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In a certain sense, yes. Continued fractions provide an infinite number of pairs $m,n$ such that $\big|\frac{\log a}{\log b} - \frac{m}{n}\big| < \frac{1}{n^2}$. For such pairs we have $$\left|1 - b^m/a^n\right| = \left| 1 - \big(b^{\frac{m}{n} - \frac{\log a}{\log b}}\big)^n\right| = \left| 1 - b^{O(1/n)} \right| = O(1/n),$$ with constants depending on ...

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Partial answer In general $x^2-1=y^p$. For $x$ even, $x-1,x+1$ are odd and relatively prime, and this it would mean $x-1=y_0^p$ and $x+1=y_1^p$. This is impossible unless $x=0$ and $p$ odd. If $y$ is even, then $x$ is odd and $x-1$ and $x+1$ have only a factor of $2$ in common. Thus, one of them must be of the form $2^{p-1}y_0^p$ and the other of the ...

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You want to find out when $m$ is close to $2^{n/2}$. If $n$ is even then $m = 2^{n/2}$ is the best you can do, and the next best are $2^{n/2} \pm 1$. If $n$ is odd then the best you can do is $m = \lceil 2^{n/2} \rceil$ or $m = \lfloor 2^{n/2} \rfloor$. Now the interesting question is how close $m$ can get to $\sqrt{2}$ times a power of $2$, and this is a ...

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