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 \( m_{i} \equiv m_{i+1} + m_{i+2} + m_{i+3} + m_{i+4} \equiv m_{i+5} \pmod{2} \). Since \( \gcd(5, 2017) = 1 \), 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.