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I am trying to solve the following equation, but I've found some difficulties.

$$\lfloor\sqrt x\rfloor=\lfloor\sqrt[3] x\rfloor\quad\quad(x\in \mathbb R)$$

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closed as off-topic by azimut, Dan Rust, Davide Giraudo, dfeuer, Thomas Oct 13 '13 at 12:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – azimut, Dan Rust, Davide Giraudo, dfeuer, Thomas
If this question can be reworded to fit the rules in the help center, please edit the question.

What is E in this case? – DanielY Oct 13 '13 at 9:55
I assume it's floor function from the title? – Lazar Ljubenović Oct 13 '13 at 9:58
instead to write [x] I 've wrote E(x) but it s the same thing – wuppertal Oct 13 '13 at 10:00
@wuppertal so far, $0$ and $1$ are the only legitimate solutions for me... – DanielY Oct 13 '13 at 10:06
All values of x belongs to this interval $ 0 \leq x < 4 $ are solution. – Shravan40 Oct 13 '13 at 10:12

If $x\lt0$, $\sqrt{x}$ is undefined hence the question makes no sense. For every $x$ in $[0,1)$, $\sqrt{x}$ and $\sqrt[3]{x}$ are both in $[0,1)$ hence $\lfloor\sqrt{x}\rfloor=\lfloor\sqrt[3]{x}\rfloor=0$.

If $x\gt1$, $\lfloor\sqrt{x}\rfloor=\lfloor\sqrt[3]{x}\rfloor$ if and only if there exists an integer $n$ such that $n\leqslant\sqrt[3]{x}\leqslant\sqrt{x}\lt n+1$. Since $x\gt1$, one knows that $n\geqslant1$. The double inequality is equivalent to $n^3\leqslant x\lt(n+1)^2$, in particular one must have $n^3\lt(n+1)^2$. For $n\geqslant1$, this means that $n=1$ or $n=2$. Thus, every $x$ such that $1^3\leqslant x\lt(1+1)^2$ or $2^3\leqslant x\lt(2+1)^2$ is a solution.

Finally, the set of solutions is $[0,4)\cup[8,9)$.

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thank you so much that's what i was searching for . :) – wuppertal Oct 13 '13 at 10:45
You are welcome. But next time, please insert your thoughts on the question you ask. "I've found some difficulties" is definitely not informative enough. – Did Oct 13 '13 at 10:48
$\large\left\lfloor 0.5\right\rfloor = 0 \not= 1$. – Felix Marin Oct 13 '13 at 10:57
@FelixMarin That is true. And? – Did Oct 13 '13 at 10:59
@Did, you had a $1$ that was supposed to be a $0$. I fixed it. It did not infect anything else. – dfeuer Oct 13 '13 at 11:45

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