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 $p$.



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(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 $p$.)



What is the status on this conjecture on arithmetic progressions of primes?
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](https://en.wikipedia.org/wiki/Prime_k-tuple) 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).


