I did a search online and found a similar notion called Sophie Germain prime, which by definition is a prime $p$ such that $2p+1$ is also prime. Sophie Germain primes are conjectured to be of infinite many. I wonder if anyone has thought of the similar question replacing $2p+1$ by $2p-1$.
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$\begingroup$These primes are tabulated at
They don't seem to have a name, though they are related to Cunningham chains.
I have managed to find a proof such that all Sophie Germain primes are of the prime form $6n-1$, but have failed to get further. My proof is:
If the primes created by the formula $2x + 1$ equals the prime type $4y + 1$, the equation of $2x + 1 = 4y + 1$ is created, which simplifies to $x = 2y$. But as $x$ is prime, $x$ cannot equal anything but $2$. Therefore, the Germain safe primes are of the prime type $4y – 1$. This creates the equation $2x + 1 = 4y – 1$, which simplifies to $x = 2y - 1$. That means both $4y – 1$ and $2y – 1$ needs to be prime.
This will never be true of the values of $y$ that are of the formula $3n + 2$, excluding $2$. This is because the first value of $y$ in the equation $2y – 1 (5)$ makes $9$, and each new term in the sequence is the last term $+ 2$. After $3$ places, you add $6$, making a new multiple of $3$.
This leads the terms in the formula $3n$. Let us substitute $y$ for $3n$. $2(3n) – 1$ simplifies to $6n – 1$. For $3n – 1$, the other formula, you produce $6n + 1$.
For hypothetical $6n + 1$ Germain primes, the safe prime would be $12n + 3$. As terms from the formula $12n + 3$ will always divide by three, the only possible Germain prime is where $n = 0$. As the Germain "prime" would be $1$, and as $1$ is not prime, there are no safe primes of the formula $12n + 3$. Therefore, all Germain primes must be from the formula $6n - 1$.
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