2021/089) Show that $2^n$ is not a factor of $3^n+1$ for n >1

We shall prove the same for 2 cases 

1) n is even

2) n is odd 

let us 1st prove for n even

case 1 :For n even say 2k $(>=2)$

$3^n + 1 =  3^{2k} + 1 = 9^k + 1 \equiv 2 \pmod 4$

so $3^n + 1$ is not divisible by 4 so cannot be divisible by $2^n$

    

case 2: for n odd  say 2k + 1 $( >=3)$

$3^n + 1 = 3^{2k+ 1} + 3 = 9^k.3 + 1 \equiv 4 \pmod 8$

so  so $3^n + 1$ is not divisible by 8 so cannot be divisible by $2^n$