Solving a floor function problem [closed]

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.

Solution 1:

Hint: Here is one case:

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

Solution 2:

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.