Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME, harder Prove it Romania

Let (K,+,)(K, +, \cdot) be a finite field with at least four elements. Prove that the set KK^* can be partitioned into two nonempty subsets AA and BB, such that
xAx=yBy. \sum_{x \in A} x = \prod_{y \in B} y.

Solution

Since the product of the elements of KK^* is 1-1, if AA and BB form a partition of KK^*, then (aAa)(bBb)=1(\prod_{a \in A} a) (\prod_{b \in B} b) = -1, so aAa=bBb\sum_{a \in A} a = \prod_{b \in B} b if and only if
(aAa)(aAa)=1.() \left(\sum_{a \in A} a\right) \left(\prod_{a \in A} a\right) = -1. \quad (*)
Let K4|K| \ge 4. If the characteristic of KK is 22, then the singleton set A={1}A = \{1\} clearly satisfies ()(*). If the characteristic of KK is odd, choose an element aa in K{±1}K^* \setminus \{\pm 1\}, and notice that the 3-element set A={1,1,a}A = \{-1, 1, a\} satisfies ()(*).

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