Hi. I want to know how many (infinitely many) pairs of primes are known.

For convinience, let me give two definitions.

For any nonconstant polynomial $f(x)\in \mathbb{Z}[x]$, define $A_{f}=\lbrace f(p) \in \mathbb{Z}|$ Both of $p,f(p)$ are primes$\rbrace$.

Also, define $P=\lbrace f(x)\in \mathbb{Z}[x] | |A_f|=\infty\rbrace$, where $|A|$ is the cardinality of set $A$.

Let me give some examples. If $f(x)=x$, then it is (trivial) prime pairs. (i.e., $f(x)=x \in P$)

If $f(x)=x+2$, then the case is that the famous twin prime conjecture. (i.e., twin prime conjecture is equivalent to determine that $f(x)=x+2$ is in $P$ or not.)

I also heard that the case of $f(x)=4x+1$ is also (famous) conjecture.

My question is that are there any nontrivial polynomial which lie in $P$?

  • $\begingroup$ You might find the survey "Equidistribution and Primes" math.princeton.edu/sarnak/EquidPrimes.pdf by Peter Sarnak interesting as it discusses results in this direction. $\endgroup$ – j.c. Dec 8 '10 at 20:04
  • $\begingroup$ If f has degree greater than 1, it's not even known if the cardinality of the set of prime values of f is infinite. If f is linear then we just get variants of the twin prime conjecture and, to my knowledge, they are all wide open (except where they are false for trivial reasons). $\endgroup$ – Qiaochu Yuan Dec 8 '10 at 20:08

As a special case of Schinzel's hypothesis H (a well-known open problem) if $f(x)$ is irreducible, has positive leading coefficient and has no "fixed divisor" then $f(x)$ should lie in what you call $P$. But nothing of this sort has been proved and I am confident that not a single polynomial (other than $x$) has been proved to be in what you call $P$. For polynomials of degree at least two, it's even worse, there is no polynomial which has been proved to take prime values at infinitely many integers (let alone primes).


  • $\begingroup$ Felipe: I guess $f(x)=x+1$ is clearly not in $P$ but it's irreducible, has positive leading coefficient, and no "fixed divisor", right? i.e. it's not quite Schinzel, it's related but it's a bit different. $\endgroup$ – Kevin Buzzard Dec 8 '10 at 20:16
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    $\begingroup$ @Kevin: the hypotheses of Schinzel's hypothesis (ha) rule that out, but Felipe has misstated them slightly; we need x f(x) to have no fixed divisor. $\endgroup$ – Qiaochu Yuan Dec 8 '10 at 20:23
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    $\begingroup$ @Qiaochu: I am confused. Schinzel is all about $f(n)$ being prime, not $f(p)$. I think you must be talking about a variant I don't know. What is the story about $xf(x)$? $\endgroup$ – Kevin Buzzard Dec 8 '10 at 21:32
  • $\begingroup$ @Qiaochu: oh, I see the trick now :-) $\endgroup$ – Kevin Buzzard Dec 8 '10 at 21:36

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