Olympiad Maths Prep

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Number theory Difficulty 5.2 AIME, harder Prove it Ukraine

Some natural numbers are placed around a circle in such a way that the product of any two neighboring numbers is a perfect square. Prove that the product of any (not necessarily neighboring) two numbers is also a perfect square.
(Arseniy Nikolaev)

Solution

Let us denote numbers as a1,a2,,aka_1, a_2, \ldots, a_k. From the problem statement it follows that for any 1i<jk1 \le i < j \le k the product (aiai+1)(ai+1ai+2)(aj1aj)=n2(a_i a_{i+1})(a_{i+1} a_{i+2}) \ldots (a_{j-1} a_j) = n^2 is a perfect square. Hence the product aiaj=n2ai+12ai+22aj12a_i a_j = \frac{n^2}{a_{i+1}^2 a_{i+2}^2 \ldots a_{j-1}^2} is also a perfect square, as the natural ratio of two perfect squares, Q.E.D.

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