Selected Problems from the Final Round of National Olympiad

First prove that $1$ and $-1$ are not expressible as the sum of at least three consecutive terms of an arithmetic sequence of integers. Let $a_1$, $a_2$, $\dots$, $a_k$ be $k$ consecutive terms of an arithmetic sequence, where $k\ge 3$. They sum up to $s=\frac{a_1+a_k}{2}\cdot k$. If $k$ is odd, then $s$ is divisible by $k$. If $k$ is even, then $s$ is divisible by $\frac{k}{2}>1$. In both cases, $s$ differs from $1$ and $-1$.

Now prove that every integer $s$ other than $1$ or $-1$ is expressible as the sum of at least three consecutive terms of an arithmetic sequence of integers. If $s=0$, then $s=-1+0+1$. If $s$ is different from zero and is even, i.e., $s=2t$, where $t\ne 0$, then $-t,0,t,2t$ sum up to $2t=s$. If $s$ is odd, i.e., $s=2t+1$, then $-t+1,\dots ,0,1,\dots ,t-1,t,t+1$ are consecutive terms of an arithmetic sequence; they sum up to $t+(t+1)=2t+1=s$, since the terms $-t+1$ through $t-1$ mutually cancel.

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Alternative solution.

Let $a_1$ be the first of the consecutive terms and $d$ be the common difference of consecutive terms. The sum of $n$ consecutive terms is $s=\frac{2a_1+d(n-1)}{2}\cdot n$. Thus $2s=(2a_1+d(n-1))n$. If $s=1$ or $s=-1$, then this equality cannot hold because $n\ge 3$ divides neither $2$ nor $-2$. If $s=0$, then choose the portion of the arithmetic progression to be $-1$, $0$, $1$. If $s$ differs from these numbers, then let $n=2|s|$, $d$ be an arbitrary odd number, and $a_1=\frac{1-(n-1)d}{2}$ if $s>0$, and $a_1=\frac{-1-(n-1)d}{2}$ if $s<0$.