Abhay's Math Page

Abhay's Math

			ab + cd = a + b + c + d + 2

			ab - a + cd - c = b + d + 2

			a(b - 1) + c(d - 1) = (b - 1 + 1) + (d - 1 + 1) + 2

			(a - 1)(b - 1) + (c - 1)(d - 1) = 4

Since 1 < = a < = b < = c < = d

So, 0 < = (a - 1) < = (b - 1) < = (c - 1) < = (d - 1)

Hence 0 < = (a - 1)(b - 1) < = (c - 1)(d - 1)

But (a - 1)(b - 1) + (c - 1)(d - 1) = 4

Therefore, 0 < = (a - 1)(b - 1) < = (c - 1)(d - 1) < = 4

This is possible only in three cases

Case I
(a - 1)(b - 1) = 0 and (c - 1)(d - 1) = 4

Case II
(a - 1)(b - 1) = 1 and (c - 1)(d - 1) = 3

Case III
(a - 1)(b - 1) = 2 and (c - 1)(d - 1) = 2

Therefore the final solution set (after considering all the cases) is { (1,1,2,5) , (1,2,2,5) , (1,1,3,3) , (1,2,3,3) , (1,3,3,3) }