<s>[INST] 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? $\lfloor \sqrt{(k + 1)\cdot2009} \rfloor = \lfloor \sqrt{k\cdot2009} \rfloor, 0 \le k \le 2009, k \in \Bbb Z$
 [/INST] 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 $k$. 
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