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 3 of 8: Primes absent from a1a_1 decrease
p∤a1⟹vp(an+1)≤vp(an)p\nmid a_1\Longrightarrow v_p(a_{n+1})\le v_p(a_n)
Detailed analysis

If p∤a1p\nmid a_1, then v(an+1a1)≥0v(\frac{a_{n+1}}{a_1})\ge0 and v(ana1)≥0v(\frac{a_n}{a_1})\ge0. Since TnT_n is an integer, the last term must also have nonnegative valuation: v(anan+1)≥0v(\frac{a_n}{a_{n+1}})\ge0. Hence vp(an+1)≤vp(an)v_p(a_{n+1})\le v_p(a_n) for every n>Nn>N.