Question

How many pairs of (m n) are there such that 4^m=n^2+15?

Answer 1

2 is there can't you count

Answer 2

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).