Question

Hw .any pairs of(m,n) of integers satisfy the equation 4^m=n^2+15

Answer 1

we have to find number of pairs of (m, n) are there such that. $4^m=n^2+15$

case 1 :- consider , n = 1
$4^m=(1)^2+15\\\\4^m=1+15\\\\4^m=16=4^2$
m = 2
so, (2, 1) is a solution of given condition.

case 2 :- consider , n = -1
$4^m=(-1)^2+15\\\\4^m=1+15\\\\4^m=16=4^2$
n = 2,
so, (2, -1) is also a solution of given condition.

case 3 :- consider , n = 7
$4^m=(7)^2+15\\\\4^m=49+15\\\\4^m=64=4^3$

m = 3
so, (3, 7) is a solution of given condition
similarly (3 , -7) is solution of given condition.

well there will be many solutions but if you talk about integers then there are four solutions e.g., (2, 1) , (2, -1) , (3, 7) and (3, -7).

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]