ giving$ x = 8, y = 5$

$y = 7$ gives $2x -y = 7$ giving $x = y = 7$

$y = 8$ gives $2x -y = 2$ giving $x = 5 y = 8$

         or $2x - y = -2$ giving $x = 3 and y = 8$

so the solutions are $(x=8,y=3), (x=8,y=5), (x=7,y=7),(x=5,y=8), (x=3,y=8)$