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Romanian Mathematical Olympiad

Algebra Difficulty 5.0 AIME, harder Prove it Romania

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

Solution

Since the product of the elements of $K^*$ is $-1$ , if $A$ and $B$ form a partition of $K^*$ , then $(\prod_{a \in A} a) (\prod_{b \in B} b) = -1$ , so $\sum_{a \in A} a = \prod_{b \in B} b$ if and only if
$\left(\sum_{a \in A} a\right) \left(\prod_{a \in A} a\right) = -1. \quad (*)$
Let $|K| \ge 4$ . If the characteristic of $K$ is $2$ , then the singleton set $A = \{1\}$ clearly satisfies $(*)$ . If the characteristic of $K$ is odd, choose an element $a$ in $K^* \setminus \{\pm 1\}$ , and notice that the 3-element set $A = \{-1, 1, a\}$ satisfies $(*)$ .

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