How to solve ${4 \over 3} ({1 \over x} - \lfloor {1 \over x} \rfloor) = x$?

Question

How to solve
${4 \over 3} ({1 \over x} - \lfloor\frac 1x\rfloor) = x$
$0<x<1$ and $x$ is rational number

$\lfloor\frac 1x\rfloor$ is the floor function

Answer 1

Let's have $\dfrac 1x=n+r$ with $n\in\mathbb Z,\ r\in[0,1)$

Equation becomes $\frac 43(n+r-n)=\frac 1{n+r}\iff 4r(n+r)=3\iff 4r^2+4nr-3=0$

Solving gives (the other root is always negative)

$$r=\dfrac{\sqrt{n^2+3}-n}{2}$$

Now using $r<1$ we get

$\sqrt{n^2+3}< n+2\implies n^2+3< n^2+4n+4\implies 4n>-1\implies n\ge 0$

This gives you all real solutions when $n\in\mathbb N$.

Now for $r$ to be rational then the square root should be integer $n^2+3=k^2\iff (k-n)(k+n)=3$

Since $3$ is prime then one factor is $\pm 1$ and the other $\pm 3$, and with $n\ge 0$ only $n=1$ is solution.

This gives $(n,r)=(1,\frac 12)\iff x=\frac 23$

Answer 2

Hint:

Let $x=\frac{m}{n}$. with $\mbox{gcd}(m,n)=1$ Then $$\frac{1}{x}=\frac{n}{m}$$

Let $n=km+r$ with $0 \leq r <m$.

Then $${4 \over 3} (\frac{r}{m}) = \frac{m}{n}\\ 4rn=3m^2$$

Now use the fact that $gcd(m,n)=1$ to deduce that $n|3$. Then, use $r<m$.