MathLabs

Problem 5

Consider an infinite sequence a1,a2,…a_1,a_2,\ldots of positive integers such that 100!(am+am+1+⋯+an)100!(a_m+a_{m+1}+\cdots+a_n) is a multiple of an−m+1an+ma_{n-m+1}a_{n+m} for all positive integers m,nm,n with m≤nm\le n. Prove that the sequence is either bounded or linear. A sequence is bounded if there is a constant NN such that an<Na_n<N for every positive integer nn, and it is linear if an=n⋅a1a_n=n\cdot a_1 for every positive integer nn.
Step 3 of 10: Propagate additivity along an arithmetic progression
akm+n=kam+an,am+kn=am+kan(k≥1)a_{km+n}=ka_m+a_n,\qquad a_{m+kn}=a_m+ka_n\quad(k\ge1)
Detailed analysis

Suppose Lm,n>c2(am+an)L_{m,n}>c^2(a_m+a_n) and write x=amx=a_m, y=any=a_n. Step 2 gives am+n=x+ya_{m+n}=x+y. Moreover, lcm⁡(x,x+y)=lcm⁡(x,y)(x+y)/y>c2(2x+y)\operatorname{lcm}(x,x+y)=\operatorname{lcm}(x,y)(x+y)/y>c^2(2x+y), and the analogous inequality holds with x,yx,y exchanged. Thus Step 2 can be applied repeatedly, yielding akm+n=kam+ana_{km+n}=ka_m+a_n and am+kn=am+kana_{m+kn}=a_m+ka_n for every k≥1k\ge1.