## Re: [PrimeNumbers] (Pn+1-Pn) mod 6 question

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• ... The two statements are not exactly equivalent , however, the reasons why both are impossible (except for the triplet containing a 3) is indeed the same.
Message 1 of 2 , Jun 25, 2001
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On Mon, 25 June 2001, omega@... wrote:> textbook question:
>
> (Pn+1-Pn) mod 6 gives
>
> 1 2 2 4 2 4 2 4 0 2 0 4 2 4 0 0 2 0 4 2 0 4 0 2 4 2 4 2 4 ...
> (3 5 7 ..)
>
> Does the following hold?
>
> n>3
> if (Pn+1-Pn) mod 6 is congr. to 4/2 then
> (Pn+2-Pn+1) mod 6 is not congruent to 4/2
>
> Yes, this is equivalent to asking if there there are prime triplets of
> the form {p,p+2,p+4}. And thats only (3,5,7) because one of them must
> be a multiple of three.

The two statements are not exactly 'equivalent', however, the reasons why both are impossible (except for the triplet containing a 3) is indeed the same. The original statement would be satisfied if there were three consecutive primes of the form {p, p+2, p+10}, for example, which for the same reason that you state is impossible.

> However does the following hold?
>
> n>3
> if (Pn+1-Pn) mod 6 is congruent to 4 then
> (Pm+1-Pm) mod 6 is congruent to 2 if
> all gaps mod 6 between n+1 and m are congruent to 0.
>
> or why cant 4 0 0 0 4 or 2 0 0 2 occur?

It's the same kind of problem as the above one. With the same conclusion and the same reason. One of the alleged primes in your sequence must be divisible by 3.

Phil

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