The number of real roots of $x\lfloor x\rfloor+187=\lfloor x^{2}\rfloor+\lfloor x\rfloor$

Question

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.

Answer 1

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 $x=(I-1)+\frac{187}{I}$.

Answer 2

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} $