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 $p$.)
Best Answer
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).