Find the number of solutions of $$\left\{x\right\}+\left\{\frac{1}{x}\right\}=1,$$ where $\left\{\cdot\right\}$ denotes Fractional part of real number $x$.
My try:
When $x \gt 1$ we get
$$\left\{x\right\}+\frac{1}{x}=1$$ $\implies$
$$\left\{x\right\}=1-\frac{1}{x}.$$
Letting $x=n+f$, where $n \in \mathbb{Z^+}$ and $ 0 \lt f \lt 1$, we get
$$f=1-\frac{1}{n+f}.$$
By Hint given by $J.G$, i am continuing the solution:
we have
$$f^2+(n-1)f+1-n=0$$ solving we get
$$f=\frac{-(n-1)+\sqrt{(n+3)(n-1)}}{2}$$ $\implies$
$$f=\frac{\left(\sqrt{n+3}-\sqrt{n-1}\right)\sqrt{n-1}}{2}$$
Now obviously $n \ne 1$ for if we get $f=0$
So $n=2,3,4,5…$ gives values of $f$ as
$\frac{\sqrt{5}-1}{2}$, $\sqrt{3}-1$, so on which gives infinite solutions.
Best Answer
Now multiply by $n+f$; solve a quadratic to express $f$ in terms of $n$. Don't forget to check negative solutions too.