*Note:*$\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$.

## APMO 2021 P1

- Anindya Biswas
**Posts:**255**Joined:**Fri Oct 02, 2020 8:51 pm**Location:**Magura, Bangladesh-
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### APMO 2021 P1

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

"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."

—

—

**John von Neumann**- Anindya Biswas
**Posts:**255**Joined:**Fri Oct 02, 2020 8:51 pm**Location:**Magura, Bangladesh-
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### Re: APMO 2021 P1

Let's assume $\lfloor x\rfloor=n, n\in\mathbb{Z}$. So we have $x=\sqrt{nr}>0\Longleftrightarrow n>0\Longleftrightarrow n\geq1$.

So, the given condition is equivalent to the inequality,

\begin{equation}

\begin{split}

&n\leq\sqrt{nr}<n+1\\

\Longleftrightarrow &n\leq r<n+2+\frac1n

\end{split}

\end{equation}

For $n\in\mathbb{N}$, we define $R_n$ to be the set \[R_n:=\left\{r\in\mathbb{R}\mid n\leq r<n+2+\frac1n\right\}=\left[n,n+2+\frac1n\right)\]

$R_n$ is defined to be the interval such that $r\in R_n$ if and only if $x=\sqrt{nr}$ is a real number satisfying $x^2=r\lfloor x\rfloor$.

It is clear that \[\bigcup_{k=2}^{\infty}R_k=[2,\infty)\]

Since $r>2$, there exists $k\geq2, k\in\mathbb{N}$ such that $r\in R_k$.

It is also clear that if $i,j$ are natural number satisfying $j\geq i+3$ then $R_i\cap R_j=\emptyset$.

$\therefore b\geq j\geq i+3\geq i+2+\frac1i>a$

$\Longrightarrow b\neq a$

$\therefore R_i\cap R_j=\emptyset$ $\square$

So, if there exists natural number $i,j$ such that $i\leq j$ and $R_i\cup R_j\neq\emptyset$ then we must have, $i\leq j\leq i+2$.

Therefore, the inequality can have at most $3$ distinct solution for $n$. It remains to prove that this inequality can't have exactly $1$ solution when $r>2$.

To prove this, let's assume $r\in R_k$ for some natural number $k\geq2$.

Now, $R_{k-1}\cup R_{k+1}=\left[k-1,k+3+\frac1{k+1}\right)$

$\therefore R_k\subseteq R_{k-1}\cup R_{k+1}$

$\Longrightarrow r\in R_{k-1}\cup R_{k+1}$

$\Longrightarrow r\in R_{k-1}$ or $r\in R_{k+1}$

So, if $n=k$ is a solution of the inequality, then either $n=k-1$ or $n=k+1$ is also a solution.

So, the number of total solutions is exactly $2$ or $3$. $\blacksquare$

So, the given condition is equivalent to the inequality,

\begin{equation}

\begin{split}

&n\leq\sqrt{nr}<n+1\\

\Longleftrightarrow &n\leq r<n+2+\frac1n

\end{split}

\end{equation}

For $n\in\mathbb{N}$, we define $R_n$ to be the set \[R_n:=\left\{r\in\mathbb{R}\mid n\leq r<n+2+\frac1n\right\}=\left[n,n+2+\frac1n\right)\]

$R_n$ is defined to be the interval such that $r\in R_n$ if and only if $x=\sqrt{nr}$ is a real number satisfying $x^2=r\lfloor x\rfloor$.

It is clear that \[\bigcup_{k=2}^{\infty}R_k=[2,\infty)\]

Since $r>2$, there exists $k\geq2, k\in\mathbb{N}$ such that $r\in R_k$.

It is also clear that if $i,j$ are natural number satisfying $j\geq i+3$ then $R_i\cap R_j=\emptyset$.

**Proof :**Let's assume $a\in R_i, b\in R_j$$\therefore b\geq j\geq i+3\geq i+2+\frac1i>a$

$\Longrightarrow b\neq a$

$\therefore R_i\cap R_j=\emptyset$ $\square$

So, if there exists natural number $i,j$ such that $i\leq j$ and $R_i\cup R_j\neq\emptyset$ then we must have, $i\leq j\leq i+2$.

Therefore, the inequality can have at most $3$ distinct solution for $n$. It remains to prove that this inequality can't have exactly $1$ solution when $r>2$.

To prove this, let's assume $r\in R_k$ for some natural number $k\geq2$.

Now, $R_{k-1}\cup R_{k+1}=\left[k-1,k+3+\frac1{k+1}\right)$

$\therefore R_k\subseteq R_{k-1}\cup R_{k+1}$

$\Longrightarrow r\in R_{k-1}\cup R_{k+1}$

$\Longrightarrow r\in R_{k-1}$ or $r\in R_{k+1}$

So, if $n=k$ is a solution of the inequality, then either $n=k-1$ or $n=k+1$ is also a solution.

So, the number of total solutions is exactly $2$ or $3$. $\blacksquare$

"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."

—

—

**John von Neumann**-
**Posts:**16**Joined:**Mon Dec 21, 2020 9:26 pm

### Re: APMO 2021 P1

Sol by Kazi Nadid

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,$a^2\le ar< (a+1)^{2}$

1st 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]

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,$a^2\le ar< (a+1)^{2}$

1st 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]