# APMO 2017 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.

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Proof:

My proof is very crooked. An obvious solution is all numbers being $$29$$. If we can show regardless of the plus and minus signs the linear system is not underdetermined, then that will be the only solution. Given any qualifying linear system with matrix $$A$$, we want to show $$det(A)\neq 0$$. Since its parity doesn't change if we flip any plus or minus sign, it suffices to show $$det(A)$$ is odd when the signs in each row are $$1, -1, 1, -1, 1$$. We notice $$AB=C$$, where $$A$$ has two consecutive ones in each row, and $$C$$ has $$1, 0, 0, 0, 0, 1$$ each each row. Since we can show that $$det(B)=det(C)$$, we have $$det(A)=1$$.
Q.E.D.