Problem: Prove that for each positive integer $n$,
there are pairwise relatively prime integers $k_0$, \dots, $k_n$,
all strictly greater than $1$, such that
$k_0k_1 \dots k_n - 1$ is the product of two consecutive integers.

Solution: In other words, if we let
\[ P(x) = x(x+1) + 1 \]
then we would like there to be infinitely many primes
dividing some $P(t)$ for some integer $t$.

In fact, this result is true in much greater generality.
We first state:
\begin{theorem}
  [Schur's theorem]
  If $P(x) \in \ZZ[x]$ is nonconstant and $P(0) = 1$,
  then there are infinitely many primes
  which divide $P(t)$ for some integer $t$.
\end{theorem}
\begin{proof}
  If $P(0) = 0$, this is clear.
  So assume $P(0) = c \neq 0$.

  Let $S$ be any finite set of prime numbers.
  Consider then the value
  \[ P\left(k \prod_{p \in S} p\right) \]
  for some integer $k$.
  It is $1 \pmod p$ for each prime $p$,
  and if $k$ is large enough it should not be equal to $1$
  (because $P$ is not constant).
  Therefore, it has a prime divisor not in $S$.
\end{proof}
\begin{remark*}
  In fact the result holds without the assumption $P(0) \neq 1$.
  The proof requires only small modifications,
  and a good exercise would be to write down a similar
  proof that works first for $P(0) = 20$,
  and then for any $P(0) \neq 0$.
  (The $P(0) = 0$ case is vacuous,
  since then $P(x)$ is divisible by $x$.)
\end{remark*}

To finish the proof, let $p_1$, \dots, $p_n$ be primes
and $x_i$ be integers such that
\begin{align*}
  P(x_1) &\equiv 0 \pmod{p_1} \\
  P(x_2) &\equiv 0 \pmod{p_2} \\
  &\vdotswithin\equiv \\
  P(x_n) &\equiv 0 \pmod{p_n}
\end{align*}
as promised by Schur's theorem.
Then, by Chinese remainder theorem,
we can find $x$ such that $x \equiv x_i \pmod{p_i}$
for each $i$, whence $P(x)$ has at least $n$ prime factor.