Simple NT, $``\text{simple}"$ algebra and (actually) simply formulated Alg-NT for P1, P2 and P5 this year! Although this is as simple as it gets for P1, the Motivation for this (at least in my experience and story in solving it) has interesting points to share; so I'll write this anyway! $\color{green} \rule{4.5cm}{2pt}$ $\color{green} \clubsuit$ $\boxed{\textbf{RHS manipulation.}}$ $\color{green} \clubsuit$ $\color{green} \rule{4.5cm}{2pt}$ We claim that there cannot exist two solutions $x_1$ and $x_2$ with $\lfloor x_1 \rfloor = \lfloor x_2 \rfloor$. $\color{green} \rule{25cm}{0.4pt}$ $\color{green} \spadesuit$ $\boxed{\textbf{Proof.}}$ $\color{green} \spadesuit$ Let their common floor-value be the natural $n$; so we can write $x_1 = n+\epsilon_1$ and $x_2 = n +\epsilon_2$. Putting these two back to the original equations, we get \[ (n+\epsilon_1)^2 = (n+\epsilon_2)^2 \]and we can infer that $n+\epsilon_1 = n+ \epsilon_2$. (Or alternatively, for a much shorter wording, $\lfloor x_1 \rfloor = \lfloor x_2 \rfloor$ implies $r\lfloor x_1 \rfloor = r\lfloor x_2 \rfloor$ implies $x_1^2 = x_2^2$ implies $x_1 = x_2$; without any need to capture $\epsilon_1 = x_1 - \lfloor x_1 \rfloor = \{x_1\}$.) $\color{green} \rule{4.4cm}{2pt}$ $\color{green} \clubsuit$ $\boxed{\textbf{LHS force-feeding.}}$ $\color{green} \clubsuit$ $\color{green} \rule{4.4cm}{2pt}$ Call $n \in \mathbb{N}$ a $\textit{hanging solution}$ of the equation $x^2 = r \lfloor x \rfloor$ iff there exists a (unique) solution $x$ so that $\lfloor x \rfloor = n$. Then, a natural $n$ is a $\textit{hanging solution}$ iff \[ n \leq r < n+2+\dfrac{1}{n} \]$\color{green} \rule{25cm}{0.4pt}$ $\color{green} \spadesuit$ $\boxed{\textbf{Proof.}}$ $\color{green} \spadesuit$ Let $x$ be a solution and $n$ be its floor/respective $\textit{hanging solution}$. Then, assuming $x^2 = n^2+a$ where $0 \leq a < (n+1)^2-n^2 = 2n+1$. Furthermore, since $x^2 = rn$, \[ n^2 \leq n^2+a = rn < n^2+2n+1 \]and when divided by $n$, we get that if $n$ is a hanging solution, $r$ must be in that aforementioned range for the bound to be viable. On the other hand, if the bound is viable, setting $0 \leq a = (r-n) n < 2n+1$ prompts the creation of $x = \sqrt{n^2+a}$. It is easily verified that this $x$ is indeed a solution (by reverse engineering or explicit substitution: when constructed that way, $\lfloor x \rfloor$ must be $n$.) $\blacksquare$ $\color{green} \rule{5.4cm}{2pt}$ $\color{green} \clubsuit$ $\boxed{\textbf{Intervals Analysis FTW.}}$ $\color{green} \clubsuit$ $\color{green} \rule{5.4cm}{2pt}$ There are two or three $\textit{hanging solutions}$ of the equation for each $r > 2$, with there existing three iff \[ r < \lfloor r \rfloor + \dfrac{1}{\lfloor r \rfloor-2} \] $\color{green} \rule{25cm}{0.4pt}$ $\color{green} \spadesuit$ $\boxed{\textbf{Proof.}}$ $\color{green} \spadesuit$ Let for a fixed $r$, $\lfloor r \rfloor = K$ for $K$ a natural number which is $2$ or greater. So, both $K$ and $K-1$ are naturals, and we can be certain that \[ K-1 \leq r < (K-1)+2+\dfrac{1}{K-1} \quad \text{and} \quad K \leq r < K+2 + \dfrac{1}{K} \]as $K \leq r < K+1$. Conclusively, $n = K-1$ and $n = K$ are hanging solutions. Now we look at the naturals which may/may not be hanging solutions, and the definitely not hanging solutions. If $n = K-2$, $n$ is a hanging solution if and only if $K-2 \leq n \leq (K-2)+2+\dfrac{1}{K-2}$; i.e. since we already know that $K \leq r < K+1$, for $r$ to be admitted, $r$ must be strictly less than $K+\dfrac{1}{K-2}$; if $n \leq K-3$ (if that number indeed is positive), then $r$ cannot be admitted to the interval $\left[n,n+2+\dfrac{1}{n}\right)$, as $r$ is too large : i.e. $r \geq K$ implies $r \geq K-3+(2+1) \geq n+2+\dfrac{1}{n}$. same goes for $n \geq K+1$: $r < K+1$ implies that $r < n$. As there are two to three hanging solutions, there are two to three solutions, too. We are done. $\blacksquare$ $\blacksquare$ $\blacksquare$

dame dame

Firstly, observe that $x^2=r[x] \leq rx$, hence $x \leq r$. In addition, $x^2=r[x]>r(x-1)$, hence $(x-\dfrac{r}{2})^2>\dfrac{r^2}{4}-r$. We now consider two Cases. Case 1: $r \geq 4$. Then, we obtain that either $x \geq \dfrac{r+\sqrt{r^2-4r}}{2}$, or $x \leq \dfrac{r-\sqrt{r^2-4r}}{2}$. If the former happens, then note that $\dfrac{r+\sqrt{r^2-4r}}{2} >\dfrac{r+r-2}{2}=r-1$, hence $r-1 <x \leq r$, implying that $[x] \in \{[r]-1,[r] \}$. $\bullet$ If $[x]=[r]-1$, then $x=\sqrt{r([r]-1)}$, which indeed satisfies $[r]-1 \leq x <[r]$, hence it is a solution. $\bullet$ If $[x]=[r]$, then $x=\sqrt{r[r]}$, which indeed satisfies $[r] \leq x <[r]+1$, hence it is a solution. Now, if the latter relation holds, then note that $x \leq \dfrac{r-\sqrt{r^2-4r}}{2} < \dfrac{r-(r-2)}{2}=1$, hence $x<1$, implying that $[x]=0$, which gives $x=0$, a contradiction. Case 2: $r < 4$. Then, $2 <r < 4$. We consider two cases. Subcase 1: $2<r \leq 3$. Then, $x \leq r \leq 3$, hence $[x] \in \{0,1,2,3 \}$. $\bullet$ If $[x]=0$ then $x=0$, a contradiction. $\bullet$ If $[x]=1$, then $x=\sqrt{r}$, which indeed satisfies $1 \leq x <2$. $\bullet$ If $[x]=2$, then $x=\sqrt{2r}$, which indeed satisfies $2 \leq x <3$. $\bullet$ If $[x]=3$, then $x=\sqrt{3r}$, which indeed satisfies $3 \leq x <4$. Subcase 2: $3 <r< 4$. Then, $x \leq r <4$, hence $[x] \in \{0,1,2,3\}$. $\bullet$ If $[x]=0$ then $x=0$, a contradiction. $\bullet$ If $[x]=1$, then $x=\sqrt{r}$, which indeed satisfies $1 \leq x <2$. $\bullet$ If $[x]=2$, then $x=\sqrt{2r}$, which indeed satisfies $2 \leq x <3$. $\bullet$ If $[x]=3$, then $x=\sqrt{3r}$, which indeed satisfies $3 \leq x <4$. In each case, we notice that the equation has either $2$ or $3$ solutions, hence we are done.

It is my contest time solution.After giving the contest I wrote up the solution.But I got 5 out of 7. But I am not able to find out where my mistake is.Can anyone please help me find out? Problem: For every real number $r>2$ there exists exactly two or three positive real number $x$ such that $x^2=r\lfloor x \rfloor$ Let $a=$ $\lfloor x \rfloor$ $a\le x< a+1$ $a^2\le x^2< (a+1)^{2}$ $a$ can't be $0$ Because then $L.H.S>0$ But $R.H.S=0$ which is a contradiction So, $a\ge 1$ $\implies a^2+3a\ge a^2+2a+1$ $\implies a^2+3a \ge (a+1)^{2}$ So, $a^2\le x^2< a^2+3a$ $ \implies a^2 \le ar<a^2+3a$ $ \implies a \le r <a+3 $ As $a$ is an positive integer the three possible values of $a$ are $\lfloor r \rfloor,\lfloor r \rfloor-1,\lfloor r \rfloor-2$ As we have increased the bound from $a^2+2a+1$ to $a^2+3a \lfloor r \rfloor$ -2 will not always work. We will show the proof later. Now we will show that $\lfloor r \rfloor , \lfloor r\rfloor -1$ always works. In the beginning we got,$a^2\le ar< (a+1)^{2}$ $1$st case: $a= \lfloor r \rfloor$ We have to show that $\lfloor r \rfloor^2\le r\lfloor r \rfloor< \lfloor r+1 \rfloor^2$ which is clearly true as, $\lfloor r \rfloor\le r$ $\implies \lfloor r \rfloor^2\le r\lfloor r \rfloor$ Again, $r\le \lfloor r \rfloor+1$ and $ \lfloor r \rfloor \le \lfloor r \rfloor+1$ So,$r\lfloor r \rfloor \le (\lfloor r \rfloor +1)^2$ So,$\lfloor r \rfloor^2 \le r \lfloor r \rfloor<\lfloor r \rfloor+1$ is always true. So,$a=\lfloor r \rfloor$ is always a valid solution. $2$nd Case: Now, for case 2 where $a=\lfloor r \rfloor-1$ we have to show that, $(\lfloor r \rfloor-1)^2 \le r (\lfloor r \rfloor-1)< \lfloor r \rfloor^2$ $\lfloor r \rfloor-1 \le r$ So,$(\lfloor r \rfloor-1)^2\le r(\lfloor r \rfloor-1)$ We know that, $r<\lfloor r \rfloor+1$ and $\lfloor r \rfloor-1<r$ $\implies(\lfloor r+1 \rfloor)(\lfloor r \rfloor-1)>r(\lfloor r \rfloor-1)$ So,$(\lfloor r \rfloor+1)(\lfloor r \rfloor-1)=\lfloor r \rfloor^2-1<\lfloor r\rfloor^2$ $\implies(\lfloor r \rfloor-1)^2 \le r (\lfloor r \rfloor-1)< \lfloor r \rfloor^2$ So,$a=\lfloor r \rfloor$ will also always work. So,There is always two valid values for $a$. Now,we will show that there are some cases where there is exactly 3 solutions for $a$ and there are some cases where there is exactly 2 solutions for $a$. $*$Proof that a=$\lfloor r \rfloor-2$ does not always work. If $b$ is an positive integer and $r=b+0.5$ that means $r-\lfloor r \rfloor=0.5$ So, $a=\lfloor r \rfloor-2=b-2$ So,$r\lfloor x \rfloor=ra=(b+0.5)(b-2)=b^2-1.5b-1$ $(a+1)^2=(b-1)^2=b^2-2b+1$ Now if we want $a$ not to be a solution, $b^2-1.5b-1>b^2-2b+1$ must be true $\implies 0.5b>2$ $\implies b>4$ So, if $r>4$ and $r-\lfloor r \rfloor=0.5$ we will get only two solutions for $a$ Now we can see if $r-\lfloor r \rfloor>0.5$ it will also work because then R.H.S will increase but L.H.S will not. We can find ranges for any $r$ by solving the inequality for any $r-\lfloor r \rfloor$ $*$Now we will show that there are cases where there exists three solutions. If $r$ is an positive integer that means $r-\lfloor r \rfloor=0$ Then $\lfloor r \rfloor-2=r-2$ We will show that then $a=\lfloor r \rfloor-2$ will always be a solution. Then, $ar=(r-2)r=r^2-r$ and $(a+1)^2=(r-2+1)^2=r^2-2r+1$ We can see that $ar<(a+1)^2$ when $r$ is positive integer. So,then $a=\lfloor r \rfloor-2$ will be a solution. We can find ranges for any $r$ by solving the inequality for any $r-\lfloor r \rfloor$ Now,for each valid $a$ we can have one solution because $x^2=ra$ and $x$ has to be positive. So,we can have at least two solutions and at most three solutions. Now,$r<2$ doesn't work because $\lfloor r \rfloor-1\le 0 $ and $\lfloor r \rfloor<0$ which is not possible as shown earlier. So, For every real number $r>2$ there exists exactly two or three positive real number $x$ such that $x^2=r\lfloor x \rfloor$ [Proved]

I hope this works. For $k \le x <k+1$, where $k$ is a positive integer, $f(x)=\frac{x^2}{[x]}$ is bijective and take all the values in the interval $ [k , k+2+\frac{1}{k} )$. Let $a\le r <a+1$ where $a$ is a positive integer. If $x_1$ is a root of the equation then $r \in [[x_1],[x_1]+2+\frac{1}{[x_1]})$. From the last sentence we deduce that $[x_1] \in \{a-2,a-1,a\}$. Thus there are at most $3$ roots. Observe that we definitely have $r \in [[x_1],[x_1]+2+\frac{1}{[x_1]})$ for $[x_1] \in \{a-1,a\}$. Thus there are at least $2$ roots.

My solution is quite similar to Tintarn. Tintarn wrote: Clearly choosing $n$ determines $q$ uniquely, so the number of solutions is equal to the number of solutions to $n^2 \le rn<(n+1)^2$ Btw, this technique is called "Chipa Trick" in our country's mo training camp

Similar to the solutions above - posting for storage.

Pretty easy, but nice enough for a P1. Harder than P3 at least. Let $x=n+f$ where $n \geq 0$ is an integer and $f \in [0,1)$. Substituting, we get a quadratic in $f$, and solving it we get $f=-n \pm \sqrt{rn}$. Since $-n-\sqrt{rn}<0$, we must have $f=-n+\sqrt{rn} \in [0,1)$ (now note that $n=0$ $\implies$ $f=0$ $\implies$ $x=0$, which is impossible, so $n$ is a positive integer). Conversely, if this holds for some $n$, we get exactly one solution for $x$. Therefore we have to prove that $0 \leq -n +\sqrt{rn} <1$ has exactly two or three solutions. Solving the inequalities, we get $$n \leq r <n+2+\frac{1}{n}$$Let $m=\lfloor r \rfloor$. Note that if $n$ is a solution, then $n \leq m$. Also, $$m-1 < m \leq r <(m-1)+2<m+2$$so both $m$ and $m-1$ are solutions (note that $r>2$ $\implies$ $m \geq 2$ $\implies$ $m-1>0$, so all is fine). Now assume some $m-k$ for $k \geq 3$ is a solution. Then $$r<m-k+2+\frac{1}{m-k} \leq m-3+2+1=m \leq r$$Contradiction! Therefore the only solutions are $m,m-1$ and possibly $m-2$ $\implies$ exactly two or three solutions. $\blacksquare$

wrong solution

Wildabandon wrote: We have \[x = \sqrt{r \left ( \lfloor r \rfloor - 2 \right )}, \sqrt{r\left (\lfloor r \rfloor -1 \right )}, \sqrt{r\lfloor r \rfloor }\]have 3 solution of $r\ge 3$... Nope. If $r=41.99999$ then the first one would be $\sqrt{39r} \approx 40.4722$ which does not work since it does not have integer part $\lfloor r\rfloor-2=39$.

For a fixed positive integer, there is a solution $t\le x<t+1$ iff $t\le r<(t+1)^2/t=t+2+1/t$. Moreover, we can ignore the case $\lfloor x\rfloor = t = 0$ because then $x=0$ contradiction. Then for any $r$, there are solutions for $t=\lfloor r\rfloor$ and $t=\lfloor r\rfloor - 1$, possibly a solution for $\lfloor r\rfloor - 2$, and certainly not a solution for $\lfloor r \rfloor + k$ with $k\le -3$ or $k\ge 1$. Done.

Define $f(x)=\frac{x^2}{\lfloor x\rfloor}=\lfloor x\rfloor+2\{x\}+\frac{\{x\}^2}{\lfloor x\rfloor}$. Now we fix $c=\lfloor x\rfloor$. And we define $g(y)=c+2y+\frac{y^2}{c}$ for $0\leq y<1$. Obviously $g$ is increasing and $\lim_{y\rightarrow 1}c+2y+\frac{y^2}{c} = c+2+\frac{1}{c}$. Thus, after fixing $c=\lfloor x\rfloor$, we get interval $[c,c+2+\frac{1}{c} )$ covered, thus for every $r>2$, we have exactly two or three positive real numbers $x$ satisfying the equation $x^2=r\lfloor x \rfloor$.

Miku3D wrote: Prove that for each real number $r>2$, there are exactly two or three positive real numbers $x$ satisfying the equation $x^2=r\lfloor x \rfloor$.

APMO 2021.1. Prove that $\forall 2<r\in\mathbb{R}$, there exists exactly two or three $x\in\mathbb{R}_+$ satisfying the equation $x^2=r\lfloor x\rfloor$. Solution. Let $\lfloor x\rfloor=a, x=a+b$. Then the equation is equivalent to $b^2+2ab+a^2-ra=0$, so $b=-a\pm\sqrt{ra}$. Note that $0\le b<1$, so $b=\sqrt{ra}-a$. Also, $x^2\ge 0$, $r>2\implies a>0$. We obtain $0\le \sqrt{ra}-a<1\implies a^2\le ra<1+2a+a^2\le a^2+3a\implies a\le r<3+a\implies a\in\{\lfloor r\rfloor,\lfloor r\rfloor-1,\lfloor r\rfloor-2\}$. From here we can already conclude that the equation has at most three positive roots, since each one in the set gives at most one positive root. To show that there are at least two positive roots, we show that $a=\lfloor r\rfloor,a=\lfloor r\rfloor-1$ must each gives a positive root. For $a=\lfloor r\rfloor$, we shall show that $\lfloor r\rfloor^2\le r\lfloor r\rfloor<\lfloor r\rfloor^2+2\lfloor r\rfloor+1$. Indeed, the left inequality is trivial and $r\lfloor r\rfloor<\lfloor r\rfloor^2+\lfloor r\rfloor<\lfloor r\rfloor^2+2\lfloor r\rfloor+1$. For $a=\lfloor r\rfloor-1$, we shall show that $\lfloor r\rfloor^2-2\lfloor r\rfloor+1\le r\lfloor r\rfloor-r<\lfloor r\rfloor^2$. For the right inequality, note that $\lfloor r\rfloor^2-r\lfloor r\rfloor=\lfloor r\rfloor(\lfloor r\rfloor-r)>-\lfloor r\rfloor\ge -r\implies \lfloor r\rfloor^2-r\lfloor r\rfloor+r>-r+r=0$. The left one is equivalent to $r\ge \lfloor r\rfloor-1$, which is trivial. Hence we conclude that the equation has at least two positive roots, which is exactly what we wanted.

It's suffice to show that the graph $y=\frac{x^2}{\lfloor x \rfloor }$ and $y=r$ intersect at three points. Solution. Let $x = a + b, 0\leq b < 1$ where $a$ is an integer. If we fix $a$, then letting $b$ varies in $[0,1)$, then $y = \frac{(a+b)^2}{a}\in [a,a+2+\frac{1}{a})$ (A parabola obviously). There are two intuitive main claims which immediately imply the conclusions: Claim 1. They can't intersect at more than three points. Proof. Assume they can. Let $a$ be the minimum value such that $r \in [a,a+2+\frac{1}{a})$. Then we must have $$r \in [a+3,a+5+\frac{1}{a+3})$$$$\implies a+3 < a+2+\frac{1}{a} \implies 1 < \frac{1}{a} \implies a < 1,$$a contradiction. Claim 2. They intersect in at least two points. Proof. Let $a$ be the minimum value such that $r \in [a,a+2+\frac{1}{a})$. It's enough to show that $r \ge a+1$. This is true since by minimality of $a$, we have $$r> a+1+\frac{1}{a-1} \ge a+1.$$ Motivation: The motivation here is to these type of "find the number of solutions" a lot of times can be solved by looking in terms of graphs. I chose $\frac{x^2}{\lfloor x \rfloor}$ since we gain important extra information: if we fix the integer part and let the fractional part varies, it gives "an increasing portion of a parabola".

Write $x=n+\varepsilon$ for a positive integer $n$ and $0 \le \varepsilon <1$ (clearly we can't have $n=0$). The equation is $$n+ \varepsilon = \sqrt{rn}.$$It suffices to solve the inequality $0 \le \sqrt{rn}-n < 1$: each such value of $n$ will give exactly one value of $x$. This inequality is equivalent to \begin{align*} n &\le r\\ n^2-(r-&2)n+1 > 0. \end{align*}If $r < 4$ then we win automatically: the quadratic $n^2-(r-2)n+1$ has negative discriminant, and hence the solutions are $1 \le n \le \lfloor r \rfloor$. We can also manually check that the solutions when $r=4$ are $n=1,2,3$, so from now on assume $r > 4$. The quadratic $n^2-(r-2)n+1$ has roots $r_1=\tfrac{r-2-\sqrt{r^2-4r}}{2}, r_2=\tfrac{r-2+\sqrt{r^2-4r}}{2}$. Due to $r>4$ we have $r_1<1$: $$(r-4)^2<r^2-4r \iff \frac{r-2-\sqrt{r^2-4r}}{2} <1.$$Hence the inequality $n^2-(r-2)n+1>0$ is equivalent to $n>r_2$. It suffices to demonstrate $r-2>r_2>r-3$, and then our solutions will be $n=\lfloor r \rfloor, \lfloor r \rfloor - 1,$ and possibly $\lfloor r \rfloor - 2$. Since $\sqrt{r^2-4r}<r-2$ it's clear that $r_2<r-2$, and $r-3<r_2$ follows from $$(r-4)^2<r^2-4r \iff r-3<\frac{r-2+\sqrt{r^2-4r}}{2}.$$

Yay! APMO 2022 is on the 14th Note that solutions $x$ must satisfy $\lfloor x \rfloor = 1,2,3,\ldots$. The interval $\lfloor x\rfloor = k$ has a solution if \[k^2\leq x^2 = r\lfloor x\rfloor = x^2 < (k+1)^2\]Or, \[k\leq r < k+2+\frac{1}{k}\]Since $r>2$, then $\lfloor r \rfloor \geq 2$. Thus, we may easily see that $k=\lfloor r \rfloor, \lfloor r \rfloor -1$ are always solutions. $\lfloor r \rfloor -2$ is a solution if $\lfloor r \rfloor-2 \geq 1 \Longrightarrow r\geq 3$, and values of $k$ that are $\geq \lfloor r \rfloor$ or $\leq \lfloor r \rfloor -3$ fail. Thus, all equations have either two or three postive real numbers, depending on if $r$ is larger or smaller than 3.

If $\lfloor x\rfloor=0$, then $x=0$ which is a solution for any $r$. Otherwise, let $x=n+s$ with $n\in\mathbb N$ and $s\in[0,1)$, suppose that $(n+s)^2=rn$. Then: $$n^2\le n^2+2ns+s^2<n^2+2n+1,$$so since $n^2+2ns+s^2=(n+s)^2=rn$ we have $n\le r<n+2+\frac1n$. If $n\ge\lfloor r\rfloor+1$ then, since $n\le r$, there are no solutions. Suppose now that $n\le\lfloor r\rfloor-3$. Note that $n\ge1$ implies $\lfloor r\rfloor\ge4$. Since $x+\frac1x$ is increasing of $x$, we have: $$r<n+\frac1n+2\le\lfloor r\rfloor-1+\frac1{\lfloor r\rfloor-3}.$$But since $r\ge\lfloor r\rfloor$ and: $$\lfloor r\rfloor-1+\frac1{\lfloor r\rfloor-3}-\lfloor r\rfloor\le0,$$this is impossible. Note that $n=\lfloor r\rfloor$ and $n=\lfloor r\rfloor-1$ always produce solutions, and $n=\lfloor r\rfloor-2$ sometimes does, so the conclusion follows.

Solved with brickybrook_25 Let $x=n+c$, where $n=\lfloor x\rfloor$. We have \[(n+c)^2=rn\implies n^2+2nc+c^2=rn\implies c^2+2nc+n^2-rn=0\] The solutions for $c$ are $c=\frac{-2n\pm \sqrt{4rn}}{2}$. Clearly $\frac{-2n-\sqrt{4rn}}{2}<0$, so $c=\frac{\sqrt{4rn}-2n}{2}=\sqrt{rn}-n$. We need $0\le \sqrt{rn}-n<1$, so $\sqrt{rn}\ge n$, which implies $rn\ge n^2\implies r\ge n$. Now also $\sqrt{rn}<n+1\implies rn<n^2+2n+1\le n^2+3n$, so $r<n+3$. This give us $r-3<n\le r$. There are at most $3$ integers in this interval, so at most three values of $x$. Now, to finish, we have to find two values of $x$ that satisfy the equation. We claim that $x=\sqrt{r\lfloor r\rfloor}$ and $x=\sqrt{r\lfloor r-1\rfloor}$ work Proof: First, we'll show $x=\sqrt{r\lfloor r\rfloor }$ works. We want to show $r\lfloor r\rfloor=r\lfloor x\rfloor$, so $\lfloor r\rfloor = \lfloor x \rfloor$. Note that $\lfloor r\rfloor ^2\le r\lfloor r\rfloor \le r^2<\lceil r \rceil^2 $, so $\lfloor r\rfloor \le x<\lfloor r\rfloor+1$, thus $\lfloor r\rfloor = \lfloor x \rfloor$ is true. Now, we'll show $x=\sqrt{r\lfloor r-1\rfloor}$ works. We want to show $\lfloor r-1\rfloor =\lfloor x \rfloor$. Note that $\lfloor r-1 \rfloor^2 < r\lfloor r-1\rfloor< \lfloor r\rfloor ^2$, so $\lfloor x\rfloor = \lfloor r-1\rfloor$ is true. $\blacksquare$

Basic. Notice $x^2 \leq rx$ so $r \geq x$. If $r \geq x+3$ then $x^2 > (x+3)(x-1) = x^2+2x-3$ so $x < \frac{3}{2}$. If $x \geq 1$ then $r = x^2 < \frac{9}{4} < 3$ and if $x < 1$ then $r = 0$ contradiction. Thus $r < x+3$. So at most three values work. Checking shows $x = \lfloor r \rfloor, \lfloor r \rfloor - 1$ both work, so we are done.

Here's a low IQ solution which literally gives all the solutions. Solution: We will show that if $r \in [2,3)$, then $x = \sqrt{r\lfloor r \rfloor}$ $x = \sqrt{r(\lfloor r \rfloor-1)}$. For $r \ge 3$, we have $x = \sqrt{r\lfloor r \rfloor}$ $x = \sqrt{r(\lfloor r \rfloor-1)}$. $x = \sqrt{r(\lfloor r \rfloor - 2)}$. They clearly work and now we will show that these are the only possible solutions of $\mathbb{R}^+$. It is easy to see that $\lfloor x \rfloor \ne 0$ for any $r > 2$. So $\lfloor x \rfloor\ge 1$. Call the original equation $(1)$ Claim: $r \ge \lfloor x \rfloor$ for any $r$ and $\lfloor x \rfloor > r- 3$ for $r \ge 4$. To be precise, \[\lfloor x \rfloor \in X = \{\lfloor r \rfloor, \lfloor r \rfloor -1, \lfloor r \rfloor -2.\}\] Proof: Substitute $x = I + f$ for $I \in \mathbb{Z}^+$ and $f \in [0,1)$ in $(1)$. Solving for $f$ by the Quadratic Formula we get \[0 \le f = \sqrt{rI} - I < 1\]Solving $0 \le f$ would give us $\lfloor x \rfloor \le r$. Solving the other side for $I$, we would get \[I > \frac{r-2+\sqrt{r^2-4r}}{2} \ge r-3\]where $ \ge r - 3$ could be easily shown by working backwards. Also do note that we skipped the negative version of square root since it will give $I < 1$ which is absurd. Here's some magic! For $r <4$, we get $1 \le \lfloor x \rfloor \le \lfloor r \rfloor \le 3$. This proves the very last statement of the claim. $\square$ For $r \ge 3$, $\lfloor x \rfloor$ can take any value from $X$. This would give the claimed solutions. But for $r \in [2,3)$, $\lfloor x \rfloor \ne \lfloor r \rfloor - 2$ otherwise $\lfloor x \rfloor = 0$ which is a contradiction. We are done. $\blacksquare$

For any $x$ that is a real number satisfying $x^2=r \lfloor x \rfloor$, it holds equivalently that ${\lfloor x \rfloor}^2 \leq r \lfloor x \rfloor$ and $(\lfloor x \rfloor + 1)^2 \geq r \lfloor x \rfloor$, as $y = x^2$ is strictly increasing for $x > 0$ and $r \lfloor x \rfloor$ is a horizontal line. Note that ${\lfloor x \rfloor}^2 \leq r \lfloor x \rfloor$ or $\lfloor x \rfloor \leq r$ and $$(\lfloor x \rfloor + 1)^2 \geq r \lfloor x \rfloor$$$$(\lfloor x \rfloor + 1)^2 \geq r (\lfloor x \rfloor + 1 - 1) $$Substituting $x' = \lfloor x \rfloor + 1$, $$ (x')^2 \geq r (x'-1) \Longleftrightarrow (x' - \frac{r}{2})^2 - \frac{r^2-4r}{4} \geq 0$$ Thus, this inequality holds when $0 \leq r \leq 4$ and for $r > 4$ if $\lfloor x \rfloor + 1 \geq \frac{r + \sqrt{r^2 - 4r}}{2}$ or $\lfloor x \rfloor + 1 \leq \frac{r - \sqrt{r^2 - 4r}}{2}$. Case 1 $(r \leq 4)$ In this case, the second inequality is satisfied by $r \leq 4$. From the first inequality $\lfloor x \rfloor \leq r$, and since $2 < r \leq 4$, $\lfloor x \rfloor$ satisfies the inequality if it is $\lfloor r \rfloor, \lfloor r \rfloor -1, \lfloor r \rfloor - 2$. $\lfloor x \rfloor$ cannot be $\lfloor r \rfloor -3$ as it cannot be less than $0$ and $r > 2$. Note that as formulated, $x^2$ intersects (and only once) with the horizontal line $r \lfloor x \rfloor$ for some $x \in [\lfloor x \rfloor, \lfloor x \rfloor + 1)$ iff the two inequalities hold for $x$. There are exactly three solutions. Case 2 $(r > 4)$ For the first case of the second inequalities, the two inequalities are $$ \frac{r + \sqrt{r^2 - 4r}}{2} - 1 \leq \lfloor x \rfloor \leq r$$ Note that as $r - 4 < \sqrt{r^2 - 4r}$ for $r > 4$, $$\frac{r + \sqrt{r^2 - 4r}}{2} - 1 > r - 3$$Furthermore, since $\sqrt{r^2 - 4r} < r - 2$ $$\frac{r + \sqrt{r^2 - 4r}}{2} - 1 < r - 2$$$$ r - 3 < \frac{r + \sqrt{r^2 - 4r}}{2} - 1 < r - 2$$ Thus, $\lfloor x \rfloor$ is guaranteed to have an intersection for $\lfloor r \rfloor, \lfloor r \rfloor - 1$ from the inequalities and potentially $\lfloor r \rfloor -2$ however it cannot be lower than this as the bound is strictly greater than $r - 3$. Therefore there are either two or three solutions. For the second case of the second inequality, the two inequalities are $$ \lfloor x \rfloor \leq \frac{r - \sqrt{r^2 - 4r}}{2} - 1 \leq r $$ Since $\sqrt{r^2 - 4r} > r - 4 $ when $r > 4$ and $\sqrt{r^2 - 4r} < r- 2$ $$1 < \frac{r - \sqrt{r^2 - 4r}}{2} < 2 \implies \lfloor x \rfloor < 1$$ $\lfloor x \rfloor$ cannot be $0$ as $x$ becomes $0$, therefore there are no intersections from this second version of the inequality. Thus when $r > 4$, there are in total two or three intersections.

Let $x= I+F$, where $F\in [0,1)$, $I\in \mathbb{Z}$ $x^2=r \lfloor x \rfloor \Rightarrow (I+F)^2=rI$ $0\leq F<1 \Rightarrow I^2\leq (I+F)^2 < (I+1)^2$ $\Rightarrow I^2\leq rI < (I+1)^2$ Clearly, $I\nleq (r-3)$, as then the inequality is contradicted. $\Rightarrow r\geq I>(r-3)$ Thus we can at most have $3$ solutions for $x$. (For each $I$ we get a corresponding $F$.) Also, $I=r$ and $I=(r-1)$ can always be a solution to the inequality, regardless of the value of $r$, giving the corresponding value of $F$, and thus fixing $x$. $\Rightarrow x$ has at least $2$ solutions and at most $3$. $\framebox{Q.E.D}$!

We claim that there are at most 3 solutions. Firstly, notice that $\lfloor x \rfloor \leq x < x + 1$. This implies that: $${\lfloor x \rfloor}^2 \leq x^2 < {\lfloor x \rfloor}^2 + 2\lfloor x \rfloor + 1 \leq (\lfloor x \rfloor)^2 + 3\lfloor x \rfloor$$$$\implies {\lfloor x \rfloor}^2 \leq r\lfloor x \rfloor < {\lfloor x \rfloor}^2 + 3\lfloor x \rfloor \implies \lfloor x \rfloor \leq r \leq \lfloor x \rfloor + 3$$This means that $r - 3 < \lfloor x \rfloor \leq r$, and since $\lfloor x \rfloor$ is a positive integer, $\lfloor x \rfloor$ and $x$ have at most 3 solutions ($\lfloor x \rfloor$ can take the values $r, r - 1, r - 2$). Plugging in $\lfloor x \rfloor$ as $r, r - 1$, they clearly work so there at least two solutions and at most 3, hence we are done.