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Q45
(CISF/2017)
Science & Technology › Basic Science (Physics, Chemistry, Biology)
What is the number of all possible positive integer values of 'n' for which n2 + 96 is a perfect square ?
Result
Your answer:
—
·
Correct:
B
Explanation
To find the number of positive integer values of n for which n2 + 96 is a perfect square, let:
n2 + 96 = k2, where k is a positive integer.
k2 - n2 = 96
(k - n)(k + n) = 96
Let x = (k - n) and y = (k + n). Then xy = 96. Since n is a positive integer, y > x. Furthermore, y - x = (k + n) - (k - n) = 2n, which implies that y and x must have the same parity. Since their product (96) is even, both x and y must be even integers.
We look for pairs of even factors (x, y) of 96 such that x < y:
- (2, 48) → 2n = 48 - 2 = 46 → n = 23
- (4, 24) → 2n = 24 - 4 = 20 → n = 10
- (6, 16) → 2n = 16 - 6 = 10 → n = 5
- (8, 12) → 2n = 12 - 8 = 4 → n = 2
Thus, there are exactly 4 possible values for n.
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