2
$\begingroup$

Are there any numbers $a$ (not a perfect square of course) for which every Sophie Germain prime $p$ $>$ $a$, $a$ will always be a quadratic residue $\pmod {2p+1}$. In other words, the Legendre Symbol ($a$ | $2p+1$) $=$ $1$ whenever $p$ is a Sophie Germain Prime greater than $a$.

The only $a$ I was able to find is $3$, and it is a quadratic residue to all "safe" primes (primes of the form $2p+1$ with $p$ prime) where $p$ $>$ $3$. The proof for this is that when $p$ $>$ $3$ and $p$ is prime, $p$ $=$ $±1$ $\pmod 6$. Then $2p+1$ is $3$ or $11$ $\pmod {12}$. $p$ $=$ $1$ $\pmod 6$ cannot be a Sophie Germain Prime, otherwise it is divisible by $3$. This leaves $p$ $=$ $5$ $\pmod 6$, and $2p+1$ $=$ $11$ $\pmod {12}$. $3$ is a quadratic residue to any prime $±1$ $\pmod {12}$, as mentioned in my previous posts.

Anyways, despite all of my work above, my question here is weather or not there exists another integer $a$ with the same properties as $3$. For example, $a$ $≠$ $2$, $5$, $7$, $8$, $10$, or $11$ as shown with ($2$ | $11$) $=$ $-1$, ($5$ | $23$) $=$ $-1$, ($6$ | $59$) $=$ $-1$, ($7$ | $23$) $=$ $-1$, ($8$ | $59$) $=$ $-1$, ($10$ | $23$) $=$ $-1$, ($11$ | $47$) $=$ $-1$. $12$ is my next possible choice, (since perfect squares are excluded) but is anyone able to demonstrate a proof/disproof that $3$ is the only such number? Thanks for help.

$\endgroup$
  • $\begingroup$ I doubt the question can be solved, since it is actually an open question whether there are infinitely many Sophie Germain primes or not. Indeed, if there are only finitely many of them, let's say all smaller than $M$, then every $a>M$ would work trivially. $\endgroup$ – Paolo Leonetti Aug 31 '16 at 11:07
  • $\begingroup$ I know that, but I am assuming there are infinitely many. Weather there are infinitely many or not, $a$ $=$ $3$ is implied. Again, my real question is weather any other $a$ (as I described above) could exist. I will give you 50 rep if you can find another such $a$ as I described or prove that $3$ is the only one. $\endgroup$ – J. Linne Sep 1 '16 at 0:37
  • 1
    $\begingroup$ If you assumed that there are infinitely many, then please write it in the OP $\endgroup$ – Paolo Leonetti Sep 1 '16 at 13:07
  • $\begingroup$ You brought up a different question which I answered independently. Now here is another lemma for you: Suppose there are finitely many Sophie Germain primes $p$ which are no greater than $x$. Suppose we have an integer integer $a$ $<$ $p$ $<$ $x$ for which $a$ is a quadratic residue to all safe primes $2p+1$ up to $2x$. (With the assumption there are finitely many Sophie Germain primes $p$). If we cannot prove the modular congruence pattern that shows that $a$ is always a quadratic residue to $2p+1$, then the $a$ I just explained does not count. $\endgroup$ – J. Linne Sep 1 '16 at 13:59

Your Answer

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

Browse other questions tagged or ask your own question.