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• ## Re: {Spam?} [PrimeNumbers] 2^(2^m) + 1 mod prime

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• ... .. ... This result is (a) well-known and (b) easy to prove. Let F_i = 2^(2^i)+1 where i = 1. Then F_{i+1} = 2^(2^{i+1}) + 1 = (2^(2^{i+1}) - 1) + 2 =
Message 1 of 2 , Jul 29, 2007
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On Sun, 2007-07-29 at 13:16, Kermit Rose wrote:
> 2^(2^m) + 1 mod prime
..
> In general no number in this series divides any other number in the
> series

This result is (a) well-known and (b) easy to prove.

Let F_i = 2^(2^i)+1 where i >= 1.

Then F_{i+1} = 2^(2^{i+1}) + 1
= (2^(2^{i+1}) - 1) + 2
= (2^(2^(i) + 1) * (2^(2^(i) - 1) + 2
= F_i * (F_i - 2) + 2

So F_{i+1} == 2 mod F_i

The general result follows by induction.

Incidentally, this provides another nice proof of the infinity of the
primes. No F_n is divisible by any other F_n and n may take on any
integer value.

Paul

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