We call a 5-tuple of integers arrangeable if its elements can be labeled $a$, $b, c, d, e$ in some order so that $a-b+c-d+e=29$. Determine all 2017-tuples of integers $n_{1}, n_{2}, \ldots, n_{2017}$ such that if we place them in a circle in clockwise order, then any 5-tuple of numbers in consecutive positions on the circle is arrangeable.

Answer: $n_{1}=\cdots=n_{2017}=29$.
A valid 2017-tuple is $n_{1}=\cdots=n_{2017}=29$. We will show that it is the only solution.

We first replace each number $n_{i}$ in the circle by $m_{i}:=n_{i}-29$. Since the condition $a-b+$ $c-d+e=29$ can be rewritten as $(a-29)-(b-29)+(c-29)-(d-29)+(e-29)=0$, we have that any five consecutive replaced integers in the circle can be labeled $a, b, c, d, e$ in such a way that $a-b+c-d+e=0$. We claim that this is possible only when all of the $m_{i}$ 's are 0 (and thus all of the original $n_{i}$ 's are 29).

We work with indexes modulo 2017. Notice that for every $i, m_{i}$ and $m_{i+5}$ have the same parity. Indeed, this follows from $m_{i} \equiv m_{i+1}+m_{i+2}+m_{i+3}+m_{i+4} \equiv m_{i+5}(\bmod 2)$. Since $\operatorname{gcd}(5,2017)=1$, this implies that all $m_{i}$ 's are of the same parity. Since $m_{1}+m_{2}+m_{3}+m_{4}+m_{5}$ is even, all $m_{i}$ 's must be even as well.

Suppose for the sake of contradiction that not all $m_{i}$ 's are zero. Then our condition still holds when we divide each number in the circle by 2 . However, by performing repeated divisions, we eventually reach a point where some $m_{i}$ is odd. This is a contradiction.