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I came up with this (I admit I'm probably not the first one to have this thought but I haven't been able to find anyone else with the same question) while reading about semiprimes.

Clearly $y$ is always odd which suggests to me that it could also be prime a lot of the time. Additionally, There are infinitely many semiprimes so it seems somewhat likely (for purely intuitive reasons) that for semiprime $n$ it could be that $2n+1$ could also be prime.

Daan Koning
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    OEIS A063640 says there are $1000$ examples up to $45971$, so small examples are not uncommon – Henry Mar 06 '23 at 17:48
  • With the exception of Dirichlet Theorem, almost any question like "Is there an infinite prime numbers of the form ..." is unsolved, so my bet is that this is not known. – jjagmath Mar 06 '23 at 17:54
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    It might be possible to show this using some density estimates for primes in arithmetic progressions, e.g. if $4p + 1$ was somehow only prime for finitely many primes $p$, then there might be "too few" primes congruent to $1$ modulo $4$. At any rate, that would probably require some quite heavy analytic-number-theory-machinery. – sbares Mar 06 '23 at 18:08
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    I think in the title you mean to ask if there are infinitely many primes of the form... – J. W. Tanner Mar 06 '23 at 18:34
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    @J.W.Tanner you're right. I updated it. English Isn't my first language :) – Daan Koning Mar 06 '23 at 19:31
  • A theorem of Chen implies that if $h$ is an even positive integer, there exist infinitely many primes $p$ for which $p+h$ is a prime or a semiprime. I wouldn't be surprised if the proof techniques can be modified to show something like "there are infinitely many primes $p$ for which $(p-1)/2$ is a prime or a semiprime," which is pretty close to what you're asking. – Carl Schildkraut Mar 06 '23 at 19:43
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    I second @CarlSchildkraut's comment: "There are infinitely many primes $p$ for which $(p-1)/2$ has at most two prime factors" is almost surely an established theorem, and even if it weren't I know it could be established using an adaptation of Chen's method. However, we can't prove "exactly two prime factors" (basically for the same mathematical reason we don't yet know how to prove "exactly one prime factor"!). – Greg Martin Mar 06 '23 at 20:01
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    The OEIS link is a bit misleading - it says it's primes of the form $pqr + 1$ where $p, q, r$ are primes. But one of $p, q, r$ must be 2 in order for $pqr + 1$ to be odd, so the primes listed there are in fact of the form $2 p_1 p_2 + 1$ as desired. – Michael Lugo Mar 06 '23 at 21:24
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    Not sure whether we can actually prove the conjecture that there are infinite many such primes. Schinzel's hypothesis implies that there are even infinite many such primes for any given prime $p_1$ (including $p_1=2$). If we have infinite many solutions just for one prime $p_1$ , the conjecture would already be true. – Peter Mar 06 '23 at 22:04
  • Related: https://math.stackexchange.com/questions/4523193/are-there-infinitely-many-primes-of-the-form-x-we-probably-dont-know – Eric Snyder Mar 07 '23 at 09:10

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Comment: Some experiments:

We know all primes are of the following forms:

$30k+r; r=1,7, 11, 13, 17, 19, 23, 29$

with some conditions we can construct primes of the form $y=2p_1p_2+1$, for example:

$p_1=30k+1$, $p_2=30k+11$, for $k=1$ gives $y=2543$

$p_1=3ok+11$, $p_2=30k+13$, for $k=1$ gives $y=3527$

some conditions for $k=1$ are:

$\begin{cases}r_{p_1}\not\equiv 3\bmod 10\\r_{p_1}\not\equiv 9\bmod 10\end{cases}$

because:

$y\equiv (2\cdot 3\cdot 9+1=55)=0\bmod 5$

That is y will be divisible by 5.

Or $p_1= 30k+11$, $p_2=30k+23$ for $k=1$ gives $y=4347=3\cdot 1449$

Now suppose there exist some $k$ for which we can construct $y$ taking primes of the form $p=3ok+r$, then we have at least $n$ primes , where:

$n={8\choose 2}=28$

We do not know what magnitudes of $r_{p_1}$ and $r_{p_2}$ with how many $k$ make primes $y$, but there is no limit for our choices of these three parameters.

$p_1$ and $p_2$ must not satisfy following identity:

$(5x+13)^2+1=(3x+7)^2+(4x+11)^2$

$p_1\neq(3x+7)$ and $p_2\neq(4x+1)$

because:

$(5x+13)^2+1=(3x+7)^2+(4x+11)^2$

$\Rightarrow (5x+13)^2+1+2(3x+7)(4x+11)=(3x+7+4x+11)^2$

$\Rightarrow 1+2p_1p_2=(7x+18)^2-(5x+13)^2=(2x+5)(12x+31)$

which is composite.

sirous
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