I have a problem with this:
Proove, that number $1\,997^{2^{n}}-1$ is divisible by $2^{n+2}$ for every $n \in \Bbb N$.
I tried mathematical induction, but I have a problem to proove the second step
$2^{n+2}$ / $1\,997^{2n}-1$
$2^{n+2}2$ / $1\,997^{2}\, 1\,997^{2n}-1$
I just don`t know, how can I extract this expression $1\,997^{2n}-1$ from this $1\,997^{2}\, 1\,997^{2n}-1$
I think, that this might help:
$1\,997^{2n}-1$ = $(1\,997^{n}-1)(1\,997^{n}+1)$
$1\,997^{2}\, 1\,997^{2n}-1$ = $(1\,997^{n+1}-1)(1\,997^{n+1}+1)$
Thank you for your time.
Your mistake seems to stem from mistaking $1997^{2^n}$ for $1997^{2n}$.
To answer the question, besides from using Euler's theorem, induction will also work. For $n=0$ we have
$$1997^1-1=2^2\cdot499$$
Now assume that $2^{n+2}k|1997^{2^n}-1$, that is $1997^{2^n}-1=2^{n+2}k$ for some integer $k$. Then we have
$$1997^{2^{n+1}}-1=\left(1997^{2^n}\right)^2-1=\left(1997^{2^n}-1\right)\left(1997^{2^n}+1\right)=\left(2^{n+2}k\right)(2^{n+2}k+2)=2^{n+3}k(2^{n+1}k+1)$$
So $2^{n+3}|1997^{2^{n+1}}-1$