What is the number of all possible positive integer values of ‘n’ for which n² + 96 is a perfect square ?
2
4
5
Infinite
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC CISF-AC-EXE – 2017
Let n² + 96 = k², where k is an integer.
Since n is a positive integer, n² is positive, so k² must be greater than n². This implies k > n (assuming k is positive, which it must be since k² = n² + 96).
Rearranging the equation, we get k² – n² = 96.
This is a difference of squares, which can be factored as (k – n)(k + n) = 96.
Since k and n are integers, (k-n) and (k+n) must be integer factors of 96.
Also, (k+n) – (k-n) = 2n. Since n is an integer, 2n is an even integer. This means (k-n) and (k+n) must have the same parity. Since their product (96) is even, both factors must be even.
Furthermore, since n is positive, k+n > k-n. Also, k+n > 0 (as n>0 and k>n).
The factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96.
The even factors are 2, 4, 6, 8, 12, 16, 24, 32, 48, 96.
Pairs (a, b) where a and b are even, a*b=96, and b > a:
1. (2, 48) -> k – n = 2, k + n = 48. Adding gives 2k=50, k=25. Subtracting gives 2n=46, n=23. (n=23 is a positive integer)
2. (4, 24) -> k – n = 4, k + n = 24. Adding gives 2k=28, k=14. Subtracting gives 2n=20, n=10. (n=10 is a positive integer)
3. (6, 16) -> k – n = 6, k + n = 16. Adding gives 2k=22, k=11. Subtracting gives 2n=10, n=5. (n=5 is a positive integer)
4. (8, 12) -> k – n = 8, k + n = 12. Adding gives 2k=20, k=10. Subtracting gives 2n=4, n=2. (n=2 is a positive integer)
We found 4 distinct positive integer values for n: 23, 10, 5, and 2.
Therefore, there are 4 possible positive integer values of ‘n’ for which n² + 96 is a perfect square.