How many pairs of positive integers (x,y) satisfy the equation x^2-y^2=18?
Because x and y > 0, x must be greater than y.
We factor x^2-y^2=18 into
Because x and y both must be positive integers, x+y and x-y are also both positive integers. Using this, we also find that x+y >x-y
18 factors into
1*18 -> x+y = 18, x-y = 1. No integer solution
2*9 -> x+y = 9, x-y = 2, No integer solution.
3*6 -> x+y = 6, x-3 = 3, No integer solution.
Thus, we find that there are no solutions.
0 pairs total.