Question

Solve the equation ${\left(x\right)}^2={\left[x\right]}^2+2x$
when [x] and (x) are the integer just less than or equal to x and just greater than or equal to x respectively.
find the number of integral solutions

Answer 1

Given equation
${\left(x\right)}^2={\left[x\right]}^2+2x$
Case I: If $x\in I$
then $\left(x\right)=\left[x\right]$
${\left(x\right)}^2={\left[x\right]}^2+2x$
$\Rightarrow {\left(x\right)}^2={\left[x\right]}^2+2x$
$\therefore x=0$
Case II: If $x\notin I$
then $\left(x\right)=\left[x\right]+1$
$\because {\left(x\right)}^2={\left[x\right]}^2+2x$
$\Rightarrow {\left[x\right]}^2+1+2\left[x\right]={\left[x\right]}^2+2x$
$\therefore x=\left[x\right]+\frac{1}{2}=n+\frac{1}{2},nϵI$
Hence the solution of the original equation is $x=0,n+\frac{1}{2},nϵI$
So, the only integral solution is 1