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- May 31st 2007, 07:12 AM #16

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- June 4th 2007, 11:00 AM #17

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The only answer is the trivial answer .

Follow, ecMathGeek's approach and arrive at:

Hence, is the only possible constant.

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This problem was taken from my math book, which said it appeared as a problem in the 1990 Putnam. The problem I posed was a variation of that problem, not exactly the same.