Find the total number of solution of the equation

Question

Find number of solution of :- $$[x]^2 = x + 2\{x\}$$

where $[.]$ represents greatest integer function and $\{.\}$ represents fractional part.

Answer 1

The problem is equivalent to finding solutions to $$[x]^2=x+2(x-[x]).$$ This equation is, in turn, equivalent to $[x]^2=3x-2[x]$. Let's write $x=n+\varepsilon$ for the unique $n\in\mathbb{Z}$ and $\varepsilon\in[0,1)$. We arrive at $$n^2=3n+3\varepsilon-2n$$ which is equivalent to $$n(n-1)=3\varepsilon.$$ We note that $3\varepsilon\in[0,3)$ so the only possibilities for $n$ are $n=-1,0,1,2$. The corresponding values for $\varepsilon$, then, are easily computed: $2/3$, $0$, $0$, and $2/3$. The solutions then are a subset of $$\begin{split} x&=-1+2/3=-1/3,\\x&=0+0=0,\\x&=1+0=1,\\x&=2+2/3=8/3.\\\end{split}$$ You can substitute all of the above numbers to determine all solutions.

Answer 2

Let $x=i+f$ where $i=\lfloor x\rfloor$, $f=\{x\}$. The equation says

$$i^2=i+3f,$$ which has solutions for

$$0\le i^2-i<3.$$

There are only four possibilities, $i=-1,0,1,2$ and we are done.