1
$\begingroup$

Continuation to this previous question.

According to Lemke-Oliver, an irreducible polynomial $G$ of degree $g$ with positive leading coefficient and $\Gamma_G\neq0$ (with $\Gamma_G$ a certain factor defined in the reference) is conjectured to represent primes with density $\Gamma_G/\log x$. (What is the name of this conjecture? Is it Bouniakowsky? Or Hardy-Littlewood's F?).

This conjecture is wide-open. In fact, according to MathWorld, it seems that it is not even known whether the density is non-zero: such polynomials, while expected to represent primes infinitely often, are not proven to represent primes at least once.

If we strengthen the hypotheses, such as $g\equiv2$ (and, say, negative discriminant), is there any known result concerning the density of primes? Is it known whether quadratic polynomials of the form above represent primes at least once?

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ No. Typically we expect that if we can show a polynomial that doesn't obviously represent a prime represents a prime at all then it represents infinitely many primes. At least, that's a corollary of all successful proofs so far. $\endgroup$
    – Stanley Yao Xiao
    Commented Nov 13, 2018 at 16:49
  • $\begingroup$ @StanleyYaoXiao Thank you for your comment, but I must obviously missing something. "It represents a prime" cannot be equivalent to "it represents infinitely many primes". For one thing, $x^2+1$ clearly represents $2^2+1=5$, one prime, but it is not known if it represents infinitely many. So "representing one prime" is a much weaker statement than "it represents infinitely many". Or did I misunderstand your comment? $\endgroup$
    – Delmastro
    Commented Nov 13, 2018 at 17:48
  • $\begingroup$ $x^2 + 1$ obviously represents 5, because you can find an explicit $x$ such that $x^2 + 1$ is equal to 5. This is not doable for a generic quadratic polynomial, say $ax^2 + bx + c$. So outside of obvious 'eyeballing', one can't prove that a quadratic polynomial represents a prime at all. If one can in fact show that a generic quadratic polynomial represents a prime, then very likely the argument will in fact produce infinitely many primes that it can represent. $\endgroup$
    – Stanley Yao Xiao
    Commented Nov 13, 2018 at 17:54
  • 2
    $\begingroup$ To say the same thing another way: the strategies that number theorists have come up with for trying to show that a given polynomial must represent at least one prime (other than brute-force computation) are all actually strategies for showing that it must represent infinitely many primes. It seems hard to think of a theoretical approach to proving "one prime" that doesn't also prove "infinitely many primes". $\endgroup$
    – Greg Martin
    Commented Nov 13, 2018 at 18:26
  • $\begingroup$ @StanleyYaoXiao Oh, I now I see what you meant. Kind of anticlimactic, but I guess that's the way it goes. Anyway, thank you very much for your insight! $\endgroup$
    – Delmastro
    Commented Nov 13, 2018 at 22:18

0

Reset to default

You must log in to answer this question.