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.