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  1. $12|(p + p+2)$, where $p,p+2$ are primes and $> 3$. Why $12$ divides the sum of twin primes?
  2. $a, ar, ar^2, \ldots $ is a Geometric series. I would like to place $a = 1$ and $r =2$. Then this series becomes $1, 2, 4, \ldots$ Now, $1 + 2 + 4 = 7$ (prime); $1 + 2 + 4 + 8 = 15$ (co-prime);$1 + 2 + 4 + 8 + 16 = 31$ (prime); $1 + 2 + 4 + 8 + 16 + 32 = 63$ (co-prime);$1 + 2 + 4 + 8 +16 + 32 + 64 = 127$ (prime) and so on...
    Why this Geometric series is acting like this? I have checked for only few values. So if there is anything wrong, explain. If it is true. Why it is true for in this series.
    Also, I would like to know, is there any such kind of Geometric series, which gives us only primes or primes and co-primes alternatively?
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coprime? one number cannot be coprime. Two are needed for that. –  Golob Nov 14 '12 at 9:28
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2 Answers 2

up vote 5 down vote accepted

Let $p$ be a prime $>3$ such that $p+2$ is also prime . $12|5+7$ . So , consider $p>5$. Then $p$ is either of the forms $6k + 1$ or , $6k + 5$ , now since $p+2$ got to be prime , $p$ is of the form $6k + 1$ then $3|p+2$ and since $p>5$ , $3$ is a nontrivial divisor so $p+2$ can not be prime . Hence if $p>5$ is a prime such that $p+2$ is also prime then $p$ is of the form $6k + 5$and then $p+p+2 = 6k + 5 + 6k + 7 = 12k+12$ , this answers your first question.

For the second question , you are getting as the sum , numbers of the form $2^n - 1$ , these are Mersenne numbers and are prime for many values of $n$ , but not for all values of $n$ , and it is still an open question whether there are infinitely many such primes

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@Thanks a lot. Very good explanation. –  madfellow Nov 14 '12 at 9:54
    
@ madfellow: You are always welcome. –  Souvik Dey Nov 14 '12 at 9:56
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(1) since twine primes must be in a form of $6k-1, 6k+1$, so the divisibility is obvious.

(why $6k-1, 6k+1$? check it manually, pretty straightforward result)

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