Prime to Own Power minus 1 over Prime minus 1 being Prime

Theorem
Let $n \in \Z_{>1}$ be an integer greater than $1$.

Then $\dfrac {n^n - 1} {n - 1}$ is a prime for $n$ equal to:
 * $2, 3, 19, 31$

Proof
Note that if $4 p + 1$ is prime for prime $p$, then $\dfrac {p^p - 1} {p - 1}$ is divisible by $4 p + 1$:

Let $q = 4 p + 1$ be prime.

By First Supplement to Law of Quadratic Reciprocity:
 * $\paren {\dfrac {-1} q} = 1$

that is, there exists some integer $I$ such that $I^2 \equiv -1 \pmod q$.

Then:

and thus:

Hence $q \divides \paren {p^p - 1}$.

Obviously $q > p - 1$.

Therefore $q \divides \dfrac {p^p - 1} {p - 1}$.