What is the number of all possible positive integer values of n for which n^{2} + 96 is a perfect square?

- 2
- 4
- 5
- Infinite

**Answer**

Let n^{2} + 96 = m^{2} where m is a positive integer.

m^{2} - n^{2} = 96

(m - n)(m + n) = 96 = 2^{5} × 3

Divisors of 96 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

Pairs of (m-n) and (m+n) can be 1,96 ; 2,48 ; 3,32 ; 4,24 ; 6,16 ; 8,12

Subtracting first value from the second,

(m+n) - (m-n) = 2n

2n = 95, 46, 29, 20, 10, 4

n = 47.5, 23, 14.5, 10, 5 and 2

Required integer values of n are 23, 10, 5 and 2.

**The correct option is B.**