Question

How many pairs (m'n) of integer the equation 4^m=n^2+15

Answer 1

$4 {m}^{2} - {n}^{2} = 15 \\ (2m) {}^{2} - {n}^{2} = 15 \\ (2m + n)(2m - n) = 15$
now make pairs of m and n satisfying this relation

...the no. of pairs will be answer

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.

Thus 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]