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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?

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    $\begingroup$ 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, 2018 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, 2018 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, 2018 at 11:53
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    $\begingroup$ Note; "irreducible" does not solve the problem. $x^2+x+2$ is also always even. $\endgroup$
    – lulu
    Sep 7, 2018 at 11:55
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    $\begingroup$ 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, 2018 at 11:58

1 Answer 1

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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.

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  • $\begingroup$ I have change the question into irreducible polynomials. $\endgroup$ Sep 7, 2018 at 11:55
  • $\begingroup$ @SteveCheng鄭宗弘 updated the answer. $\endgroup$
    – freakish
    Sep 7, 2018 at 11:59
  • $\begingroup$ Irreducible is not sufficient: $x^2+x+2$ is always even, so almost never prime. $\endgroup$
    – xarles
    Sep 7, 2018 at 13:04
  • $\begingroup$ @xarles Yes, this is explained in the wiki article. $\endgroup$
    – freakish
    Sep 7, 2018 at 13:05

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