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