the question
Solve the equation $\{x\}+\frac{1}{\{x\}}=2+\left\{\frac{1}{[x]}\right\}$, where $[a]$ and $\{ a\}$ represent the whole and fractional parts of the real number $a$ respectively.
The idea
I tried getting $\{\frac{1}{[x]}\}$ to a more acceptable form so I tried these cases:
case 1 : if $[x]=0,1,-1$ then the equation would look like this:
$\{x\}+\frac{1}{\{x\}}=2$ which means that $\{x\}=1$ which isn't possible.
case 2 : if $[x]$ isn't $0,1,-1$ then $\frac{1}{[x]} <1 \implies \left\{ \frac{1}{[x]}\right\}=\frac{1}{[x]}$
From here we simplified the equality to:
$$\{x\}+\frac{1}{\{x\}}=2+\frac{1}{[x]} \implies \frac{{{(\{x\}-1})}^2}{\{x\}}=\frac{1}{[x]}$$
From here I tried using the fact that $\{x\}+ [x]=x$ or get into some inequality but got to nothing useful. Hope one of you can help me! Thank you!
Best Answer
Let
$$ x = a + b \quad$$
Where $a \in \mathbb{Z}, b \in \mathbb{R} \quad 0 \leq b \lt 1$, We have integer part $a$ and fractional part $b$
Now we get
$$b + \frac{1}{b} = 2 + \{\frac{1}{a}\}$$
$$ \implies a,b \neq 0 \quad$$
Case 1
When $a=\{-1, 1\}$, we have $\{\frac{1}{a}\} = 0$, which implies $b=1$ but we have $0 \lt b \lt 1$, hence no solutions exists when $a=\{-1, 1\}$
Case 2
When $a \geq 2$, we have $\{\frac{1}{a}\} = \frac{1}{a}$
$$b + \frac{1}{b} = 2 + \frac{1}{a}$$
$$ ab^2 + a = 2ab + b $$
$$ ab^2 + b(-2a-1) + a = 0$$
Now this a quadratic in terms of $b$
$$ b = \frac{2a+1 \pm \sqrt{4a + 1}}{2a} $$
Now we use the inequality by which $b$ was defined
$$ 0 \lt 1 + \frac{1 \pm \sqrt{4a + 1}}{2a} \lt 1 $$
Solving which yields
$$0 \lt 4a^2$$
Hence there exists a value of $b$ s.t. $0 \lt b \lt 1$ for all $a \geq 2$
Case 3
When $a \leq -2$, we have $\{\frac{1}{a}\} = \frac{1}{a} - \lfloor \frac{1}{a} \rfloor$
As $a \leq -2$, we get $\lfloor \frac{1}{a} \rfloor = -1$
Therefore $\{\frac{1}{a}\} = \frac{1}{a} + 1$
Now we have
$$b + \frac{1}{b} = 2 + \frac{1}{a} + 1$$
$$ab^2 + a = 3ab + b$$
$$ab^2 + b(-3a - 1) + a = 0$$
Solving the quadratic again we get
$$ b = \frac{3a+1 \pm \sqrt{5a^2 + 6a + 1}}{2a} $$
Using the inequality again we get
$$ 0 \lt \frac{3a+1 \pm \sqrt{5a^2 + 6a + 1}}{2a} \lt 1 $$
Solving the inequality you get
$$5a^2 + 5a + 2 > 0$$
Which is true for all values of $a$, hence there exists $b$ s.t. $0 \lt b \lt 1$ for all $a \leq -2$
Answer
There exists a solution for all $x \leq -2$ and $2 \leq x$
Method
A common technique to solve problems involving questions about whole part and fractional part of a number is to write the number in the form $x=a+b$ and then proceed to solve some diophantine equation as done above