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How many perfect cubes are between $2^8+1$ to $2^{18}+1$ ( inclusively)

$2^9,2^{12},2^{15},2^{18}$ are all perfect cube.there are many other. I try to use modulo 2 .but it won't work, and no other methods i tried get me nowhere

Any ideas?

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    $\begingroup$ Hint : Which is the smallest and which the largest number , such that its cube is in the given range ? $\endgroup$
    – Peter
    Jan 8, 2019 at 16:33

2 Answers 2

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Hint:

For every positive integer $x$: $$2^8+1\le x^3\le2^{18}+1\iff\sqrt[3]{2^8+1}\le x\le \sqrt[3]{2^{18}+1}\iff 7\le x\le 2^6$$

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  • $\begingroup$ Thanks i got it $\endgroup$
    – Cloud JR K
    Jan 8, 2019 at 16:41
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One way to get around your very limited human sense of numbers is to take advantage of logarithms and roots whenever you can.

For problems like the one you're dealing with today, what you need is the computation of cubic roots. For example, $10$ is not a perfect cube, but we see that $\root 3 \of {10} \approx 2.15$, and then we check that $2^3 < 10 < 3^3$. What's more, this "$\approx 2.15$" should tell you that $10$ is much closer to the next lower perfect cube than it is to the next higher cube.

So for your particular problem, you need to find that $\root 3 \of {2^8 + 1} \approx 6.35$ and $\root 3 \of {2^{18} + 1} \approx 64.00008$.

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