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 7 of 8: The tail is a divisibility chain
an+1∣anfor all sufficiently large na_{n+1}\mid a_n\quad\text{for all sufficiently large }n
Detailed analysis

For primes p∤a1p\nmid a_1, Step 3 gives vp(an+1)≤vp(an)v_p(a_{n+1})\le v_p(a_n). For primes p∣a1p\mid a_1, Step 6 makes the valuations equal after a common point. Therefore, for all sufficiently large nn, every prime exponent in an+1a_{n+1} is at most its exponent in ana_n, so an+1∣ana_{n+1}\mid a_n.