TITLE: Could prime numbers be defined like this?
QUESTION [8 upvotes]: While playing with prime numbers, I found the following definition.
Let $p$ be an integer.
Then $p$ is a prime number if and only if
there is some integer $b \neq 1$ such that
$$ \frac{b^p - 1}{b - 1} $$
is also a prime number.
It is easy to show that the primality of $p$ is a necessary condition for primality of $(b^p - 1)/(b - 1)$.
I am however stuck to prove that, for given $p$, there is always at least one prime number of the form $(b^p - 1)/(b - 1)$.
Is my definition correct, and if so, how to prove the second part?
Any suggestion is welcomed.

REPLY [7 votes]: This is probably hopeless to prove unconditionally, but at least it is true assuming the Bunyakovsky conjecture saying that any non-constant polynomial $P$ with integer coefficients takes infinitely many prime values provided that

the leading coefficient of $P$ is positive;
$P$ is irreducible;
$\mathrm{gcd}\{P(z)\colon z\in\mathbb Z\}=1$.  

Your assertion follows readily by applying the conjecture to the polynomial $P(x)=x^{p-1}+\dotsb+x+1$.