<s>[INST] Floor function equation $⌊x + 1/2⌋ + ⌊x⌋ = \frac12 x^6$ 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?
 [/INST] 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.
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