Can anyone help finish this proof? Induction proof $2^{n+2} \mid k^{2^n} -1$ for all $n \geq 1$ and odd $k$

Question

The question is prove $2^{n+2} \mid k^{2^n} -1$ for all $n \geq 1$ and odd $k$. I got as far as:

Notice for $n=1$, $2^{1+2}= 8 \mid k^2-1.$ Also notice for an odd $k$ when $n=1$, thus $k^{2}-1=(k-1)(k+1)$. Hence, $k-1$ and $k+1$ is divisible by $4$ and the other of those expressions is divisible by $2$. Since $2$ and $4$ are integers, we can then conclude that the equality is true for $n=1$.

Now suppose it's true for some $n=a \in \mathbb{Z}$. Then

\begin{align*} 2^{a+2} \mid k^{2^a} -1 \\ 2^{a}\cdot 4 \mid k^{2^a} -1\\ .\\ .\\ . \\ 2^{a+3} \mid k^{2^{a}}\cdot k^{2} -1 \\ 2^{a+3} \mid k^{2^{a+1}} -1 \\ \end{align*}

Thus, by induction $2^{n+2} \mid k^{2^n} -1$ for all $n \geq 1$ and odd $k$.

I Need help solving for the "..." Any tips or guidance would be appreciated.

Answer 1

It seems to me that the induction step is $2^{n+2} \mid k^{2^n} -1\Rightarrow 2^{n+3} \mid k^{2^{n+1}} -1$

But $k^{2^{n+1}} -1=k^{2^n\cdot2} -1=(k^{2^n})^2 -1=(k^{2^n}-1)(k^{2^n}+1)$

The first factor is divisible by $2^{n+2}$ by using the hypothesis of induction and the right factor is even. Which was to be proved.

Answer 2

I would prove first this factorisation, valid for any $k$: $$k^{2^n-1}=(k^2-1)(k^2+1)(k^{2^2}+1)(k^{2^3}+1)\dotsm(k^{2^{n-1}}+1),$$ based on the relation $\;(k^{2^{n-1}})^2=k^{2^n}$.

Now each of the last $n-1$ factors is divisible by $2$, hence their product is divisible by $2^{n-1}$.

As to the first factor, it is divisible by $2^3\;$ if $k$ is odd. Indeed, if $k=2i+1$, we have $$(2i+1)^2-1=4i^2+4i=4i(i+1),$$ and $i(i+1)$ is even.