The number of real roots of $x\lfloor x\rfloor+187=\lfloor x^{2}\rfloor+\lfloor x\rfloor$ Find the number of real roots of the equation $x\lfloor x\rfloor+187=\lfloor x^{2}\rfloor+\lfloor x\rfloor$.
I search other methods except checking case by case.
 A: For $x$ an integer the only solution is $x=187$.
Note that $x\lfloor x \rfloor$ must be an integer since all the other quantities are integers. Let $x=I+f$, where $I$ is an integer and $0 < f  <1$. Then $x\lfloor x \rfloor = I^2+If$. So for this to be an integer $f=\frac{t}{I}$, where $t$ is an integer such that $0<t \leq I-1$. 
Using $x=I+\frac{t}{I}$, we get
$$I^2+t+187=I^2+2t+\left\lfloor \frac{t^2}{I^2}\right \rfloor +I$$
But $\left\lfloor \frac{t^2}{I^2} \right\rfloor =0$.
Thus we have $I+t=187$. 
However $0<t \leq I-1$ gives $94 \leq I <187$.
Now you can get $x=(I-1)+\frac{187}{I}$.
A: For
$x[x]+n=[x^{2}]+[x]
$,
the solutions are
$x = m+\frac{n-m}{m}$
for
$n/2 < m \le n$.
This problem is $n=187$.
Let
$[x] = m$
and
$y = x-m$,
$0 \le y < 1$.
$\begin{array}\\
(m+y)m+n
&=[(m+y)^2]+m\\
\text{or}\\
m^2+n+my
&=[m^2+2my+y^2]+m\\
&=m^2+m+[2my+y^2]\\
&=m^2+m+2my\\
\text{or}\\
n
&=m+my\\
\text{or}\\
my
&=n-m\\
\end{array}
$
Since
$0 \le y < 1$,
$y = k/m$
where
$0 \le k < m$.
Therefore
$k
= n-m
$.
Since $0 \le k < m$,
$n \ge m$
and
$n < 2m$,
so
$n/2 < m \le n$.
For any such $m$,
let
$x = m+\frac{n-m}{m}$.
Then
$\begin{array}\\
x[x]+n
&=m(m+\frac{n-m}{m})+n\\
&=m^2+(n-m)+n\\
&=m^2+2n-m\\
\text{and}\\
[x^2]+[x]
&=[(m+\frac{n-m}{m})^2]+[m+\frac{n-m}{m}]\\
&=m^2+2(n-m)+m\\
&=m^2+2n-m\\
\end{array}
$
