ab + cd = a + b + c + d + 1
ab - a + cd - c = b + d + 1
a(b - 1) + c(d - 1) = {(b - 1) + 1} + {(d - 1) + 1} + 1
a(b - 1) + c(d - 1) - (b - 1) - (d - 1) = 1 + 1 + 1
(a - 1)(b - 1) + (c - 1)(d - 1) = 3
Where have we landed up ? Where to go ? Can we solve this to get a finite set of solutions ? The answer is no. We will have to take cases to solve this equation. Example we can have a case like (a - 1)(b - 1) = 1 and (c - 1)(d - 1) = 2 , or yet another case as (a - 1)(b - 1) = -5 and (c - 1)(d - 1) = 8. This way we can have infinite cases and infinite solutions. So as to get a finite number of solutions, there must be atleast one more condition.
Let the condition be 1 < = a < = b < = c < = d