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Is there any way to solve this equation (or to tell how many solutions are there), other than checking all 2009 possibilities? $\lfloor \sqrt{(k + 1)\cdot2009} \rfloor = \lfloor \sqrt{k\cdot2009} \rfloor, 0 \le k \le 2009, k \in \Bbb Z$

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  • $\begingroup$ My feeling is that it is going to happen sporadically after a certain number $k$. Since, this is the floor function we are talking about, it does not matter if the actual difference keeps decreasing. So, in short, yes, you will have to check all values after a certain $k$. The first occurence is at $524$ and there are $503$ such values. $\endgroup$ – Shailesh Aug 3 '15 at 17:42
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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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  • $\begingroup$ Very nice solution +1 $\endgroup$ – N. S. Aug 3 '15 at 19:03
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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 $n=2,3,4,..$.

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