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I'm reading the alternative proof of the MO problem: http://koopakoo.wordpress.com/2008/09/03/cgmo-2007-problem-7-and-liouvilles-theorem/ . However, I have a problem, namely that in the alternative proof (just right below the official proof), I don't understand why the assumption will imply $(15+4a+b)(c-a-4) \not=0$, please help.

Chung Hei Wu
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1 Answers1

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Suppose $(15+4a+b)(c-a-4)=0$. Then one of $15+4a+b$ or $c-a-4$ is zero.

If $15+4a+b$ is zero, then $f(\alpha)=c-a-4$ is an integer, but by hypothesis $|f(\alpha)|<10^{-4}$, so $f(\alpha)=0$ which contradicts the hypothesis made just before.

If $c-a-4$ is zero, then $\frac{f(\alpha)}{\alpha}=15+4a+b$ is an integer, but by hypothesis

$$ |\frac{f(\alpha)}{\alpha}| \leq \frac{10^{-4}}{2+\sqrt{3}} \leq 10^{-4}, $$

so again we have $f(\alpha)=0$.

Ewan Delanoy
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