Choose the smallest number in this list is a, and b = a+1, c = a+2, d = a+3

So abcd + 1 = a(a+1)(a+2)(a+3) + 1 = a(a+3).(n+1)(n+2) + 1

= (a^{2} + 3a)(n^{2} + 3n +2) + 1

= [a^{2} + 3a][(n^{2} + 3n) + 2] + 1

= (a^{2} + 3a)^{2} + 2.(a^{2} + 3a).1 + 1^{2} = (a^{2} + 3a + 1)^{2 }is a square number