Problem 1547

The given problem can be generalized as follows:
Given three positive integers $k, m, n$. Let $k_{1}, k_{2}, \ldots, k_{m}$ be any positive integer. Prove that there are infinitely many pair of positive integers $(a, b)$ such that
$$\prod_{k=1}^{m}\left(a+k_{i}\right) \mid b(b+k) \text{ but } \prod_{k=1}^{m}\left(a+k_{i}\right) \nmid b \text{ and } \prod_{k=1}^{m}\left(a+k_{i}\right) \nmid b+k .$$
Proof. Let $k < p_{1} < p_{2} < \ldots < p_{m}$ be any $m$ distinct primes and denote
$$M = \prod_{k=1}^{m}\left(a+k_{i}\right) .$$
By Chinese remainder theorem, there are infinitely many positive integers $a > n$ such that
$$a \equiv -k_{i} \pmod{p_{i}}, \forall i = 1, \ldots, m .$$
Hence $M \equiv 0 \pmod{p_{1} p_{2} \ldots p_{m}}$.
Then we write
$$M = p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{m}^{\alpha_{m}} q_{1}^{\beta_{1}} q_{2}^{\beta_{2}} \ldots q_{s}^{\beta_{s}}$$
with $\alpha_{i} \geq 1, i = 1, \ldots, m ; \beta_{j} \geq 1, j = 1, \ldots, s$ and $q_{1}, q_{2}, \ldots, q_{s}$ are $s$ prime divisors of $M$ which are different from $p_{1}, p_{2}, \ldots, p_{m}$.
By Chinese remainder theorem, there exist infinitely many positive integers $b > m$ such that
$$\begin{cases} b \equiv 0 \pmod{\frac{M}{p_{m}^{\alpha_{m}}}} \\ b \equiv -k \pmod{p_{m}^{\alpha_{m}}} \end{cases}$$
This implies that
$$M \nmid b, \quad M \nmid b+k \text{ and } M \mid b(b+k) .$$
$\square$