Problem 1

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.

Step 4 of 4: Infinite descent forces every

m_i=0

Previous step Next step

m_i \text{ all even} \implies m_i/2 \text{ satisfies the same condition} \implies m_i=0\ \forall i

Detailed analysis

Since every $m_i$ is even, replacing each $m_i$ by $m_i/2$ preserves the property that any $5$ consecutive terms can be relabeled to satisfy $a-b+c-d+e=0$ , as this condition is linear and homogeneous. If some $m_i\ne0$ , this halving could be repeated indefinitely, but repeated halving of a nonzero integer eventually yields an odd number, contradicting that all terms stay even at every stage. Hence every $m_i=0$ , so $n_1=\cdots=n_{2017}=29$ .