Let $P(m)$ denote the greatest prime factor of $m$. Is it true that, for any two primes $p,q$, there exists some integer $n$ such that $P(n)=p$ and $P(n+1)=q$?

In fact, the answer to this question as written is easily seen to be no, since there are no solutions to $2^k\equiv -1\pmod{7}$, and hence this fails with $p=2$ and $q=7$. It is possible that Erdős meant to exclude such obstructions, by amending this to 'odd primes' or 'all sufficiently large primes' or such.

In fact, the answer to this question as written is easily seen to be no, since there are no solutions to $2^k\equiv -1\pmod{7}$, and hence this fails with $p=2$ and $q=7$.

Even with such amendments, this problem is false in a strong sense: Alan Tong has provided the following elegant elementary proof that, for any given prime $p$, there are infinitely many primes $q$ such that this statement is false: let $m$ be the product of all primes $\leq p$, and choose a prime $q$ congruent to $-1$ modulo $4m$. If $p$ is the greatest prime divisor of $n$ then, using quadratic reciprocity, every prime divisor of $n$ is a quadratic residue modulo $q$, and hence $n$ is a quadratic residue modulo $q$. On the other hand, since $q\equiv 3\pmod{4}$ we know that $-1$ is not a quadratic residue modulo $q$, and hence $n\not\equiv -1\pmod{q}$, so it is impossible for $q\mid n+1$.

Tong asks whether, for any given odd prime $q$, there are infinitely many primes $p$ such that there is no integer $n$ with $P(n)=p$ and $P(n+1)=q$.

Sampaio independently observed that the answer to Erdős' original problem is no if one of the primes can be $2$ - for example this is false with $p=19$ and $q=2$, since if $n+1=2^k$ and $19\mid n$ then (since $2$ is a primitive root modulo $19$) we must have $18\mid k$, and hence $73\mid 2^{18}-1\mid n$.

Problem 6 in the 12th Romanian Master of Mathematics Competitions in 2020 was to prove that there exist infinitely many odd primes $p$ such that, for every $n$, $P(n)P(n+1)\neq 2p$.