Problem 3

Suppose that

s_1, s_2, s_3, \dots

is a strictly increasing sequence of positive integers such that the subsequences

s_{s_1}, s_{s_2}, s_{s_3}, \dots

and

s_{s_1+1}, s_{s_2+1}, s_{s_3+1}, \dots

are both arithmetic progressions. Prove that the sequence

s_1, s_2, s_3, \dots

is itself an arithmetic progression.

Step 2 of 5: Bound the consecutive differences by D

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d_{s(n)}=B-A\ \text{for every } n;\qquad d_n\le D\ \text{for every } n

Detailed analysis

Let $d_n=s(n+1)-s(n)\ge1$ . Since $s(n)$ and $s(n)+1$ are consecutive integers, $s(s(n)+1)-s(s(n))$ is exactly $d_{s(n)}$ ; but by Step 1 this difference equals $(Dn+B)-(Dn+A)=B-A$ . So $d_{s(n)}=B-A$ for every $n\ge1$ — a fixed constant. Separately, $s(s(n+1))-s(s(n))=D$ telescopes as the sum of the $d_n$ consecutive differences $d_{s(n)}+d_{s(n)+1}+\dots+d_{s(n+1)-1}$ (there are $s(n+1)-s(n)=d_n$ terms). Since every term is $\ge1$ , we get $D\ge d_n$ for every $n$ . Hence $d_n$ is bounded, so $m:=\min_n d_n$ and $M:=\max_n d_n$ both exist.