THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS

Question

a) Determine all 4-tuples (x_0, x_1, x_2, x_3) of pairwise distinct integers such that each x_k is coprime to x_k+1 (indices are reduced modulo 4), and the cyclic sum x_0/x_1 + x_1/x_2 + x_2/x_3 + x_3/x_0 is an integer. b) Show that there are infinitely many 5-tuples (x_0, x_1, x_2, x_3, x_4) of pairwise distinct integers such that each x_k is coprime to x_k+1 (indices are reduced modulo 5), and the cyclic sum x_0/x_1 + x_1/x_2 + x_2/x_3 + x_3/x_4 + x_4/x_0 is an integer.

Accepted Answer

a) The only 4-tuples satisfying the required conditions are

(1, 2, - 1, - 2) and (1, - 2, - 1, 2)

along with their cyclic permutations; the cyclic sum in question is

- 3

for the former and its cyclic permutations, and

3

for the latter and its cyclic permutations. The condition that the cyclic sum of the

x_{k} / x_{k + 1}

be integral is equivalent to the cyclic sum of the

x_{k}^{2} x_{k + 2} x_{k + 3}

being of the form

N x_{0} x_{1} x_{2} x_{3}

for some integral

N

. Since the

x_{k}

are pairwise distinct, and each

x_{k}

is coprime to

x_{k + 1}

, it follows that

x_{2} = - x_{0}

and

x_{3} = - x_{1}

, so

2 (x_{0}^{2} - x_{1}^{2}) = N x_{0} x_{1}

. Since

x_{0}

and

x_{1}

are relatively prime, so are

x_{0}^{2} - x_{1}^{2}

and

x_{0} x_{1}

, hence the latter is a divisor of

2

. The required 4-tuples are now easily derived.

b) If

k

is an integer greater than

1

, then

(1, - k, k + 1, - 1, - k^{2} - k)

is a 5-tuple satisfying the required conditions; the corresponding cyclic sum is

- k^{2} - 2 k - 2

.