Value of x - where [x] is greatest integer function

[man111]

how can i find [tex]x[/tex] in [tex]\left[x^2+2x\right] = \left[x\right]^2+2\left[x\right][/tex]

where [tex]\left[x\right]=[/tex] greatest integer function

[vel.angelov]

It's not hard to see that if [tex]x\in Z[/tex] the equation hold
Now let [tex]x\notin Z[/tex]
Then [tex]x=[x]-r[/tex] ,[tex]r\in (0,1)[/tex]
[tex][x^{2}+2x]=[([x]-r)^{2}+2([x]-r)]=[[x]^2+2[x]-2[x]r+r^{2}-2r]=[x]^2+2[x]+[-2[x]r+r^{2}-2r][/tex]
[tex]=>[/tex]
[tex][x]^2+2[x]+[-2[x]r+r^{2}-2r]=[x]^2+2[x][/tex]
[tex][-2[x]r+r^{2}-2r]=0[/tex]
[tex]-1<-2[x]r+r^{2}-2r\le 0[/tex]
[tex]\frac{r}{2 } -1\le [x]<\frac{(1-r)^2}{2r }[/tex]
but [tex]r\in (0,1)[/tex]
we put [tex][x]=n[/tex] where [tex]n\in Z[/tex]
and solve inequalities for r
[tex]r\in (0, 1+n-\sqrt{n^2+2n} )[/tex]for [tex]n\ge 0[/tex]
Answer:
[tex]x\in Z \cup x\in (\sqrt{n^2+2n} -1,n)[/tex] for [tex]n\in N_{ 0}[/tex]

[man111]

Thanks vel.angelov