Question

How many pairs (m,n) of integers satisfy the equation 4^m=n^2+15?

Answer 1

Hello,

Solution:

let m = 2, n=1

4² = 1²+15

16 = 1+15

16=16

m= 3 , n= 7

4³ = 7²+15

64 = 49+15

64 = 64

by this way you can analyse more,but according to me there are two such pairs ,

these are (2,1) and (3,7)

hope it helps you.

Answer 2

M = 2; N = 1

15 can be written as product of 3 and 5.

i.e. 15 = (4-1) * (4+1) = (4² - 1²)

or, 4² = 1² + 15

Now comparing it with 4^m=n^2+15 we get,

m = 2; n = 1.

But this isn't the answer.

You've to find out the number of possible combinations out of this.

Thuis we have only two options:

64 - 49 = 15 and 16 - 1 = 15.

So, two possible combinations, where (m,n) = (3,7) & (2,1) [4³=64]