<s>[INST] Find $x > 0$ for which $\int_{0}^{x} [t]^2 \ dt = 2 (x-1)$. What are all possible $x > 0$ for which the following equation is satisfied? 
$$\int_{0}^{x} [t]^2 \ dt = 2 (x-1),$$ where $[.]$ denotes the bracket (or floor) function. 
I guess we will have to consider different cases depending on whether $\sqrt{n-1} \leq x \leq \sqrt{n}$ for each positive integer $n$. 
For $0 \leq x \leq 1$, the integral on the left-hand side equals $1$; so the solution in this case is clearly $x = 1$ alone. 
What are all possible solutions for $x > 0$? 
 [/INST] Differentiate both sides, we get $[x]^2 = 2$. Le us denote by $\{x\}$ the fractional part of the number $x$, so that $\{x\}=x-[x]$. It is clear that $0\le \{x\} <1$ and $[x]=x-\{x\}$. Thus, the equation becomes $$[x]^2=(x-\{x\})^2=2$$ that is $$x^2-2x\{x\}+\{x\}^2=2.$$ Setting $p=\{x\}$, and write $$x^2-2px+p^2-2=0.$$ Now, using the general law we have
\begin{align}
x_{1,2}  &= \frac{{ - b \pm \sqrt {b^2  - 4ac} }}{{2a}} 
\\
&= \frac{{2p \pm \sqrt {4p^2  - 4\left( 1 \right)\left( {p^2  - 2} \right)} }}{2}
\\
&= \frac{{2p \pm \sqrt 8 }}{2}.
\end{align}
Thus, $x_1 =p+\sqrt{2}$ and $x_2 =p-\sqrt{2}$ however, $a\le p<1$ which means that $x_2<0$ (reject), therefore the required solution is $x=p+\sqrt{2}$, for all $p\in[0,1)$. If you want write as $$\left\{ {x:x = p + \sqrt 2 ,0 \le p < 1} \right\}.$$
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