## PATCH: 2^n+1 = np?

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• Hello! Sorry, there was a little error. At first there was a proof for such fact that p is composite; then I obtained yet stronger result (namely 3|p), but I
Message 1 of 1 , May 30, 2001
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Hello!

Sorry, there was a little error. At first there was a proof
for such fact that p is composite; then I obtained yet
stronger result (namely 3|p), but I didn't update the
proof... :-)

***
>Proof:
>
>Let p=(2^n+1)/n is prime greater than 3. n must be odd, so
>2^n+1 is divisible by 3; p is prime, hence, n is divisible
>by 3.
***

***
Proof:

Let p=(2^n+1)/n isn't divisible by 3. n must be odd, so
2^n+1 is divisible by 3; hence, n is divisible by 3.
***

Best wishes,

Andrey
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