Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to prove that for $n=k+1$ that,
$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$
We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.
Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot  2^{k}}+2^{2\cdot  2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$
I get stuck here, please help me.

$b-a=4^{2\cdot2^k}-4^{2^k}+2^{2\cdot2^k}-2^{2^k}$
$=4^{2^k}(4^{2^k}-1)+2^{2^k}(2^{2^k}-1)$
$=4^{2^k}(2^{2^k}-1)(2^{2^k}+1)+2^{2^k}(2^{2^k}-1)$
$=(2^{2^k}-1)(8^{2^k}+4^{2^k}+2^{2^k})$
$=(2^{2^k}-1)2^{2^k}\color{blue}{(4^{2^k}+2^{2^k}+1)}$
