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 2 of 4: Parity is

5

-periodic, hence constant

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m_i-m_{i+1}+m_{i+2}+m_{i+3}-m_{i+4}\equiv m_i+m_{i+1}+m_{i+2}+m_{i+3}+m_{i+4}\pmod2 \implies m_i\equiv m_{i+5}\pmod2

Detailed analysis

Reducing the arrangeable condition modulo $2$ (where signs are irrelevant), any $5$ consecutive $m_i$ 's satisfy $m_i+m_{i+1}+m_{i+2}+m_{i+3}+m_{i+4}\equiv0\pmod2$ . Comparing this for the window starting at $i$ with the window starting at $i+1$ shows $m_i\equiv m_{i+5}\pmod2$ for every $i$ . Since $\gcd(5,2017)=1$ , repeatedly stepping by $5$ cycles through all $2017$ indices, so all $m_i$ share the same parity.