How to solve $\lfloor \sqrt{(k + 1)\cdot2009} \rfloor = \lfloor \sqrt{k\cdot2009} \rfloor$

Question

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

Answer 1

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 $k$.

Answer 2

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 $n=2,3,4,..$.