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A natural number $x$ is far from squares and cubes if the inequalities $\left|x-k^{2}\right| > 10^{6}$ and $\left|x-k^{3}\right| > 10^{6}$ hold for every natural number $k$. Prove that there exist infinitely many natural numbers $n$ such that $2^{n}$ is far from squares and cubes.

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    $\begingroup$ Honestly, I don't even know where to start, Mr. Clement. $\endgroup$ – beautyofmath Dec 21 '15 at 7:43
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This is not a full proof, but I would start like this..

A power of 2 that is neither square nor cube must be $2^{6n+1}$ or $2^{6n-1}$, let's take $x=2^{6n+1}$. Now let as find the integers nearest to the square root, which are $\lfloor2^{3n+\frac{1}{2}}\rfloor$ and $\lfloor2^{3n+\frac{1}{2}}\rfloor+1$. First you have to prove that for infinitely many $n$, the squares of these numbers are far enough from $x$. In other words, for infinitely many $n$, $2^{3n}\sqrt{2}$ are far enough from integers -- it is enough to prove that they are contained in $\Bbb Z+[\frac{1}{8}, \frac{7}{8}]$, for example. Here you need to use the irrationality of $\sqrt{2}$: in particular the binary representation of $\sqrt{2}$ contains infinitely many times the string "10". So, for infinitely many $n$, $2^{3n}\sqrt{2}$ looks, in binary, like $\mathrm{something}.*10\ldots$ where $*$ stands for a (possibly empty) string of at most 2 digits.

Further, you need to prove that for infinitely many $n$, both $2^{3n}\sqrt{2}$ and $2^{2n} \sqrt[3]{2}$ are far enough from integers $\ldots$

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  • $\begingroup$ I would write that this is a sketch of the full proof at the start, since this is the way to solve the question. $\endgroup$ – Eric Naslund Dec 21 '15 at 12:03

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