Find total number of positive real values of $x$ such that $x, \lfloor x\rfloor, \{x\} $ are in harmonic progression. Where $\lfloor. \rfloor =$g.i.f and $\{. \} =$ fractional part.
Try
As $x, \lfloor x\rfloor, \{x\} $ are in harmonic progression hence $\lfloor x\rfloor =\frac {2x\{x\}}{x+\{x\}}$ then after some manipulation I get $x^2 +\lfloor x\rfloor \{x\} =3x\{x\} $.
Unable to proceed further. Need some help. Any independent process is also welcome. Thanks in advance.
Best Answer
Setting $x=n+a$ helps where $n$ is a non-negative integer with $0\leqslant a\lt 1$.
You already have $\lfloor x\rfloor =\frac {2x\{x\}}{x+\{x\}}$, so one gets $$n=\frac{2(n+a)a}{(n+a)+a}\implies n(n+2a)=2a(n+a)\implies n^2=2a^2\implies a=\frac{n}{\sqrt 2}$$
Now, you can determine $n$ by using $0\leqslant (a=)\frac{n}{\sqrt 2}\lt 1$. Note that $x$ is positive.