So I would like to prove that $ a_n \; | \; a_{n+1} - 2 $, where $a_n = 6^{2^n} + 1 $ Now I know I need to do this by induction and so I begin by showing this is true for the base case.
(Proposition): P(n) := $ 6^{2^{n+1}} -1 = (6^{2^n} + 1)q \;$, where $q \in \mathbb{Z}$
(Base Case): P(1) is true as $ 6^{2^{2}} -1 = (6^{2} + 1)35 \;$
(Hypothesis): We assume that P(k) is True, where P(k):= $ 6^{2^{k+1}} -1 = (6^{2^k} + 1)q \;$
(Induction Step): Need to show that $P(k) \implies P(k+1)$ Now I know I need to write $ 6^{2^{k+2}} -1$ in terms of my hypothesis and end up with something along the lines of $(6^{2^{k+1}} + 1)r$, where $r \in \mathbb{Z}$. My issue: I have and the reason I'm here is I don't feel comfortable working with powers of powers. I'm not sure how to use my hypothesis. In short looking for some guidance to get me started.
Best Regards and thanks for your time.
You're on the right track. The key thing to use here is the difference of squares formula. In particular,
$$6^{2^{k + 1}} - 1 = (6^{2^k})^2 - 1 = (6^{2^k} - 1)(6^{2^k} + 1)$$
from which your result should follow easily.
However, induction isn't really necessary here... this gives a direct proof of divisibility.