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Q.

Let $A = {k | k \in W & \sqrt{k^{2} - k p} = n}$ ; where n is Natural number, W is a set of whole numbers and p is an odd prime number, then which of the following is/are true ?

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a

$n (A) = 8$ for $k \in [4, 200]$

b

$n (A) = 13$ for $k \in [4, 200]$

c

k is a perfect square number

d

k is a square of natural number greater than or equal to 2

answer is A, C, D.

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Detailed Solution

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$\sqrt{k^{2} - p k} = n \Rightarrow k = \frac{p \pm \sqrt{p^{2} + 4 n^{2}}}{2} \Rightarrow p^{2} + 4 n^{2} = m^{2} \Rightarrow m - 2 n = 1 & m + 2 n = p^{2}, n \geq 3 \Rightarrow m = \frac{p^{2} + 1}{2}, n = \frac{p^{2} - 1}{4}$
& $k = \frac{p \pm m}{2}, = \frac{2 p \pm (p^{2} + 1)}{4} \Rightarrow k = \frac{{(p + 1)}^{2}}{4}$
(negative value is committed )

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