Abhay's Math Page

Abhay's Math

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) = 3

Therefore, 0 < = (a - 1)(b - 1) < = (c - 1)(d - 1) < = 3

This is possible only in two cases

Case I
(a - 1)(b - 1) = 0 and (c - 1)(d - 1) = 3
implies a = b = 1 , c = 2 , d = 4
or a = 1 , b = c = 2 , d = 4

Case II
(a - 1)(b - 1) = 1 and (c - 1)(d - 1) = 2
implies a = b = c = 2 , d = 3

So there are 3 sets of solutions { (1,1,2,4) , (1,2,2,4) , (2,2,2,3) }