# Number of Prime Numbers Such $p$ such that $p, 10 + p, 10p + 1$ and their sum are primes.

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.

• 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. Jun 26, 2021 at 4:45
• 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 . Jun 26, 2021 at 4:53
• Although this question is not exactly a prime $k$-tuple, the literature on that subject could be enlightening. Jun 26, 2021 at 4:56
• Thanks, @Infinity_hunter I think I had liked a 3blue1brown video related to that. Jun 26, 2021 at 5:01
• 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. Jun 26, 2021 at 5:49

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.
• 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. Jun 26, 2021 at 7:53