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Ukrainian National Mathematical Olympiad

Algebra Difficulty 5.3 AIME, harder Prove it Ukraine

Determine all possible pairs of integers $a, b$ so that exactly one of them is even and so that there are non-integer $x, y$ , such that both $x+y$ and $ax+by$ are integer?

Solution

Since $ax+by = a(x+y) + (b-a)y$ , the value of $(b-a)y$ is integer. If $|a-b|=1$ , then $y$ has to be integer – contradiction.

For $|a-b|>1$ we can let $y = \frac{1}{b-a}$ and $x = -y$ . Then $x+y=0$ , $(b-a)y$ is integer, hence $ax+by$ is also integer.

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