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All-Soviet-Union Mathematical Olympiad

Algebra Difficulty 4.7 AIME Prove it Soviet Union

Problem:

The $n^2$ numbers $\mathbf{x}_{ij}$ satisfy the $n^3$ equations: $\mathbf{x}_{ij} + \mathbf{x}_{jk} + \mathbf{x}_{ki} = 0$ . Prove that we can find numbers $\mathbf{a}_1, \ldots, \mathbf{a}_n$ such that $\mathbf{x}_{ij} = \mathbf{a}_i - \mathbf{a}_j$ .

Solution

Solution:

Taking $i = j = k$ , we have that $\mathbf{x}_{ii} = 0$ .

Now taking $j = k$ , we have that $\mathbf{x}_{ij} = -\mathbf{x}_{ji}$ .

Define $\mathbf{a}_i = \mathbf{x}_{i1}$ .

Then we have $\mathbf{x}_{i1} + \mathbf{x}_{ij} + \mathbf{x}_{ji} = 0$ .

Hence $\mathbf{x}_{ij} = \mathbf{a}_i - \mathbf{a}_j$ .

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