Find all integer pairs (x, y) that satisfy the equation
x y
----- + ----- = 1
y + 7 x + 7
Multiply through to clear fractions:
x(x+7) + y(y+7) = (x+7)(y+7)
Distribute and simplify:
x^2 - xy + y^2 = 49
Now I will multiply through by 4 and convert this into a sum of the form a*(x+y)^2 + b*(x-y)^2:
4*(x^2 - xy + y^2) = 196
3*(x^2 - 2xy + y^2) + 1*(x^2 + 2xy + y^2) = 196
3*(x-y)^2 + (x+y)^2 = 196
Without loss of generality assume x>=y
We will need x+y >= x-y and x+y<=14. Together these will limit the range of values to 7 <= x+y <= 14. Then it is simple enough to test by hand to get four viable answers:
x+y=14 and x-y=0 which makes (x,y) = (7,7).
x+y=13 and x-y=3 which makes (x,y) = (8,5).
x+y=11 and x-y=5 which makes (x,y) = (8,3).
x+y=7 and x-y=7 which makes (x,y) = (7,0).
Then since originally, x and y are symmetric the ordered pairs can also be reversed leading to seven solutions: (x,y) = (0,7), (3,8), (5,8), (7,0), (7,7), (8,3), and (8,5).
Solution
