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## Re: [PrimeNumbers] Question why are 2 and 5 unlike any of the other primes?

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• In a message dated 28/02/04 00:28:21 GMT Standard Time, tftn@earthlink.net ... Your conjecture doesn t stand scrutiny. 2^(2^n)-1 = (2^(2^(n-1)))^2-1 which has
Message 1 of 2 , Feb 28, 2004
In a message dated 28/02/04 00:28:21 GMT Standard Time, tftn@...
writes:

> In my work with Primes (where 1 isn't taken as the zeroth Prime),
> I've found that the subset {2,5} has unique properties.
> Conjecture:
> Primes of the sort should form 8 unique sets if I'm right about this:
> 2^n+/-1
> 2^(2^n)+/-1
> 5^n+/-2
> 5^(5^n)+/-2
> and 4 new types like:
> 5^(2^n)+/-2
> 2^(5^n)+/-1
> are also indicated.
>

Your conjecture doesn't stand scrutiny.

2^(2^n)-1 = (2^(2^(n-1)))^2-1
which has (2^(2^(n-1)))+/-1 as algebraic factors and so is never prime.

2^(5^n)+/-1 = (2^(5^(n-1)))^5+/-1
which has (2^(5^(n-1)))+/-1 as a factor and so is never prime.

-Mike Oakes

To Mr Bagula: as already requested, please do not post identically to so many
groups (4 in this case); I already sent the above reply yesterday to
numbertheory, and it now has to go to primenumbers as well; people get fed up with
seeing the same post several times.

To the primenumbers moderator: since he ignored your request, cannot you
please from now on suppress these multiple postings?

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