IMO 2018 · Problem 5

Let \(a_1,a_2,\ldots\) be an infinite sequence of positive integers. Suppose that there is an integer \(N>1\) such that, for each \(n\geq N\), the number \[ \frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots+\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1} \] is an integer. Prove that there is a positive integer \(M\) such that \(a_m=a_{m+1}\) for all \(m\geq M\).

Official text of the problem. Official problem paper (PDF)

Solution

Denote the given sum by \(S_n\). For every \(n\geq N\), both \(S_n\) and \(S_{n+1}\) are integers, so \[ k_n:=S_{n+1}-S_n =\frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_1}-\frac{a_n}{a_1} \] is an integer.

For each \(n\geq N\), write \[ \frac{a_n}{a_1}=\frac{p_n}{q_n}, \qquad p_n,q_n\in\mathbb Z_{>0}, \qquad \gcd(p_n,q_n)=1. \] We will show that \[ p_{n+1}\mid p_n \quad\text{and}\quad q_{n+1}\mid q_n. \]

Indeed, substituting the reduced fractions into the expression for \(k_n\) gives \[ \frac{p_nq_{n+1}}{q_np_{n+1}} +\frac{p_{n+1}}{q_{n+1}} -\frac{p_n}{q_n}=k_n. \] Multiplying by \(q_np_{n+1}q_{n+1}\) and rearranging, we obtain \[ p_nq_{n+1}^{\,2}+q_np_{n+1}^{\,2} =(p_n+k_nq_n)p_{n+1}q_{n+1}. \] It follows that \[ p_{n+1}\mid p_nq_{n+1}^{\,2}, \qquad q_{n+1}\mid q_np_{n+1}^{\,2}. \] Since \(p_{n+1}\) and \(q_{n+1}\) are coprime, these imply the claimed divisibilities.

Thus both \((p_n)_{n\geq N}\) and \((q_n)_{n\geq N}\) are nonincreasing sequences of positive integers. Each can decrease only finitely many times, so there is an integer \(M\geq N\) such that \[ p_m=p_{m+1} \quad\text{and}\quad q_m=q_{m+1} \qquad(m\geq M). \] Consequently, \[ a_m=a_1\frac{p_m}{q_m} =a_1\frac{p_{m+1}}{q_{m+1}} =a_{m+1} \qquad(m\geq M), \] as required. \(\square\)

The solution was checked against the official IMO shortlist solutions.