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.

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.
