I was just looking at a sequence of primes and suddenly I got this thought that
$p_2 +p_3 −p_4 = 1$ since $p_2 = 3, p_3 = 5, p_4 = 7$. Also for $p_3 = 5,
p_4 = 7, p_5 = 11$ one has $p_3 +p_4 −p_5 = 1$.
So is it possible to find all the prime triples $(p_n,p_{n+1},p_{n+2})$ such that
$$p_{n} +p_{n+1} −p_{n+2} = 1$$ I know that could be a very difficult thing to ask . I would also like to ask about what we can know about these types of primes using current methods and how?
Note that if $p_{n} +p_{n+1} −p_{n+2} = 1$ then
$$p_{n+2} >2 p_n-1$$
By the Prime Number Theorem this fails for $n$ large enough, and the lower bound for $n$ large enough should be pretty low. And techniques from Analityc number theory should yield such bounds.
I also think that Erdos' proof for Bertrand Postulate can be changed to find an pretty small $N$ such that for all $n>N$ we have $$p_{n+2} <2 p_n-1$$ I would expect this $N$ to be in single-double digits range, but one has to calculate it.
Once you prove that all the solutions are smaller than some $N$, the rest is just test and error.
P.S. The stronger versions of Bertrand Postulate listed here yield easily such $N$, but I expect that there are much simpler ways of getting it.
For example, the result of Nagura implies that any solution satisfies $p_n \leq 25$.
This means that the only solutions are the ones you listed.