Question

Find all ordered pairs of(m,n) natural numbers that satisfy the equation 9^m +3^m - 2=2p^n
where p is a prime number.​

Answer 1

Given : 9^m +3^m  - 2=2p^n

To Find :  all ordered pairs of(m,n) natural numbers

Solution:

9^m +3^m  - 2=2p^n

=> (3²)m +3^m  - 2=2p^n

=> (3^m)² + 3^m  - 2=2p^n

=> (3^m)² + 2.3^m - 3^m  - 2=2p^n

=> 3^m(3^m + 2) - 1(3^m + 2) = 2p^n

=> (3^m + 2)(3^m - 1) = 2p^n

3^m - 1 is always even

3^m + 2 is always odd

p is prime number hence it must be odd  as only even prime is  2 then right side will have only even factors while LHS has odd and even  both

=> 3^m - 1 = 2

=> 3^m  = 3

=> m = 1

3^m + 2  = 5

p^n = 5

=> p = 5  and n = 1

m = 1  and n = 1

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