## Re: Primality with respect to multiplication modulo n.

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• ... No problem, Dimeter. ... Thanks. Where you write ... which avoids having to use your detailed checks, value by value. ... and then we have only to write
Message 1 of 7 , Feb 4, 2011
"Dimiter Skordev" <skordev@...> wrote:

> I decided to submit the statement as a problem to the American
> Mathematical Monthly. I actually submitted there a little more
> precise form of that statement, but yesterday I withdraw my
> submission, since I observed that already on January 11
> David Broadhurst has posted a proof of the statement from my
> message (I am sorry for observing David's message so late!).

No problem, Dimeter.

> I am now free to post my proof

Thanks. Where you write

> Suppose n is some of the numbers 2, 3, 4, 6. It easy to check
> that, whenever x and y are natural numbers less than n and
> different from 1, the number x*y is not congruent to n-1 modulo n.

> For n = 3, 4, or 6, the group of units (Z/Zn)* has
> precisely 2 elements

which avoids having to use your detailed checks, value by value.

That gave the key to my proof of the main theorem:

> Theorem 3: Lemmas 1 and 2 exhaust the situations in which a
> prime p < n is a "prime mod n", as defined by Skordev.

> Proof: First, suppose that n is coprime to p. If n = 5 or
> n > 6, then the group of units (Z/Zn)* has at least 4 elements

and then we have only to write tidy words, as you did.

Best regards

David
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