Here is what I observed:
Let $R_{p}$ = $(n^p-1)/(n-1)$ or $(n^p+1)/(n+1)$ when $n$ is a positive integer $> 1$ and $p$ a prime number $> 2$ and $p > n$.
$R_{p}\text{ is a prime iff: } \ s_{p} \equiv s_{1} \pmod{R_{p}}$ when $s_{1}$ = $p^n$ and $s_{i+1}=s_{i}^n$
For example with $(2^5+1)/3 = 11$ and $(5 > 2)$ :
- $s_0$ = $5^2$ = $25$
- $s_1$ = $5^2$ mod $11 = 3$
- $s_2$ = $3^2$ mod $11 = 9$
- $s_3$ = $9^2$ mod $11 = 4$
- $s_4$ = $4^2$ mod $11 = 5$
- $s_5$ = $5^2$ mod $11 = 3$
Then $11$ is prime.
Another example with $(3^7-1)/2 = 1093$ :
- $s_0$ = $7^3$ = $343$
- $s_1$ = $7^3$ mod $1093 = 343$
- $s_2$ = $343^3$ mod $1093 = 47$
- $s_3$ = $47^3$ mod $1093 = 1081$
- $s_4$ = $1081^3$ mod $1093 = 458$
- $s_5$ = $458^3$ mod $1093 = 491$
- $s_6$ = $491^3$ mod $1093 = 1057$
- $s_7$ = $1057^3$ mod $1093 = 343$
Then $1093$ is prime
Another exemple with a composite number $(3^5-1)/2$ = $121$
- $s_0$ = $5^3$ = $125$
- $s_1$ = $5^3$ mod $121 = 4$
- $s_2$ = $4^3$ mod $121 = 64$
- $s_3$ = $64^3$ mod $121 = 58$
- $s_4$ = $58^3$ mod $121 = 60$
- $s_5$ = $60^3$ mod $121 = 15$
$121$ is not prime because 4 =/= 15
Is there a way to explain that ? I don't know how to start for proving it.