How to solve $\lfloor \sqrt{(k + 1)\cdot2009} \rfloor = \lfloor \sqrt{k\cdot2009} \rfloor$ Is there any way to solve this equation (or to tell how many solutions are there), other than checking all 2009 possibilities? $\lfloor \sqrt{(k + 1)\cdot2009} \rfloor = \lfloor \sqrt{k\cdot2009} \rfloor, 0 \le k \le 2009, k \in \Bbb Z$
 A: Partial answer
We know $ \sqrt{(k + 1)\cdot2009}  >  \sqrt{k\cdot2009}$.
Moreover, 
$$\lfloor \sqrt{(k + 1)\cdot2009} \rfloor = \lfloor \sqrt{k\cdot2009} \rfloor$$
implies
$$ \sqrt{(k + 1)\cdot2009}   <   \sqrt{k\cdot2009}+1 \Rightarrow \\
(k + 1)\cdot2009   <  2009k+2\sqrt{k\cdot2009}+1 \Rightarrow \\
1004   <  \sqrt{k\cdot2009} \Rightarrow \\
1004^2   <  k\cdot2009 \Rightarrow \\
k > \frac{1004^2}{2009} \sim 501.75\Rightarrow \\
$$
This that there is no solution up to $k=501$, so you only need to worry about
$$502 \leq k \leq 2009$$
P.S. Note that for each $ n \geq 1$ you can explicitly find like above some $k_0$ so that for each $k > k_0$ you have
$$ \sqrt{(k + 1)\cdot2009}   <   \sqrt{k\cdot2009}+\frac1n$$
If that is the case, then for any $n$ consecutive integers larger than $k_0$ at least $n-1$ are solutions to your equation. 
This means that if you can find one which is not, the next $n-1$ are for sure. So you need to seek the non-solutions, and then skip few integers looking for the next non-solution...
I would recommend you to test what values you get for $n=2,3,4,..$.
A: As shown in N.S.'s answer, there are no solutions for $k \le 501$ since $\sqrt{2009(k+1)} > \sqrt{2009k}+1$ for all $k \le 501$. Also, for all $502 \le k \le 2009$, we have $\sqrt{2009(k+1)} < \sqrt{2009k}+1$. 
Thus, for each $502 \le k \le 2009$, we have either $\lfloor\sqrt{2009(k+1)}\rfloor - \lfloor\sqrt{2009k}\rfloor = 0$ or $\lfloor\sqrt{2009(k+1)}\rfloor - \lfloor\sqrt{2009k}\rfloor = 1$.
Since $\displaystyle\sum_{k = 502}^{2009}\left(\lfloor\sqrt{2009(k+1)}\rfloor - \lfloor\sqrt{2009k}\rfloor\right)$ $= \lfloor\sqrt{2009 \cdot 2010}\rfloor - \lfloor\sqrt{2009 \cdot 502}\rfloor$ $= 2009 - 1004 = 1005$, we have that $\lfloor\sqrt{2009(k+1)}\rfloor - \lfloor\sqrt{2009k}\rfloor = 1$ for exactly $1005$ values of $k$ in the range $502 \le k \le 2009$. 
Since there are $2009-502+1 = 1508$ values of $k$ in the range $502 \le k \le 2009$, we have that $\lfloor\sqrt{2009(k+1)}\rfloor - \lfloor\sqrt{2009k}\rfloor = 0$ for exactly $1508 - 1005 = 503$ values of $k$. 
