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 $P(n+1)=q$?

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 $q\mid n+1$.