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

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?

• Hint : Which is the smallest and which the largest number , such that its cube is in the given range ? Jan 8, 2019 at 16:33

For every positive integer $$x$$: $$2^8+1\le x^3\le2^{18}+1\iff\sqrt[3]{2^8+1}\le x\le \sqrt[3]{2^{18}+1}\iff 7\le x\le 2^6$$
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 $$\root 3 \of {2^{18} + 1} \approx 64.00008$$.