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 4 of 8: If a valuation reaches vp(a1)v_p(a_1), it descends
vp(ak)≥c=vp(a1)⟹c≤vp(an+1)≤vp(an)(n≥k)v_p(a_k)\ge c=v_p(a_1)\Longrightarrow c\le v_p(a_{n+1})\le v_p(a_n)\quad(n\ge k)
Detailed analysis

Now suppose p∣a1p\mid a_1 and put c=vp(a1)>0c=v_p(a_1)>0. If vp(ak)≥cv_p(a_k)\ge c for some k>Nk>N, induction gives vp(an)≥cv_p(a_n)\ge c and vp(an+1)≤vp(an)v_p(a_{n+1})\le v_p(a_n) thereafter. Indeed, if the next valuation were below cc, the first and third terms of TnT_n would have unequal valuations (the possible equality would force vp(an)<cv_p(a_n)<c), leaving a unique negative valuation; and if it exceeded the current one, the third term would be uniquely negative. Thus the integer sequence of valuations is eventually constant.