# Floor function equation $⌊x + 1/2⌋ + ⌊x⌋ = \frac12 x^6$ [closed]

So in this floor equation $$⌊x + 1/2⌋ + ⌊x⌋ = \frac12 x^6$$, I've tried putting $$x = n + e$$, where $$0 \le e < 1$$, but I didn't get anything useful. What should be an approach in these situations?

• $x=0$ and $x=\sqrt[3]{2}$ are trivial solutions. Given the right side of the equation, there really is just a limited number of things to try. Commented Jul 12, 2020 at 20:11
• I think he means $e= \{x\}$ and $n= ⌊x⌋$ Commented Jul 12, 2020 at 20:11
• Well, you know that the left hand can't exceed $2x+2$ which limits the range considerably.
– lulu
Commented Jul 12, 2020 at 20:12
• Note: typo in my earlier comment. Meant to say the left hand can't exceed $2x+\frac 12$.
– lulu
Commented Jul 12, 2020 at 20:20

Suppose $$x = n + e$$ and consider two cases $$e < 0.5$$ and $$e \geq 0.5$$.

• $$0\leq e < 0.5$$

\begin{alignat*}{2} &2\cdot⌊n + e + 0.5⌋ + 2\cdot⌊n + e⌋ = (n + e)^6\\ &2\cdot n + 2\cdot n = (n + e)^6\\ &4\cdot n = (n + e)^6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\ &2\cdot \sqrt{n} = (n + e)^3\\ &(\sqrt{n} + 1)^2 - (n + 1) = (n + e)^6\\ &-(n + 1) = (n + e)^3 - (\sqrt{n} + 1)^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \end{alignat*}

Notice, that LHS is negative, while RHS positive for $$n \in \mathbb{Z}_{>1}$$. Hence, we need to consider only $$n \in \{0,1 \}$$. Considering $$n = 0$$ and using $$(2)$$ yields $$x = 0$$. While $$n = 1$$ yields $$e = 4^{1/6} - 1 < 0.5$$, hence $$x = 1 + 4^{1/6} - 1 = 4^{1/6}$$.

• $$0.5\leq e < 1$$

The second case can be worked out similarly, but there are no solutions.

Hints:

• $$[x+{1\over 2}] =[2x]-[x]$$
• $$2x-1<[2x]\leq 2x$$

So you have to solve $$4x-2