Largest prime divisors of consecutive numbers
For example, $2$ and $23$ won't work. One of the consecutive integers must be a power of $2$. Since the order of $2$ mod $23$ is $11$, $2^k + 1$ is never divisible by $23$, while $2^k - 1$ is divisible by $23$ iff $k$ is divisible by $11$. But then $2^k - 1$ is also divisible by $89$.
EDIT: I suspect that $3$ and $89$ won't work, but I don't have a proof.