0
$\begingroup$

I have read about Dirichlet's theorem recently, that is, for relative prime positive integers $a,b$, there exists infinitely primes with the form $ax+b$.

What I want to ask is the situation when the $ax+b $ is changed as any irreducible polynomial with relative prime positive integer coefficients. Is there still infinitely many primes?

$\endgroup$
7
  • 1
    $\begingroup$ This is entirely unknown, see this. We don't even know a single polynomial (of degree $>1$) that can be shown to take infinitely many prime values. $\endgroup$
    – lulu
    Sep 7 '18 at 11:47
  • $\begingroup$ Note: it's clear that what you wrote is too broad. The polynomial $x^2+2x+1$ passes your tests, but clearly can't take prime values. Similarly, every value of $x^2+x$ is even. $\endgroup$
    – lulu
    Sep 7 '18 at 11:50
  • $\begingroup$ Talking about polynomials in two variables, there is this : www.michaelnielsen.org/polymath1/index.php?title=Friedlander-Iwaniec_theorem $\endgroup$ Sep 7 '18 at 11:53
  • 1
    $\begingroup$ Note; "irreducible" does not solve the problem. $x^2+x+2$ is also always even. $\endgroup$
    – lulu
    Sep 7 '18 at 11:55
  • 1
    $\begingroup$ So far the problem appears to be astonishingly intractable. Even a concrete example, like $x^2+1$ seems to be beyond existing methods. $\endgroup$
    – lulu
    Sep 7 '18 at 11:58
2
$\begingroup$

No, $x^2+2x+1=(x+1)^2$ obviously is not prime for natural $x>0$.

Edit: for irreducible case you may want to read about the (unsolved) Bunyakovsky conjecture.

$\endgroup$
4
  • $\begingroup$ I have change the question into irreducible polynomials. $\endgroup$ Sep 7 '18 at 11:55
  • $\begingroup$ @SteveCheng鄭宗弘 updated the answer. $\endgroup$
    – freakish
    Sep 7 '18 at 11:59
  • $\begingroup$ Irreducible is not sufficient: $x^2+x+2$ is always even, so almost never prime. $\endgroup$
    – xarles
    Sep 7 '18 at 13:04
  • $\begingroup$ @xarles Yes, this is explained in the wiki article. $\endgroup$
    – freakish
    Sep 7 '18 at 13:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.