The List of unsolved problems in mathematics contains varies conjectures of exotic primes like:

to name just a few. Each of these primes is provided with special property. Sitting enthroned over the list on the wiki page you'll find the Twin Prime Conjecture (ignoring Catalan's Mersenne conjecture for the moment). So here is my

Question: Assuming the infinitude of all exotic primes, for which is it possible to disprove the infinitude of twin prime pairs containing one exotic prime?


It could be done for e.g. primes of the form $p_n=(6n)^2+1$, where it's obvious to show that $p_n\pm 2$ are composite and $p_n$ is called isolated prime. And the weaker $5$th Hardy-Littlewood conjecture asserts that $a^2+$1 is prime for an infinite number of integers $a>1$ [from Bouniakowsky Conjecture].

Don't get me wrong: This is not what I'm looking for! The question is on the primes given in the list above: Fermat/SophieGermain/Mersenne/... and how to disprove that they are infintely often one of the twins in a pair. Mea culpa, if this is misleading.

Collected partial results

  • Percentages given in the list above, show the ratio of exotic primes having a twin (some where counted twice, specially in the regular case).

  • For Fermat primes it seems promising ($100\% !$) to prove that every prime has a twin, but then this restricts to Fermat primes of the form $2^{2^{2n}}+1$, since $7 \mid 2^{2^{2n+1}}+3$ (see coment below) and $2^{2^{2n+1}}-1$ is obviously composite.

  • An analogous analysis might done for Mersenne primes, but I haven't yet.

  • The Wiki page on Twin Primes gives some more general ways to tackle the problem, but I'm not sure, if they are really useful:

    1. Every twin prime pair except $(3, 5)$ is of the form $(6n - 1, 6n + 1)$ for some $n$, and with $n \neq 1$, $n$ must end in $0, 2, 3, 5, 7 \text{ or } 8$. -- This seems related to my example given above, since $1,4,6$ and $9$ are missing, which show up as end digits of square numbers, see here. $0$ and $5$ seem to be exceptions.

    2. The pair $(m, m+2)$ is twin prime, iff $4((m-1)! + 1) \equiv -m \pmod {m(m+2)}$.

So if you feel that you can disprove the twin prime conjecture on any of these exotic primes, I'd be every so happy to read your answer here. If you think you can prove it for a kind of primes, where the infinitude is also proven, send me an eMail.

  • $\begingroup$ $7 \mid 2^{2^{2n+1}}+3$ can be written as $2^{2^{2n+1}}-4 \equiv 4\left(2^{2(4^n-1)}-1\right)\equiv 0\bmod 7$. With $4^n-1\equiv 0\bmod 3$ we have $4\left(2^{2\cdot 3k}-1\right)\equiv 0\bmod 7$, proven here. $\endgroup$ – draks ... May 23 '12 at 18:15
  • $\begingroup$ For the interested reader:Regularities of Twin, Triplet and Multiplet Prime Numbers by H. J. Weber $\endgroup$ – draks ... May 24 '12 at 17:32
  • $\begingroup$ What are the percentages in your list of exotic primes? $\endgroup$ – Vincent May 25 '18 at 14:53
  • $\begingroup$ @Vincent: Percentages given in the list above, show the ratio of exotic primes having a twin... $\endgroup$ – draks ... May 29 '18 at 10:48
  • $\begingroup$ Interesting. The percentages are uncannily large. Perhaps because they are only calculated over small primes? $\endgroup$ – Vincent May 29 '18 at 14:17

The standard conjectures imply that there are infinitely many primes of the form $p=n^4+2$. No such prime can be part of a twin pair, since $p-2=n^4$ and $p+2=n^4+4=n^4+4n^2+4-4n^2=(n^2+2)^2-(2n)^2=(n^2+2n+2)(n^2-2n+2)$.

A simpler example is $p=n^2-2$.

An even simpler example is $p=21n+5$, and here we don't need any conjectures --- we know that there are infinitely many such primes $p$. Many more examples can be constructed along the same lines, e.g., $15n+7$, $15n+8$, $77n+9$, $39n+11$, etc., etc.

EDIT: The above was written before OP edited the question to indicate interest only in the five types of prime at the top of the question. So, let's look at those. In all cases, please ignore tiny counterexamples to generally true statements.

Mersenne primes $q=2^p-1$, $p$ prime. Trivially $q$ can't be the smaller of a pair of twin primes, so we are asking about $2^p-3$ and $2^p-1$ both being prime. Apparently this does happen from time to time, so there's no simple reason why it shouldn't happen infinitely often. On the other hand, we don't even know there are infinitely many Mersenne primes, so we're not going to prove it does happen infinitely often. In short: hopeless.

Sophie Germain primes: $p$ such that $p$ and $2p+1$ are both prime. $p-2$ is a multiple of 3, so we are asking whether there are infinitely many $p$ such that $p$, $p+2$, and $2p+1$ are all prime. The standard conjectures (e.g., Schinzel's Hypothesis H) say yes, but no one has a clue as to how to prove this. In short: hopeless.

Fermat primes. Maybe there are some congruences to show $2^{2^{2n}}+3$ can't be prime for sufficiently large $n$. It's worth a look. On the other hand, maybe there are only 5 Fermat primes anyway. There are heuristic arguments suggesting that there are only finitely many.

Regular primes. Another set that hasn't been proved infinite, although the smart money is leaning that way. I can't imagine any relation between the regularity of $p$ and the primality of $p\pm2$. Possibly just ignorance on my part, but I'm going to call this one: hopeless.

Fibonacci primes. The Fibonacci numbers grow exponentially, just like the powers of 2 (only not quite as fast), so the situation here is comparable to that with the Mersenne numbers. Hopeless.

  • $\begingroup$ Nice, but how does this relate to Fermat/SophieGermain/Mersenne/... primes? $\endgroup$ – draks ... May 24 '12 at 13:06
  • 2
    $\begingroup$ I didn't know that those were the only primes that interested you. Indeed, you brought up the $(6n)^2+1$ primes, so I see your $(6n)^2+1$ and raise you an $n^4+2$, an $n^2-2$, and a $21n+5$. $\endgroup$ – Gerry Myerson May 24 '12 at 13:34
  • $\begingroup$ Ah, I see you've edited the question to clarify. It's a very odd collection of types of primes to ask about, but de gustibus non disputandum est. Anyway, I suspect the problem is hopeless for those families of primes, so my advice is, take what you can get. $\endgroup$ – Gerry Myerson May 24 '12 at 13:38
  • $\begingroup$ The example obviously was misleading. I didn't image that it draws so much attention, sorry for that. Now that you know, what do you think? $\endgroup$ – draks ... May 24 '12 at 13:39
  • $\begingroup$ So much attention? 19 hours before it got a single comment/answer, and you consider that so much attention? You didn't downvote my answer, by any chance, did you? Well, none of my business, that's why these things are anonymous after all, but, heck, I gave a good-faith answer to the question as it was posted, and the only answer anyone has bothered to give. Where's the love? $\endgroup$ – Gerry Myerson May 24 '12 at 13:43

There are almost surely infinitely many regular primes which are part of a twin prime pair. The first few are

3, 5, 7, 11, 13, 17, 19, 29, 31, 41, 43, 61, 71, 73, 107, 109, 137, 139, 151, 179, 181, 191, 193, 197, 199, 227, 229, 239, 241, 269, 281, 313, 349, 419, 431, 521

and there are 1513 below $10^5.$

I disagree with Gerry on Mersenne primes. While we expect that there are infinitely many, there should be only finitely many which are part of a twin prime pair since you'd need $2^p-3$ and $2^p-1$ to both be prime and this happens with probability $\sim k/p^2$ and $\int dx/x^2$ converges.

For Sophie Germain primes, this would follow from Dickson's conjecture and so this one is perhaps the closest of all these questions to actual resolution. This is related to A045536 but sadly no information is present there.

Fermat primes are not expected to be infinite in number, so there should not be infinitely many twins.

Fibonacci primes are spread pretty thin, so like with the Mersenne primes I expect there are only finitely many which are twins.


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