Solving $x – \dfrac{1}{ \lfloor x \rfloor } = \lfloor x \rfloor$

Question

Determine the three smallest positive solutions to the equation $$x –
\dfrac{1}{ \lfloor x \rfloor } = \lfloor x \rfloor$$

This is my attempt:
If $x$ is an integer, we have:

$$x^2-1=x^2,$$
which has no solutions. Therefore, $x$ cannot be an integer. We can therefore assume that $x=n+\delta$, where $n$ is an integer and $0<\delta<1$. We thence obtain:

$$(n+\delta) – \dfrac{1}{ \lfloor n+\delta \rfloor } = \lfloor n+\delta \rfloor,$$
which gives
$$(n+\delta) – \dfrac{1}{ n} = n$$
$$\delta=\frac1n$$

Since $0<\delta<1$, $n_1=1,\ n_2=2\, n_3=3$. Hence, $$x=(1+\delta), \ x_2=(2+\delta), \ x_3=(3+\delta)$$

But can this be right? I have the feeling something is odd here. Do I have to specify $\delta$? That can be as small as we like…

Thanks

Answer 1

A quicker way to find the solution would be notice that the original equation can be rewritten as

$$\frac{1}{\lfloor x\rfloor} = \{x\}$$

where $\{\cdot\} = \cdot - \lfloor \cdot \rfloor$ denotes the fractional part function. This means the solutions are of the form

$$x = \lfloor x \rfloor + \frac{1}{\lfloor x \rfloor}$$

or

$$x_n = \frac{n^2+1}{n}$$

for $n\geq 2$.