# For every prime $p$, are there infinitely many integers $k$, such that $p$, $p+k$, and $kp+1$ are all primes?

For every (fixed) prime $p$, there are infinitely many integers $k$ such that $p$, $p+k$, and $kp+1$ are all prime?

I wasn't exactly sure if this was implied by Dickinson's Conjecture, since two variables are required. It seems likely that it is the case, and is proven true that there are infinitely many primes for $p$ and $p+k$ for a fixed prime $p$ (Euclid's Theorem) as well as $kp+1$ (Dirichlet's Theorem), for any prime $p$. Would a combination of both theorems prove the conjecture above?

• I do not see any congruence issue, so it would be covered by DIckson's Conjecture. – André Nicolas Jun 28 '16 at 3:11
• I don't think the answer is known even for $p=2$. It's similar to the conjecture on the infinitude of Sophie-Germain primes, so I would expect that a proof would be quite difficult, if possible. – Batominovski Jun 28 '16 at 3:14
• @Batominovski I think you mean $k = 2$ (generally speaking of a fixed $k$, not a prime) as well, $p = 2$ will be fine for $2+k$ and $2k+1$. – J. Linne Jun 28 '16 at 3:27
• I mean $p=2$. There is no reason to fix $k$ if you want to show infinitude of $k$. – Batominovski Jun 28 '16 at 3:30
• The least $k$ for $p=2$ is $3$. Likewise, the least $k$ for $p=3$ is is $2$. – J. Linne Jun 28 '16 at 3:34

By setting $p+k=q$, we just need to prove that there exist an infinitude of primes of the form $$p(q-p)+1.$$ For a fixed $p$, by removing the primality constraint on $q$ we have that $p(q-p)+1$ is an AP through $1$ with common difference $p$: by Dirichlet's theorem, it contains an infinite number of primes. If we restrict $q$ to prime numbers, it looks unlikely that the previous property does not hold anymore, but that may be (extremely) hard to prove since primes do not have a positive upper density in $\mathbb{N}$ (a necessary condition for applying Szemeredi's theorem, for instance).