MathLabs

Problem 5

Let a1,a2,…a_1,a_2,\ldots be an infinite sequence of positive integers. Suppose there is an integer N>1N>1 such that, for every n≥Nn\ge N, the number a1a2+a2a3+⋯+an−1an+ana1\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 MM such that am=am+1a_m=a_{m+1} for all m≥Mm\ge M.
Step 8 of 8: A descending chain of positive integers stabilizes
an+1∣an and an∣an0⟹an is eventually constanta_{n+1}\mid a_n\text{ and }a_n\mid a_{n_0}\Longrightarrow a_n\text{ is eventually constant}
Detailed analysis

Choose n0n_0 after the common stabilization point. Then an∣an0a_n\mid a_{n_0} for every n≥n0n\ge n_0. Only finitely many positive divisors of an0a_{n_0} exist, while an+1∣ana_{n+1}\mid a_n; the chain can therefore change only finitely often. Thus an=an+1a_n=a_{n+1} for all sufficiently large nn, as required.