A question which could naturally arise when studying or thinking on prime number is: does gaps between prime numbers of arbitrary length always exists? In other words, given a positive integer $m$, and indicating with $s(p)$ the prime number which follows $p$, there always exists a prime number $p$ such that $s(p) - p \ge m$?
This question has already a (positive) well known response: for example, given $m$ we can construct the sequence of $m$ composite integers $(m+1)! + 2, (m+1)!+3, \ldots, (m+1)!+m$.
A natural follow-up question can be: does a prime surrounded by arbitrary large prime gaps always exists? In other words, given $n$ and $m$ positive integers, there always exist a prime number $p$ such that both sets $\{p-1, p-2, p-3, \ldots, p-n\}$ and $\{p+1, p+2,\ldots, p+m\}$ doesn't contain any prime number? Put it in another way, indicating with $s^{-1}(p)$ the prime number which precedes a given odd prime number $p$, there always exists a prime number $p$ such that $p - s^{-1}(p) \ge m$ and $s(p) - p \ge n$?
You can set up a series of congruences you want to solve - for example $$p+2\equiv 0 \bmod 3$$ $$p-2\equiv 0 \bmod 5$$ $$p+4\equiv 0 \bmod 7$$$$p-4\equiv 0 \bmod 11$$$$p+6\equiv 0 \bmod 13$$$$p-6\equiv 0 \bmod 17$$
The Chinese Remainder Theorem finds you a solution modulo the product of the primes you use as moduli. And then you can use Dirichlet's Theorem to guarantee a prime in that congruence class.