Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $P$?

share|improve this question
    
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. –  j.c. Dec 8 '10 at 20:04
    
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). –  Qiaochu Yuan Dec 8 '10 at 20:08

1 Answer 1

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).

http://en.wikipedia.org/wiki/Schinzel%27s_hypothesis_H

share|improve this answer
    
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. –  Kevin Buzzard Dec 8 '10 at 20:16
2  
@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. –  Qiaochu Yuan Dec 8 '10 at 20:23
1  
@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 $xf(x)$? –  Kevin Buzzard Dec 8 '10 at 21:32
    
@Qiaochu: oh, I see the trick now :-) –  Kevin Buzzard Dec 8 '10 at 21:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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