<s>[INST] 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.
 [/INST] 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}
$
</s>