How can i prove that for any $ n \geq 1$ exist that (a is natural): $ 3^{(3^n)} + 8 = 7a$?

Question

How can i prove that for any $ n \geq 1$ exist that (a is natural):

$$ 3^{(3^n)} + 8 = 7a$$

How can i prove it with induction?

Answer 1

$$3^{3^n}+8=3^{3^n}+1+7=(3^3+1)(3^{3^n-3}+...+1)+7=7\left(4(3^{3^n-3}+...+1)+1\right).$$ I used that for all odd positive $k$ we have $$x^k+1=(x+1)(x^{k-1}-x^{k-2}+...-x+1).$$

The proof by induction.

For $n=1$ it's true.

Let $3^{3^n}+1$ is divisible by $7$.

Thus, $$3^{3^{n+1}}+1=\left(3^{3^n}\right)^3+1=\left(3^{3^n}+1\right)\left(3^{2\cdot3^n}-3^{3^n}+1\right)$$ is divisible by $7$ and we are done.