Considering Prime Numbers $p\in P$ such that $(p^2+4)\in P, (p^2+4)^2+4\in P, ((p^2+4)^2+4)^2+4\in P, (((p^2+4)^2+4)^2+4)^2+4\in P,((((p^2+4)^2+4)^2+4)^2+4)^2+4\in P$, seems to be either very rare, or logically impossible. $ 10806580157\in P$ is the largest Prime I've found so far, which successfully results in a new Prime 4 times. So that:
- $10806580157^2+4 \in P, (10806580157^2+4)^2+4\in P, ((10806580157^2+4)^2+4)^2+4\in P, (((10806580157^2+4)^2+4)^2+4)^2+4\in P,$
So I wonder if it can be proofen that 4 times is the limit here.
Working modulo $13$, the effect of squaring and adding $4$ is as follows:
$4$ and $9$ go to $7$ (that is, $4^2+4\equiv9^2+4\equiv7\pmod{13}$).
$7$ and $6$ go to $1$.
$1$ and $12$ go to $5$.
$5$ and $8$ go to $3$.
$3$ and $10$ go to zero, which means you get a multiple of $13$, not a prime.
Also, $2$ and $11$ go to $8$, which has already been shown to lead to a multiple of $13$.
So, it's impossible to get a chain of more than five primes this way.