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

Question

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?

Answer 1

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.

Answer 2

Hints:

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

So you have to solve $$4x-2<x^6\leq 4x$$