Let $R=$  the set of real numbers,

$Z=$ the set of integers,

$N=$ the set of natural numbers.

If S be the solution set of the equation ${\left(x\right)}^2+{\left[x\right]}^2={\left(x-1\right)}^2+{\left[x-1\right]}^2,$  

where $\left(x\right)=$  the least integer greater than or equal to $x$  

and $\left[x\right]=$ the greatest integer less than or equal to$x,$ then

Given that $R=$  the set of real numbers,

$Z=$ the set of integers,

$N=$ the set of natural numbers.

The given set of the equation ${\left(x\right)}^2+{\left[x\right]}^2={\left(x-1\right)}^2+{\left[x-1\right]}^2$

Here $\left(x\right)=$  least integer $\le x$

For example, $\left(2,3\right)=\left[2,3\right]=2$

$\therefore$  $\left(x\right)-\left[x\right]=1,$   if $x$ is not  integer and $\left[x\right]=\left(x\right)$

$\therefore$  ${\left(x\right)}^2+{\left[x\right]}^2={\left(x-1\right)}^2+{\left[x+1\right]}^2$

$\Rightarrow {\left(x\right)}^2+{\left[x\right]}^2={\left(x\right)}^2-2\left(x\right)+1+{\left[x\right]}^2+2\left[x\right]+1$

$\Rightarrow$   $-1+1=0$   if $x\notin Z$

and $0+1\ne 0$    if $x\in Z$

$\therefore$    The solution set $S=R-Z$ .