The solution set of ${\left(x\right)}^2+{\left(x+1\right)}^2=25$ , where $\left(x\right)$  is the least integer greater than or equal to $x$ , is

The given equation is ${\left(x\right)}^2+{\left(x+1\right)}^2=25$ 

When$x=n\in Z$

Then given equation becomes $\Rightarrow 2x^2\mathit{+}2x-24 =0$

 $\Rightarrow x^2+x-12=0$

$\Rightarrow \left(x+4\right)\left(x-3\right)=0$

$\Rightarrow x=3,-4$ .

Thus$x=3,-4$ .

When $x=n+k,n\in Z,0<k<1$ ,

then we have, ${\left(n+1\right)}^2+{\left(n+2\right)}^2=25$ $\Rightarrow 2n^2+6n-20=0$

$\Rightarrow n^2+3n-10=0$ $\Rightarrow n=2,-5$

$\therefore x=2+k,-5+k$ , where $0<k<1$

Hence$x\in (-5,-4]\cup (2,3]$ .