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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.

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    $\begingroup$ Doesn’t this require you to already know what the primes are? $\endgroup$
    – Randall
    Feb 24, 2019 at 18:36
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    $\begingroup$ It seems like this is implausible just based on a density argument. For large primes, heuristically, we shouldn't expect to see any primes of the form $(b^p-1)/(b-1)$, since the set of such numbers has very low density. It might be difficult to explicitly find a counterexample, though. $\endgroup$
    – Noah Stephens-Davidowitz
    Feb 24, 2019 at 18:43
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    $\begingroup$ @Randall: Yes. However, this is similar to the traditional definition. You must know all primes below $p$ or at least below $\sqrt{p}$ to state the $p$ is a prime. $\endgroup$
    – DaBler
    Feb 25, 2019 at 10:19
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    $\begingroup$ @DaBler However, the definition involving the prime counting function that you mention goes downwards: that is, there are always finitely many things to check for a given prime, and there's a finite base case that you can work from. Yours goes upwards, so there are always infinitely many things to check, and there is no such base case. In particular, both the empty set and the set of all integers greater than 1 both have this property. $\endgroup$
    – user3482749
    Feb 26, 2019 at 13:47
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    $\begingroup$ @user3482749 You are right. Now I see the definition in the question is surely not a definition. Sorry for that. Perhaps conjecture would be better name for it. $\endgroup$
    – DaBler
    Feb 27, 2019 at 10:12

1 Answer 1

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

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  • $\begingroup$ Interesting! This is something I have missed... $\endgroup$
    – DaBler
    Feb 25, 2019 at 10:37

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