Subfield: Algebra


Question: Suppose that $s_{1}, s_{2}, s_{3}, \ldots$ is a strictly increasing sequence of positive integers such that the subsequences

$$
s_{s_{1}}, s_{s_{2}}, s_{s_{3}}, \ldots \quad \text { and } \quad s_{s_{1}+1}, s_{s_{2}+1}, s_{s_{3}+1}, \ldots
$$

are both arithmetic progressions. Prove that $s_{1}, s_{2}, s_{3}, \ldots$ is itself an arithmetic progression.

Solution:
Let $D$ be the common difference of the progression $s_{s_{1}}, s_{s_{2}}, \ldots$. Let for $n=$ $1,2, \ldots$

$$
d_{n}=s_{n+1}-s_{n} \text {. }
$$

We have to prove that $d_{n}$ is constant. First we show that the numbers $d_{n}$ are bounded. Indeed, by supposition $d_{n} \geq 1$ for all $n$. Thus, we have for all $n$

$$
d_{n}=s_{n+1}-s_{n} \leq d_{s_{n}}+d_{s_{n}+1}+\cdots+d_{s_{n+1}-1}=s_{s_{n+1}}-s_{s_{n}}=D .
$$

The boundedness implies that there exist

$$
m=\min \left\{d_{n}: n=1,2, \ldots\right\} \quad \text { and } \quad M=\max \left\{d_{n}: n=1,2, \ldots\right\} \text {. }
$$

It suffices to show that $m=M$. Assume that $m<M$. Choose $n$ such that $d_{n}=m$. Considering a telescoping sum of $m=d_{n}=s_{n+1}-s_{n}$ items not greater than $M$ leads to

$$
D=s_{s_{n+1}}-s_{s_{n}}=s_{s_{n}+m}-s_{s_{n}}=d_{s_{n}}+d_{s_{n}+1}+\cdots+d_{s_{n}+m-1} \leq m M
\tag{1}
$$

and equality holds if and only if all items of the sum are equal to $M$. Now choose $n$ such that $d_{n}=M$. In the same way, considering a telescoping sum of $M$ items not less than $m$ we obtain

$$
D=s_{s_{n+1}}-s_{s_{n}}=s_{s_{n}+M}-s_{s_{n}}=d_{s_{n}}+d_{s_{n}+1}+\cdots+d_{s_{n}+M-1} \geq M m
\tag{2}
$$

and equality holds if and only if all items of the sum are equal to $m$. The inequalities (1) and (2) imply that $D=M m$ and that

$$
\begin{aligned}
d_{s_{n}}=d_{s_{n}+1}=\cdots=d_{s_{n+1}-1}=M & \text { if } d_{n}=m \\
d_{s_{n}}=d_{s_{n}+1}=\cdots=d_{s_{n+1}-1}=m & \text { if } d_{n}=M
\end{aligned}
$$

Hence, $d_{n}=m$ implies $d_{s_{n}}=M$. Note that $s_{n} \geq s_{1}+(n-1) \geq n$ for all $n$ and moreover $s_{n}>n$ if $d_{n}=n$, because in the case $s_{n}=n$ we would have $m=d_{n}=d_{s_{n}}=M$ in contradiction to the assumption $m<M$. In the same way $d_{n}=M$ implies $d_{s_{n}}=m$ and $s_{n}>n$. Consequently, there is a strictly increasing sequence $n_{1}, n_{2}, \ldots$ such that

$$
d_{s_{n_{1}}}=M, \quad d_{s_{n_{2}}}=m, \quad d_{s_{n_{3}}}=M, \quad d_{s_{n_{4}}}=m, \quad \ldots
$$

The sequence $d_{s_{1}}, d_{s_{2}}, \ldots$ is the sequence of pairwise differences of $s_{s_{1}+1}, s_{s_{2}+1}, \ldots$ and $s_{s_{1}}, s_{s_{2}}, \ldots$, hence also an arithmetic progression. Thus $m=M$
$\quad$ Let the integers $D$ and $E$ be the common differences of the progressions $s_{s_{1}}, s_{s_{2}}, \ldots$ and $s_{s_{1}+1}, s_{s_{2}+1}, \ldots$, respectively. Let briefly $A=s_{s_{1}}-D$ and $B=s_{s_{1}+1}-E$. Then, for all positive integers $n$,

$$
s_{s_{n}}=A+n D, \quad s_{s_{n}+1}=B+n E .
$$

Since the sequence $s_{1}, s_{2}, \ldots$ is strictly increasing, we have for all positive integers $n$

$$
s_{s_{n}}<s_{s_{n}+1} \leq s_{s_{n+1}}
$$

which implies

$$
A+n D<B+n E \leq A+(n+1) D
$$

and thereby

$$
0<B-A+n(E-D) \leq D
$$

which implies $D-E=0$ and thus

$$
0 \leq B-A \leq D
\tag{3}
$$

Let $m=\min \left\{s_{n+1}-s_{n}: n=1,2, \ldots\right\}$. Then

$$
B-A=\left(s_{s_{1}+1}-E\right)-\left(s_{s_{1}}-D\right)=s_{s_{1}+1}-s_{s_{1}} \geq m
\tag{4}
$$

and

$$
D=A+\left(s_{1}+1\right) D-\left(A+s_{1} D\right)=s_{s_{s_{1}+1}}-s_{s_{s_{1}}}=s_{B+D}-s_{A+D} \geq m(B-A) .
\tag{5}
$$

From (3) we consider two cases.

Case 1. $B-A=D$.

Then, for each positive integer $n, s_{s_{n}+1}=B+n D=A+(n+1) D=s_{s_{n+1}}$, hence $s_{n+1}=s_{n}+1$ and $s_{1}, s_{2}, \ldots$ is an arithmetic progression with common difference 1 .

Case 2. $B-A<D$. Choose some positive integer $N$ such that $s_{N+1}-s_{N}=m$. Then

$$
\begin{aligned}
m(A-B+D-1) & =m((A+(N+1) D)-(B+N D+1)) \\
& \leq s_{A+(N+1) D}-s_{B+N D+1}=s_{s_{s_{N+1}}}-s_{s_{s_{N}+1}+1} \\
& =\left(A+s_{N+1} D\right)-\left(B+\left(s_{N}+1\right) D\right)=\left(s_{N+1}-s_{N}\right) D+A-B-D \\
& =m D+A-B-D,
\end{aligned}
$$

i.e.,

$$
(B-A-m)+(D-m(B-A)) \leq 0 .
\tag{6}
$$

The inequalities (4)-(6) imply that

$$
B-A=m \quad \text { and } \quad D=m(B-A) \text {. }
$$

Assume that there is some positive integer $n$ such that $s_{n+1}>s_{n}+m$. Then $\left.m(m+1) \leq m\left(s_{n+1}-s_{n}\right) \leq s_{s_{n+1}}-s_{s_{n}}=(A+(n+1) D)-(A+n D)\right)=D=m(B-A)=m^{2}$, a contradiction. Hence $s_{1}, s_{2}, \ldots$ is an arithmetic progression with common difference $m$.