Question

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

Answer 1

Final Answer : (2,1),(2,-1),(3,7),(3,-7)

Steps:
1) Given : $m, n \in I$
We have,
${4}^{m} = {n}^{2} + 15 \\ = > { ({2}^{m} )}^{2} - ( {n}^{2} ) = 15 \\ = > ( {2}^{m} - n)( {2}^{m} + n) = 15$

2) Cases possible :
$15 \times 1 \\ 1 \times 15 \\ 3 \times 5 \\ 5 \times 3$
3) Solving all these cases, we obtain
=> m =2 , n = ±1
=> m=3 ,n=±7



Hence, Possible integral pairs :
(2,1),(2,-1),(3,7),(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]