1
$\begingroup$

In the book of Guy on unsolved problem in number theory, problem B 48 it is stated that

$S_n=\prod_{1}^{n} \frac{p_n +1}{p_n - 1}$ is never an integer after $n=8$. I just realized that this is related to the Sophie german conjecture on primes $q$ for which $2q+1$ is also a prime. If we know that there are infinitely many Sophie German primes then we are done.

Does this observation appear somewhere?

I reached this by realizing that if $q=2p+1$ is a Sophie Germain prime and is the $n$ th prime, then $S_n=S_{n-1} \cdot \frac{p+1}{p}$, and $p$ does not appear anywhere in the numerator of $S_n$ (we can show $2p-1$ wont be a prime), so then $S_n$ must be a fraction, and will remain so till $p$ appears in the numerator of $S_m$ for some $m\geq n$. The smallest such $p_m$ for which p appears in the numerator of $S_m$ is $4p-1$(we can only get even numbers on top, and $2p$ does not appear as if it did $2p-1$ would be a prime, but it is impossible for $p,q$ and $2p-1$ to be primes simultaneously by mod $3$ conditions), so if we can get another Sophie German prime $q' = 2p'+1$ between $q$ and $4p-1$, then $q'>q$ so $p'>p$ so $4p'-1 > 4p-1$ by repeating the same argument $p'$ appears in the denominator and we can say that the product is not an integer till the case of $m$ when $p_m$ is not more than $4p'-1$. Now We again pick a SG prime in between $q'$ and $4p'-1$ and proceed. Thus we cover all the integers. But by the Sophie German conjecture such a prime always exists.

$\endgroup$
5
  • 1
    $\begingroup$ Are you claiming that an infinitude of Germain primes, no matter how sparse they become, is enough to show that the given product is never again an integer? $\endgroup$ May 11 '17 at 19:58
  • $\begingroup$ If $\frac{p_m-1}{2}$ is a prime $>3$ then $\prod_{n=1}^m \frac{p_n+1}{p_n-1}$ is not an integer. But it doesn't say anything of $\prod_{n=1}^{k} \frac{p_n+1}{p_n-1}$ with $k > m$ $\endgroup$
    – reuns
    May 11 '17 at 20:00
  • $\begingroup$ pls see the edit $\endgroup$ May 11 '17 at 20:04
  • $\begingroup$ So you meant assuming a Bertrand's postulate for the Sophie Germain's prime. And $$S_m=\prod_{n=1}^m \frac{p_n+1}{p_n-1}$$ $\endgroup$
    – reuns
    May 11 '17 at 20:46
  • $\begingroup$ That probably follows from the conjectured density of the SG primes $\endgroup$ May 11 '17 at 20:57
1
$\begingroup$

In the third edition of UPNT, Guy writes that Jud McCranie "notes the connexion with Sophie Germain primes." There is, however, no further elaboration or reference.

$\endgroup$
1
  • $\begingroup$ I take it you have a first or second edition UPNT ;-) $\endgroup$ May 11 '17 at 20:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.