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- May 31st 2007, 06:12 AM #16

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- Jun 4th 2007, 10:00 AM #17

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The only answer is the trivial answer $\displaystyle f(x)=0 \mbox{ on }\mathbb{R}$.

Follow, ecMathGeek's approach and arrive at:

$\displaystyle A^2e^x = A^2(e^x-1)$

Hence, $\displaystyle A=0$ 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.