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 1 of 8: Take consecutive differences
In plain words

The long cyclic sums differ in only three terms.

Sn=∑i=1n−1aiai+1+ana1S_n=\sum_{i=1}^{n-1}\frac{a_i}{a_{i+1}}+\frac{a_n}{a_1}
Detailed analysis

For n>Nn>N, both SnS_n and Sn+1S_{n+1} are integers. Their difference is Tn=Sn+1−Sn=an+1a1−ana1+anan+1T_n=S_{n+1}-S_n=\frac{a_{n+1}}{a_1}-\frac{a_n}{a_1}+\frac{a_n}{a_{n+1}}, so Tn∈ZT_n\in\mathbb Z.