Could prime numbers be defined like this?

Question

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 welcome.

Answer 1

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 $P(x)=x^{p-1}+\dotsb+x+1$.