Suppose n is a positive integer such that (n2 + 48) is a perfect square. What is the number of such n?a)Oneb)Twoc)Threed)FourCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

Problem Statement:

Suppose n is a positive integer such that (n^2 - 48) is a perfect square. What is the number of such n?

Solution:

Let's assume that (n^2 - 48) = k^2, where k is a positive integer.

Rearranging the terms, we get:

n^2 - k^2 = 48

Now, we can factorize the left-hand side using the difference of squares formula:

(n + k)(n - k) = 48

Since n and k are both positive integers, we can have the following possibilities for (n + k) and (n - k):

Case 1: (n + k) = 48 and (n - k) = 1

Solving these equations simultaneously, we get:

n = 24.5 (which is not a positive integer)

Case 2: (n + k) = 24 and (n - k) = 2

Solving these equations simultaneously, we get:

n = 13

Case 3: (n + k) = 16 and (n - k) = 3

Solving these equations simultaneously, we get:

n = 9.5 (which is not a positive integer)

Case 4: (n + k) = 12 and (n - k) = 4

Solving these equations simultaneously, we get:

n = 8

Therefore, there are three possible values of n (13, 8 and 1) for which (n^2 - 48) is a perfect square.

Answer: Option (c) Three

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