Solving a floor function problem

Question

I am trying to solve the following question: find all real numbers $x$ such that:

$$\Big \lfloor\frac{x^2+1}{10}\Big \rfloor+\Big \lfloor\frac{10}{x^2+1}\Big \rfloor=1$$

If I substitute $n=\frac{x^2+1}{10}$ I will get:

$$ \lfloor n \rfloor+\Big \lfloor\frac{1}{n}\Big \rfloor=1$$

But, from now on I am completely lost.

Answer 1

Well, let's solve a more general case of your problem. We want to find all $x\in\mathbb{R}$ such that:

$$\left\lfloor\frac{x^2+\beta}{\epsilon}\right\rfloor+\left\lfloor\frac{\epsilon}{x^2+\beta}\right\rfloor=1\tag1$$

Where $\beta$, and $\epsilon$ are all real and positive integers, $\beta<\epsilon$ and $\left\lfloor\cdot\right\rfloor$ is the floor function.

First, let's substitute:

$$\text{y}:=\frac{x^2+\beta}{\epsilon}\tag2$$

From $(2)$ we can note that $\text{y}\ge\frac{\beta}{\epsilon}$ because $x^2\ge0$ for all real numbers $x$.

This leads to:

$$\left\lfloor\text{y}\right\rfloor+\left\lfloor\frac{1}{\text{y}}\right\rfloor=1\tag3$$

From $(3)$ we can note that we can't have $\text{y}\ge2$ because will get $\left\lfloor\text{y}\right\rfloor\ge2$ and that leads to $2=1$ which is wrong. This leads to the fact that:

$$\text{y}\in\left[\frac{\beta}{\epsilon},2\right)\tag4$$

We can be more exact by noticing that we also can't have $\text{y}=1$ (it is easy to check why). So we can write:

$$\text{y}\in\left[\frac{\beta}{\epsilon},1\right)\space\cup\space\left(1,2\right)\tag5$$

Now, we can work this out in cases:

• Case 1: when we are on the interval $\text{y}\in\left[\frac{\beta}{\epsilon}\space,\space1\right)$, we can note that $\left\lfloor\text{y}\right\rfloor=0$ because $\beta<\epsilon$. So this leads to $\left\lfloor\frac{1}{\text{y}}\right\rfloor=1$, but in order to get this we need $\frac{1}{\text{y}}\in\left(1,2\right)$. This implies that $\text{y}\in\left(\frac{1}{2},1\right)$;
• Case 2: when we are on the interval $\text{y}\in\left(1,2\right)$, which leads to the fact that $\frac{1}{\text{y}}\in\left(\frac{1}{2},1\right)$. This will give $\left\lfloor\text{y}\right\rfloor=1$ and $\left\lfloor\frac{1}{\text{y}}\right\rfloor=0$. But this all numbers are possible in this region;
• Conclusion: so we conclude that: $$\text{y}\in\left(\frac{1}{2},1\right)\space\cup\space\left(1,2\right)\tag6$$

We can break this up into two cases again and solve it:

• Case 1: $$\frac{1}{2}<\frac{x^2+\beta}{\epsilon}<1\space\Longleftrightarrow\space\frac{\epsilon}{2}-\beta<x^2<\epsilon-\beta\tag7$$
• Case 2: $$1<\frac{x^2+\beta}{\epsilon}<2\space\Longleftrightarrow\space\epsilon-\beta<x^2<2\epsilon-\beta\tag8$$

I will let you finish and don't forget the negative values of $x$.

Answer 2

Hint: Here is one case:

If $r>1$, then $\lfloor \frac{1}{r}\rfloor=0$ and a solution is given if $\lfloor r\rfloor =1$.

Answer 3

Note: $x^2 \ge 0$ so $x^2 + 1 >0; \frac {10}{x^2+1} > 0; \frac {x^2 + 1}{10} > 0$ and therefore we have three fairly broad options:

Case 1: $x^2 + 1= 10$ and then we have $\lfloor\frac{x^2+1}{10}\rfloor+\lfloor\frac{10}{x^2+1}\rfloor = \lfloor 1\rfloor + \lfloor 1\rfloor = 1 + 1 = 2 \ne 1$ so that will yield no solutions.

Case 2: $x^2 + 1 > 10$. then $\frac {x^2+ 1}{10} > 1$ and $0 < \frac {10}{x^2 + 1}< 1$ so $\lfloor\frac{10}{x^2+1}\rfloor = 0$ and we must solve $\lfloor\frac{x^2+1}{10}\rfloor = 1$

Okay. $\lfloor\frac{x^2+1}{10}\rfloor = 1 \iff$

$1 \le \frac {x^2+1}{10} < 2$.

I'll leave that to you to solve.

Case 3: $0 < x^2 + 1 < 10$. Then $0< \frac {x^2 + 1}{10} < 1$ and $\frac {10}{x^2 + 1} > 1$. Thus $\lfloor \frac {x^2 + 1}{10}\rfloor = 0$ and we must solve $\lfloor \frac {10}{x^2 + 1} \lfloor = 1$

Okay. $\lfloor \frac {10}{x^2 + 1} \lfloor = 1\iff$

$1 \le \frac {10}{x^2+1} < 2$.

That's a little trickier than Case 2, but I'll still leave you to solve it.