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Algebra Difficulty 5.3 AIME, harder Prove it Ukraine

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

Solution

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

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

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