Romanian Mathematical Olympiad

a) If $a_1 < a_2 < \dots < a_n$ are consecutive positive integers, then we have
$$\left(1 + \frac{1}{a_1}\right) \left(1 + \frac{1}{a_2}\right) \dots \left(1 + \frac{1}{a_n}\right) = \frac{a_n + 1}{a_1}.$$
Choosing $a_k = an + k - 1, \forall k = \overline{1, n}$, we get
$$\left(1 + \frac{1}{a_1}\right) \left(1 + \frac{1}{a_2}\right) \dots \left(1 + \frac{1}{a_n}\right) = \frac{an + n}{an} = 1 + \frac{1}{a}.$$

b) We prove by induction on $n \in \mathbb{N}^*$, the following property:
$P(n)$: "For any rational number $q > 1$, we can choose only finitely many possible $n$ positive integers $a_1 \le a_2 \le \dots \le a_n$ such that $\prod_{k=1}^{n} \left(1 + \frac{1}{a_k}\right) = q$."
It is easy to see that $P(1)$ is true.
Suppose now that $P(n)$ is true for a positive integer $n$.
Let $q \in \mathbb{Q}$, $q > 1$. From the previous point, we know that there exist $n + 1$ positive integers $a_1 \le a_2 \le \dots \le a_{n+1}$ such that $\prod_{k=1}^{n+1} \left(1 + \frac{1}{a_k}\right) = q$. Using the inequalities $1 + \frac{1}{a_1} < q \le \left(1 + \frac{1}{a_1}\right)^{n+1}$, we get that $a_1 \in A$, where $A = \left(\frac{1}{q-1}, \frac{1}{\sqrt[n]{q-1}}\right] \cap \mathbb{N}^*$ is a finite set. For a fixed number $a_1 \in A$, let $q_1 = \frac{q}{1+1/a_1}$. It is clear that $q_1 \in \mathbb{Q}$, $q_1 > 1$. Using the induction hypothesis, we can choose only finitely many possible $n$ positive integers $a_2 \le \dots \le a_{n+1}$, such that $\prod_{k=2}^{n+1} \left(1 + \frac{1}{a_k}\right) = q_1$. We deduce from here that we can choose only finitely many possible $n + 1$ positive integers $a_1 \le a_2 \le \dots \le a_{n+1}$, such that $\prod_{k=1}^{n+1} \left(1 + \frac{1}{a_k}\right) = q$, that is $P(n + 1)$ is also true.