Determine all possible pairs (x,y) of positive integers that satisfy this equation:
- x2y2 - 4(x+y) = n2, where n is a nonnegative integer.
Lets start by creating an inequality (xy)^2 > x^2*y^2-4x-4y > (xy-1)^2.
For any (x,y) where this is true then it is impossible for x^2*y^2-4x-4y to be a perfect because it is bounded between two consecutive perfect squares.
So one part of the inequality must be false. Given x and y are positive then the left side inequality is always true.
Therefore any solutions must come from when the right side is false:
x^2*y^2-4x-4y <= (xy-1)^2.
Expand and move everything to the left side except the constants:
2xy-4x-4y <= 1
Now divide by 2, add 4 and factor the left side:
(x-2)*(y-2) <= 9/2
This inequality will have three families of solutions.
1: one of x or y equals 1. (The left side will be negative in most cases)
2: one of x or y equals 2. (The left side is zero)
3: specific cases (x,y) = (3,3), (3,4), (3,5), (3,6), (4,4), (4,3), (5,3), (6.3).
Case 1. Without loss of generality let x=1 and substitute into the original equation:
y^2 - 4y - 4 = n^2
Complete the square and rearrange a bit
(y-2)^2 - n^2 = 8
Then we need a pair of squares differing by 8, which is only given by 8=3^2-1^2. Then y-2=3, which yields y=5.
So we have (x,y)=(1,5) and its reflection (5,1).
Case 2. Without loss of generality let x=2 and substitute into the original equation:
4y^2 - 4y - 8 = n^2
Complete the square and rearrange a bit
(2y-1)^2 - n^2 = 9
Then we need a pair of squares differing by 9, which is given by 9=5^2-4^2 and 9 = 3^2-0^2. Then 2y-1=5 or 2y-1=3, which yields y=3 or y=2.
So we have (x,y)=(2,3) and its reflection (3,2) and (x,y)=(2,2).
Case 3: By direct evaluation none of (x,y) = (3,3), (3,4), (3,5), (3,6), (4,4), (4,3), (5,3), (6,3) give valid values for n^2.
So the set of all possible pairs (x,y) of positive integers that satisfy x^2*y^2-4x-4y = n^2 are (1,5), (5,1), (2,3), (3,2), (2,2).
Solution
