floor function

[man111]

find all [tex]n\in\mathbb{N}[/tex] which satisfy [tex]\displaystyle \lfloor \sqrt{n}\rfloor = \lfloor \sqrt{n+2011}\rfloor[/tex]

* [tex]\lfloor x \rfloor=[/tex] floor function

[martin123456]

Let [tex]r^2 \le n < (r+1)^2[/tex]. Then we can write [tex]n=r^2+k[/tex], where [tex]k \in \{0,1,\ldots 2r\}[/tex].
We have that [tex]n+2011=r^2+k+2011[/tex]. It is a must [tex]n+2011 < (r+1)^2[/tex] so [tex]k+2011<2r+1 \Leftrightarrow k \le 2r-2011[/tex]. So we must have [tex]2r-2011 \ge 0 \Leftrightarrow r \ge 1006[/tex]. So [tex]n \in \[r^2,r^2+2r-2011\][/tex].

Now lets check that if [tex]n \in \[r^2,r^2+2r-2011\][/tex] and [tex]r\ge 1006[/tex] we have all conditions fulfilled.
1) [tex]r^2 \le n[/tex] and [tex]n < r^2+2r-2010 =(r+1)^2-2011<(r+1)^2[/tex].
2) [tex]r^2<r^2+2011 \le n+2011[/tex] and [tex]n+2011 \le r^2+2r<(r+1)^2[/tex].