Floor equation, determine solutions

Question

there is a equation $\lfloor\sqrt{6}x\rfloor=\lfloor2.5x\rfloor$, determine how many integer solutions and real solutions has the equation.

I know that $\{-1,0,1\}$ satisfies the equation, but I don't know if there isn't any other solution.

Answer 1

Since $\sqrt6<2.5$, $0\le\left\lfloor\sqrt6 x\right\rfloor=\lfloor 2.5x\rfloor$ if and only if there is an integer $n$ such that $$n\le\sqrt6 x<2.5x<n+1\;.$$

Suppose that $n=\sqrt6x$; then $x=\frac{n}{\sqrt6}$, so $2.5x=\frac{2.5}{\sqrt6}n$, and we require that this be less than $n+1$:

$$\frac{2.5}{\sqrt6}n<n+1\;,\tag{1}$$

or $$n<\frac{\sqrt6}{2.5-\sqrt6}\approx48.5\;,$$ i.e., $n\le 48$.

For example, if you take $n=48$, then $x=\frac{48}{\sqrt6}\approx19.59591794227$, and $$\lfloor 2.5x\rfloor\approx\lfloor48.98979485566\rfloor=48\;.$$ You can increase $x$ up to but not including $\frac{49}{2.5}=19.6$, so $n=48$ gives you solutions $$\frac{48}{\sqrt6}\le x<19.6\;.$$

If you take $n=0$, at the other extreme, then $0\le x<0.4$.

I’ll leave it to you to look at the other integers and to do the analysis for the case $\left\lfloor\sqrt6 x\right\rfloor=\lfloor 2.5x\rfloor<0$.

Answer 2

You know $\sqrt{6} < 2.5$. So, $x\cdot (2.5 -\sqrt{6}) \longrightarrow \infty$. Thus, there are only finitely many integer $x$'s that satisfy the relationship. Also $20=\lceil \frac{1}{2.5-\sqrt{6}}\rceil$. So that $\frac{1}{2.5-\sqrt{6}}<20$, thus, $20 \cdot (2.5 - \sqrt{6})>1$. So, if $|x| \geq 20$ the statement is false. Since the difference between $\sqrt{6}x$ and $2.5x$ will always be larger than 1.

Hence it is only a matter of checking integers $x \in [-19,19]$. I leave it to you to solve the rest, this should help, though.