Re: Fermat Numbers help please
- From: Muhammad <mzafrullah@xxxxxxx>
- Date: Wed, 14 Nov 2007 22:11:06 -0800 (PST)
On Nov 14, 10:03 pm, "BUFFE...@xxxxxxx" <BUFFE...@xxxxxxx> wrote:
Notation: F(n) = nth Fermat # = 2^2^n+1
I need to prove that if n < m then F(n) | (F(m) - 2).
I dont even know where to start.
I could multiple F(n) by a specific constant to get F(m) - 2. But how
do I find this constant?
I cant think of any other way to show divisibility.
I've tried letting m = nk+r and working with the definitions but that
didnt help me much.
I dont know what else to do.
Any suggestions and help is much appreciated.
Thank you.
Hint: Note that F(m)-2 =2^{2^m} - 1.
Also note that (2^{2^n} + 1)(2^{2^n}- 1) = 2^{2^{n+1}} - 1 and (2^{2^{n
+1}} + 1)(2^{2^{n+1}} - 1) = 2^{2^{n+2}} - 1.
So F(n) | (F(n+1)-2) and F(n) | (F(n+2) - 2) and you have an easy to
see induction step.
Muhammad
.
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