Simple statement on number theory Say $k$ is an odd natural number and $n$ is a positive natural number and also say that $2^{n+2}$ $|$ $(k^{2^n}-1)(k^{2^n}+1)$. We are also aware of the fact the $k^{2^n}+1$ is not divisible by $2^{n+2}$. Can we deduce that $2^{n+2}$ $|$ $k^{2^n}-1$? This statement was used on a proof i saw, but can't prove it myself.
 A: Too long for a comment:
(*)If $2^{n+3}|(k^{2^n}-1)(k^{2^n}+1)$ the answer is obviously yes: $(k^{2^n}-1), (k^{2^n}+1)$ are two consecutive even numbers, one has to be divisible by $2$ but not divisible by $4$. Therefore the other is divisible by $2^{n+2}$.
()If there exists some $k,n$ such that  $2^{n+2}|(k^{2^n}-1)(k^{2^n}+1)$ but $2^{n+3} \nmid (k^{2^n}-1)(k^{2^n}+1)$** Then, the answer to your question is no, and that example provides a counter example.
This means that your question is equivalent to the following:
Question If $k$ is an odd integer and $n$ is a natural number so that $2^{n+2}|k^{2^{n+1}}-1$ does it follow that $2^{n+3} | k^{2^{n+1}}-1$?
I suspect that the answer is no, and a counterexample can be found fast with a computer, but I could be wrong.
P.S. Doublecheck the powers over what you are trying to prove, are you sure that the obvious thing which was claimed was not (*).
A: The question is worded somewhat strangely, but it is true that $2^{n+2}\,|\,k^{2^n}-1$ (using your restrictions on $n,k$)  This is because the multiplicative group of units $\pmod {2^{n+2}}$ is not cyclic for $n≥1$ (see, e.g. this).  Thus every odd $k$ has order $2^m\pmod {2^{n+2}}$ where $m≤n$.
