Prove by mathematical induction that $2^{3^n}+1$ is divisible by $3^{n+1}$

Question

Prove by mathematical induction that $2^{3^n}+1$ is divisible by $3^{n+1}.$

I'm currently trying to finish a task that requires me to use mathematical induction to prove that $2^{3^n}+1$ is divisible by $3^{n+1}$. This is the first time that the divider is not a simple integer and I'm having trouble trying to prove it.

This is what I managed to do:

First step:

For $n=1$:

$2^3+1=9$

$3^2=9$

$L=P$

Second step: I assume that for some numbers $n>=1$ $2^{3^n}+1$ is divisible by $3^{n+1}$

Third step: (induction hypothesis): $2^{3^{n+1}}+1$ is divisible by $3^{n+1}$

I'm getting stuck here, trying to do some algebra and simplify the dividend:

$2^{3^{n+1}}+1 = 2^{3^n*3}+1$

Answer 1

You can end your work by induction.

Indeed, $$2^{3^{n+1}}+1=2^{3\cdot3^n}+1=\left(2^{3^n}\right)^3+1=\left(2^{3^n}+1\right)\left(2^{2\cdot3^n}-2^{3^n}+1\right).$$

$2^{3^n}+1$ is divisible by $3^{n+1}$ by the assumption of the induction and since $$2^{3^n}\equiv-1(\mod3),$$ we obtain that $2^{2\cdot3^n}-2^{3^n}+1$ is divisible by $3$ and we are done!

Answer 2

Hint: $2^{3^n} + 1 = 3^n x$, then $2^{3^{n+1}} = (2^{3^n})^3 = (3^n x - 1)^3$. Now expand out the cube.