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• ## Re: sufficent prove for prime numbers of the kind p:=x^2+x+1

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• From: bhelmes_1 ... You do not prove this: If a^[(p-1)/3]=x mod p or a^[(p-1)/3]=x^2 mod p, then there are exactly (p-1)/2 quadratic
Message 1 of 2 , Oct 30, 2005
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From: "bhelmes_1" <bhelmes@...>
>
> A beautifull evening,
>
> i have found a fast test to prove, if p of a certain kind is a prime
> number or not. The test needs only one Fermat-test.
>
> I hope that you will enjoy it:
> http://www.devalco.de/helmes_e.htm
>
> Please let me know, if the test is well proved.

You do not prove this:

If a^[(p-1)/3]=x mod p or a^[(p-1)/3]=x^2 mod p, then there are exactly (p-1)/2

There are, AFAIK, no known pseudoprimes that would fail your test,
but that's true about pseudoprimes on most non-linear functions.
That doesn't mean they don't exist though.

Phil

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