Question

solve the above question.......⤴️⤴️​

Answer 1

$\huge{\underline{\mathtt{\red{❥A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$

Let n^2 + 96 = x^2

⇒ x^2 – n^2 = 96

⇒ (x – n) (x + n) = 96

⇒ both x and n must be odd or both even 

on these condition the cases are

x – n = 2, x + n = 48

x – n = 4, x + n = 24

x – n = 6, x + n = 16

x – n = 8, x + n = 12

and the solution of these equations can be given as

x = 25, n = 23

x = 14, n = 10

x = 11, n = 5

x = 10, n = 2

So, the required values of n are 23, 10, 5, and 2.

Answer is 4 values of n.

Answer 2

$\huge\mathfrak\pink{Answer}$

$Let \\  k^2=n^2+96 , \: where n,k is \: a \: positive integer </p><p></p><p></p><p>$

$=> {k}^{2} - {n}^{2} = 96$

(k+n)(k-n)=96

Both k,n must have the same parity. Hence we will try to find two factors of 96 such that both are even.

96 can be factorized as 2×48

4×24

6×16

8×12

The smaller factor =k−n and the bigger factor =k+n

Hence, the values of (n,k) are (23,25),(10,14),(5,11),(2,10)

Hence, there are 4 possible values of n.

Option C is the answer.

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