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Let's denote that $k$ is an odd number and $n\in \mathbb{N}$. Prove that:

$$2^{n+1}|k^{2^n}-1$$

Could you give me any HINT how to start with this?

JosephConrad
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    Hint: $k^{2^n}-1=(k^{2^{n-1}}-1)(k^{2^{n-1}}+1).$ Apply this again, and again, and again... Additionally, note that if $k$ is odd, $k^{2^m}+1$ is $\cdots ?$ – L. F. Apr 16 '13 at 01:02
  • $k = 2m -1$. When we unfold $k^{2^n}-1$ we receive $k^{2^n}-1 = (k^{2^{n-1}}+1)(k^{2^{n-2}}+1) \ldots (k^{2^2}+1) (k^{2^1}+1)(k^{2^0}-1)$. Last factor equals to $(2m-1-1)=2(m-1)$. Well, can get from every factor one $2$. So, I have received only $2^n$. Where is a mistake? – JosephConrad Apr 16 '13 at 01:19
  • You have made a slight mistake. The tail of that expression should be $(k^2+1)(k+1)(k-1)$, not just $(k^2+1)(k-1)$. So it is indeed divisible by $2^{n+1}$. In fact, since one of $k-1,, k+1$ is a multiple of $4$ it is divisible by $2^{n+2}$. – L. F. Apr 16 '13 at 01:29
  • That's true. Thank you so much for your help! – JosephConrad Apr 16 '13 at 01:40
  • @L.F.: So, that gives an induction kind of proof. Very nice. – Inceptio Apr 16 '13 at 01:40
  • See also here – Bill Dubuque Nov 08 '21 at 20:46

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Hint:

$k=(2m-1) \implies k^{2^n}=(2m-1)^{2^n} \implies 2m^{2^n}$-$2^n \choose 2$ $2m^{2^n-1} + \dots$-$2^n \choose 2^n-1$$2^{1}+ 1$

$k^{2^n}-1=2m^{2^n}$-$2^n \choose 2$ $2m^{2^n-1} + \dots$-$2^n \choose 2^n-1$$2^{1}$

Do you see that every term is a multiple of $2^{n+1}$?, last term being $2^{n+1} \cdot m$.

Inceptio
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