To my knowledge it is open so far whether the polynomial $x^2+1 \in \mathbb{Z}[x]$ takes infinitely many prime numbers as values. Is it known so far whether there is at all any polynomial $P \in \mathbb{Z}[x]$ of degree $\geq 2$ which takes infinitely many prime values? -- Note that obtaining this result would not necessarily require to prove that any particular polynomial has this property.
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asked Jun 6, 2015 at 20:30
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4$\begingroup$ It is open whether there exists a polynomial in one variable of degree $>1$ that represents infinitely many primes. This is mentioned here. $\endgroup$– Jeremy RouseCommented Jun 6, 2015 at 20:33
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7$\begingroup$ @MicahMilinovich, how does one deduce that from Baier & Zhao's result? To the best of my knowledge, these results typically require that the number of polynomials in the family increase with other parameters in the average and thus do not imply asymptotic statements about any single (even a typical) polynomial. This is analogous to the fact that while Bombieri–Vinogradov is an averaged form of the RH, B-V does not imply that the RH holds even a single L-function. $\endgroup$– Mark LewkoCommented Jun 6, 2015 at 22:14
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3$\begingroup$ @MarkLewko: Correct, I have retracted my comments. $\endgroup$– Micah MilinovichCommented Jun 6, 2015 at 23:03
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6$\begingroup$ @Gerhard Paseman: $x^2+y^2$ even assumes all the infinitely many primes in $4\mathbb N+1$ ... $\endgroup$– Peter MuellerCommented Jun 8, 2015 at 17:11
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2$\begingroup$ Along the lines of Friedlander-Iwaniec, Heath-Brown proved that $x^3+2y^3$ represents infinitely many primes, with $x$ and $y$ positive. With Moroz, he extended this to general irreducible binary cubic forms. $\endgroup$– Gerry MyersonCommented Jun 9, 2015 at 6:27
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2 Answers
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As far as I am aware, it is unknown whether any irreducible polynomial of degree greater than one assumes infinitely many prime values. Certainly this is the case if one insists that the polynomial be given explicitly. I merely add that what is conjectured is that if an irreducible polynomial $f(x)\in\mathbb{Z}[x]$ satisfies $1=\mathrm{gcd}\{f(1), f(2), f(3), f(4), \dots\}$ then $f(n)$ is prime for infinitely many $n$. This is known as Bunyakovsky's conjecture. It has not been proven for any polynomial of degree greater than $1$. Generalizations include Schinzel's hypothesis H and the Bateman-Horn Conjecture.
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3$\begingroup$ @StefanKohl: I have edited this answer (which was actually meant more as an extended comment) so as to emphasize that no explicitly given irreducible polynomial in Z[x] of degree greater than one is known to assume infinitely many prime values and to clarify what is thought to be the truth. Of course this was not exactly what you asked for in your question, which is why I say that this is more of an extended comment. $\endgroup$– user1073Commented Jun 6, 2015 at 23:38
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The comments and the answer are related to the probably hopeless Bunyakovsky conjecture. It seems to me that Stefan Kohl had a different idea in mind, maybe something like the following: Let $A_p$ be the set of polynomials $f\in\mathbb Z[X]$ of degree $\ge2$ with $p\in f(\mathbb Z)$. The question amounts to asking if there is an infinite set $P$ of primes such that $\cap_{p\in P}A_p$ is not empty.
So in this form the question is if some density argument (with respect to some measure on $\mathbb Z[X]$) could be strong enough.
answered Jun 8, 2015 at 14:43
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$\begingroup$ Yes, of course. If you think adding this would make the question more clear, feel free to do so! $\endgroup$– Stefan Kohl ♦Commented Jun 8, 2015 at 15:16
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$\begingroup$ Another version, stronger but perhaps easier to visualize, is whether there exists a finite set of polynomials with degree >1, which overall represents infinitely many primes. $\endgroup$ Commented Jun 8, 2015 at 15:34
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1$\begingroup$ @Yaakov: Doesn't that imply that one of the finitely many polynomials represents infinitely many primes ...? $\endgroup$ Commented Jun 8, 2015 at 16:25
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$\begingroup$ @PeterMueller: yes - what I meant is that it's weaker (and easier) than proving that a specific polynomial represents infinitely many primes. But it is stronger (and harder) than your version in that it pins down that one polynomial to be in a specific finite set. $\endgroup$ Commented Jun 8, 2015 at 20:50