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 ?

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

k is a perfect square number

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

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

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

answer is A, C, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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