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I was intrigued by the reputation I had a few days before.
I had a reputation of 1209 a few days before, and out of silly curiosity I thought of just checking in Numbermatics about the specialties of the number and ended up seeing its prime factorisation : $3 \times 13 \times 31$. What I found interesting is that these prime numbers add up to another prime : $47$. This made me think : are there any more prime numbers $p$ such that $p, 10 + p, 10p + 1$ are prime and their sum (i.e., $11(p+1) + p$) is also a prime?

The apparent solution at first sight is $3$, but I doubt if more solutions exist. If I look at the final digit of the sum, it is $2p + 1 \pmod{10}$ which clearly implies that the last digit is always odd. But also note that the addends must also be also be prime, so $5$ can be eliminated. The addends are prime when $p = 7$ as well, but the sum is composite. $p = 11$ is invalid as the sum will be divisible by $11$. Thus when $p$ is $13$, the sum and the addends are prime again. Based in this observation, we can conjecture that primes of the form $10x + 3$ satisfy the conditions, but $23$ is a contradiction as one of the addends will turn composite. And so is $43$ - the sum turns composite.

I am not an advanced mathematician and has not enough background in number theory (w.r.t this problem), so please give me an answer (a sufficiently detailed one is most invited) to how I can solve this problem.

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  • $\begingroup$ May not be directly related : It is known that if $a ,b $ are relatively primes then the sequence $an+b$ has infinitely many primes (Dirichlet's theorem). I don't know whether it applies when we have multiple such sequences. $\endgroup$ Jun 26, 2021 at 4:45
  • $\begingroup$ In general it is hard to find number of primes $p$ such that $ap+b$ is prime for example it is not known that whether there are infinitely many Sophie Germain primes . $\endgroup$ Jun 26, 2021 at 4:53
  • $\begingroup$ Although this question is not exactly a prime $k$-tuple, the literature on that subject could be enlightening. $\endgroup$ Jun 26, 2021 at 4:56
  • $\begingroup$ Thanks, @Infinity_hunter I think I had liked a 3blue1brown video related to that. $\endgroup$
    – Spectre
    Jun 26, 2021 at 5:01
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    $\begingroup$ For the question you posed, we can see that $p=6k+1$ for some integer $k$, for otherwise $p+10$ won't be prime. I couldn't do anything more than this. But if you pose the question as "The primes $p$ such that $10+p,11p-2$ and their sum $13p+8$ are all primes", then we can see that $p=3$ is the only solution... but that isn't so interesting. $\endgroup$
    – Aditya
    Jun 26, 2021 at 5:49

1 Answer 1

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I wrote a quick program to search for such $p$. There are a lot of them. The first bunch are: 3,13,19,31,241,409,439,631,733,811,1009,1021,1039,1279,1483,1609,2383,2953,3319,3529,3823,4513,4933,4999,5431,5839,5851,6133,6481,6793,6949,6991,7243,7573,8161,8821,...

In general, showing whether their exist infinitely pairs of primes of the form $p$ and $ap+b$ is an unsolved problem. The particular case of $p$ and $p+2$ is known as the Twin prime conjecture. I would expect proving there are infinitely many such quadruplets of primes to be very difficult, but it's probably true.

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  • $\begingroup$ any other way by math, perhaps?? $\endgroup$
    – Spectre
    Jun 26, 2021 at 5:02
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    $\begingroup$ I don't really expect anything more substantial can be done. You can obtain some modular restrictions like $ p \not \equiv 2 \pmod 3$ and $ p \not \equiv 2,4 \pmod 7$, but they won't restrict them to finitely many primes. $\endgroup$
    – arbashn
    Jun 26, 2021 at 7:53
  • $\begingroup$ @arbashn Yes, you're right. $\endgroup$
    – Spectre
    Jun 26, 2021 at 7:56

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