Problem: Let $P$ be the set of all primes, and let $M$ be a
non-empty subset of $P$. Suppose that for any non-empty subset
$\{p_1, p_2, \dots, p_k\}$ of $M$, all prime factors of
$p_1 p_2 \dots p_k + 1$ are also in $M$. Prove that $M = P$.

Solution: The following solution was found by user \texttt{Aiscrim} on AOPS.

Obviously $|M| = \infty$.
Assume for contradiction $p \notin M$.
We say a prime $q \in M$ is \emph{sparse}
if there are only finitely many elements of $M$
which are $q \pmod p$
(in particular there are finitely many sparse primes).

Now let $C$ be the product of all sparse primes (note $p \nmid C$).
First, set $a_0 = 1$.
For $k \ge 0$, consider then the prime factorization of the number
\[ Ca_k + 1. \]
No prime in its factorization is sparse,
so consider the number $a_{k+1}$ obtained by
\textbf{replacing each prime in its factorization
with some arbitrary representative of that prime's residue class}.
In this way we select a number $a_{k+1}$ such that
\begin{itemize}
  \ii $a_{k+1} \equiv Ca_k + 1 \pmod p$, and
  \ii $a_{k+1}$ is a product of distinct primes in $M$.
\end{itemize}
In particular, $a_k \equiv C^k + C^{k-1} + \dots + 1 \pmod p$

But since $C \not\equiv 0 \pmod p$,
we can find a $k$ such that $a_k \equiv 0 \pmod p$
(namely, $k=p-1$ if $C \equiv 1$ and $k=p-2$ else)
which is clearly impossible since $a_k$ is a product of primes in $M$!