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The Green-Tao theorem states that for every $n$, there is an arithmetic sequence of length $n$ consisting of primes.

For primes, $p$, let $P(p)$ be the maximum length of an arithmetic progression of primes whose least element is $p$.

Is it known whether $P(p)=p$ for every prime?

(This clearly generalizes the Green-Tao theorem, asserting that long progressions show up "as soon as possible." Note that $P(p) \leq p$ by viewing the progression mod $p$.)

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    $\begingroup$ uhmm, I'm not sure, maybe its not well define the way I wrote it, but I guess it doesn't matter. $\endgroup$ – Jorge Fernández Jan 28 '17 at 22:14
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    $\begingroup$ I am not aware of anything in that direction. Even $P(p)\geq 3$ sounds difficult to me. $\endgroup$ – GH from MO Jan 28 '17 at 23:12
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    $\begingroup$ Regarding $p=11$: the smallest sequence is $11+n\times 210\times 7315048$ for $0\le n \le 10$. $\endgroup$ – Yaakov Baruch Jan 29 '17 at 13:47
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    $\begingroup$ A good question. If true, the proof would have to be very delicate (not at all like the regularity-lemma-like proofs of Green-Tao). If false, I think a proof would be extremely strange. Perhaps current techniques can give some lower bound on $P(p)$, but this too sounds tricky to me. See the following (especially the last page or so) for some discussion of other related results. people.maths.ox.ac.uk/~conlond/green-tao-expo.pdf $\endgroup$ – Pat Devlin Jan 31 '17 at 0:24
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    $\begingroup$ I am not sure about the complete history of the problem $\ P(p)=p?\ $ I know that Siemion Fajtlowicz proposed this conjecture in 1991/2 or earlier. At that time I've got an algorithm and coded a program which gave me $\ P(13)=13.\ $ Once again, I am not a specialist, I don't know the full history here. My feeling was that $\ P(17) < 17\ ($ perhaps $\ \le 15).\ $ I feel strongly that $\ P(p) < p\ $ for every prime $\ p>13;\ $ I'd even conjecture that $\ p-P(p)\rightarrow \infty\ $ for $\ p\rightarrow\infty$. $\endgroup$ – Włodzimierz Holsztyński Jan 31 '17 at 5:37
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Yes, this is unknown; it is even unknown (as GH from MO suspected in a comment) whether $P(p) \ge 3$ always. An equivalent statement to $P(p) \ge 3$ is that there exists an integer $x>0$ such $p+x$ and $p+2x$ are both prime. This is a twin-prime-like problem: nobody has ever proved a statement saying that two fixed linear polynomials $ax+b$ and $cx+d$ are infinitely often simultaneously prime, or even that they must generally be simultaneously prime once. (The Green-Tao theorem converts into a statement about linear polynomials $x,x+d,x+2d,...$ in two variables $x$ and $d$; when we fix $p$ here, we have only one variable.)

On the other hand, the prime $k$-tuples conjecture does imply that $P(p)=p$ for every prime $p$: the corresponding polynomials are $p+x,\dots,p+(p-1)x$, and these polynomials form an admissible set (their product is not identically zero modulo any prime).

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    $\begingroup$ It's not the $k$-tuples conjecture which you have to use here, but rather Dickson's conjecture. $\endgroup$ – Wojowu Aug 12 '18 at 10:43

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