Question

Find all positive integral values of n for which n2+96 is a perfect square

Answer 1

Let n2 + 96 = m2 where m is assumed to be a positive integer.

m2 - n2 = 96

=> (m - n)(m + n) = 96 = 25 × 3

We know that, the divisors of 96 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

=> Pairs of (m-n) and (m+n) can be 1,96 ; 2,48 ; 3,32 ; 4,24 ; 6,16 ; 8,12

Subtracting first value from the second one,we get:-

(m+n) - (m-n) = 2n

2n = 95, 46, 29, 20, 10, 4

n = 47.5, 23, 14.5, 10, 5 and 2

Hence, required integer values of n are 23, 10, 5 and 2.

So, the answer is that n2+96 is a perfect square for 4 positive integral values.

Answer 2

Answer:

Step-by-step explanation:

Let n^2 + 96 = a^2 where m is assumed to be a positive integer.

a^2 - n^2 = 96

=> (a - n)(a + n) = 96

divisors of 96 : 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

=> Pairs of (a-n) and (a+n) can be (1,96) ; (2,48) ; (3,32) ; (4,24) ; (6,16) ; (8,12)

Subtracting first value from the second one,we get:-

(a+n) - (a-n) = 2n

2n = 95, 46, 29, 20, 10, 4

n = 47.5, 23, 14.5, 10, 5 , 2

We require integer values of n that are 23, 10, 5 and 2.

So, the answer is n^2+96 is a perfect square for 4 positive integral values.