TITLE: Are there infinitely many primes $p$ such that $p^2+1$ is divisible by a prime greater than $p$?
QUESTION [5 upvotes]: I've been trying to come up with an elementary construction, but to no avail. Are more advanced tools needed? 
If we consider an analog to this question where we do not require p to be prime (i.e. if there are infinitely many integers $n$ such that $n^2+1$ is divisible by a prime greater than $n$), then the question becomes much easier: 
Pick a prime $p$ that is $1$ mod $4$, and let $g$ be a generator of mod $p$. Then choosing $n\equiv g^{\frac{p-1}{4}}$ mod $p$, with $n<p$, works.

REPLY [5 votes]: Iwaniec has shown through sieve methods the existence of infinite values of $n$ such that $n^2+1$ is the product of $\leq 2$ primes. This is, as far as I know, the closest claim to Landau's conjecture about the existence of infinite primes of the form $n^2+1$ (still an open problem), together with the results of Iwaniec, Friedlander and Heath Brown about the existence of infinite primes of the form $a^4+b^2$ (like $2017$) or $a^3+2b^3$. Inserting the constraint $n\in\mathcal{P}$ into Iwaniec' approach and exploiting Vinogradov's estimates about exponential sums over primes should be possible but quite a technical challenge. On the other hand, it looks highly unrealistic that $\omega(p^2+1)\geq 3$ for every prime large enough: I believe the opposite can be shown by proving a lack of correlation between the arithmetic functions $\omega$ and $\chi$, with $\chi$ being a Legendre symbol.