THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

a) If $a_1<a_2<\cdots <a_n$ 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 $n\in {\mathbb{N}}^{\ast }$, the following property: $P(n)$: "For any rational number $q>1$, we can choose only finitely many possible $n$ positive integers $a_1\le a_2\le \cdots \le a_n$ 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 $a_1\le a_2\le \cdots \le a_{n+1}$ 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 $A=\left(\frac{1}{q-1},\frac{1}{\sqrt[n]{q-1}}\right]\cap {\mathbb{N}}^{\ast }$ 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 $a_2\le \cdots \le a_{n+1}$, 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 $a_1\le a_2\le \cdots \le a_{n+1}$, such that ${\prod }_{k=1}^{n+1}\left(1+\frac{1}{a_k}\right)=q$, that is $P(n+1)$ is also true.