Number of solutions of $\left\{x\right\}+\left\{\frac{1}{x}\right\}=1$

Question

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.

Answer 1

Now multiply by $n+f$; solve a quadratic to express $f$ in terms of $n$. Don't forget to check negative solutions too.

Answer 2

Let $x:=n+f$. The equation is

$$f+\frac1{n+f}=1,$$

giving the solutions in $f$

$$f=\frac{\pm\sqrt{(n+1)^2-4}-n+1}2.$$

The negative sign cannot work, nor the negative $n$. Then $n\ge1$ is required, but $n=1$ yields $x=1$, which is wrong. Finally,

$$f=\frac{\sqrt{(n+1)^2-4}-n+1}2, \forall n>1.$$

Answer 3

Using continued fraction:

\begin{align} f+\frac{1}{n+f} &= 1 \tag{$n>1$, $0<f<1$} \\[5pt] n+f &= n+1-\frac{1}{n+f} \\[5pt] x &= n+\frac{n+f-1}{n+f} \\[5pt] &= n+\frac{1}{\dfrac{n+f}{n+f-1}} \\[5pt] &= n+\frac{1}{1+\dfrac{1}{n+f-1}} \\[5pt] x &= n+\frac{1}{1+\dfrac{1}{x-1}} \tag{$\star$} \\[5pt] &= \left[ n;\overline{1,(n-1)} \right] \\[5pt] \alpha &= \frac{n+1+\sqrt{(n-1)(n+3)}}{2} \tag{$\alpha=x$} \\[5pt] \beta &= \frac{n+1-\sqrt{(n-1)(n+3)}}{2} \tag{$\beta=\frac{1}{x}$} \end{align}

where $\alpha$, $\beta$ are the roots of $(\star)$.

Note that $\alpha \beta=1$, the symmetric roles for $\alpha$ and $\beta$.

Answer 4

Observations

I am assuming that $\{x\}=x-\lfloor x\rfloor$, which is in $[0,1)$.

There are no integer solutions; if $x\in\mathbb{Z}$, then $\{x\}+\left\{\frac1x\right\}=0+\left\{\frac1x\right\}\lt1$.

Since $\{x\}+\left\{\frac1x\right\}=1$ is stable under $x\leftrightarrow\frac1x$, we can get all positive solutions by looking at $x\gt1$.

Since $\{-x\}=1-\{x\}$ for all $x\not\in\mathbb{Z}$, $\{x\}+\left\{\frac1x\right\}=1$ is stable under $x\leftrightarrow-x$. Thus, we can get all solutions looking at $x\gt0$.

---

Assuming that $\boldsymbol{x\gt1}$

Let $t=\{x\}$ and $n=\lfloor x\rfloor$. This is equivalent to solving $$ t+\frac1{n+t}=1\tag1 $$ for $t\in[0,1)$. Equation $(1)$ gives the quadratic equation $$ t^2+(n-1)t-(n-1)=0\tag2 $$ which has the solution $$ t=\frac{-n+1+\sqrt{(n+1)^2-4}}2\tag3 $$ Since $x=n+t$, we get $$ x=\frac{n+1+\sqrt{(n+1)^2-4}}2\tag4 $$

---

All Solutions

As mentioned above, all solutions can be gotten by taking the reciprocal and negating $(4)$. That is, we can get all solutions, from $$ x=\frac{\pm n\pm\sqrt{n^2-4}}2\tag5 $$ where $n\in\mathbb{Z}$ and $n\ge3$.

Answer 5

It's simpler if you realize that:

EDIT (new text):

$\{a\} + \{\frac b\} = (a + b) - ([a]+ [b])$ and $([a] + [b])$ is always an integer. So $\{a\} + \{frac b\}$ is an integer if and only if $a + b$ is an integer.

Now $0\le \{a \} < 1$ so $0 \le \{a\} + \{b\} < 2$.

So $\{a \} +\{b\} = 0$ if and only if $a, b$ are both integers.

And $\{a\} + \{b\} = 1$ if and only if $ a + b$ is an integer but neither $a$ nor $b$ are integers.

Now the only way $x$ and $\frac 1x$ can both be integers is if $x=\pm 1$.

So $\{x\} + \{\frac 1x\} = 1$ means nothing more or than $\{x\} + \{\frac 1x\}$ is an integer and $x\ne \pm 1$.

OLD: $\{x\} + \{\frac 1x\} = (x + \frac 1x) - ([x] + [\frac 1x])$ and $([x] + [\frac 1x])$ is always an integer.

And as $0 \le \{k\} < 1$ and $\frac 1x \ne 0$ and if $\frac 1x$ is defined $x \ne 0$, then it is always the case that $0 < \{x\} + \{\frac 1x\} < 2$ whenever $x \ne 0$.

so $\{x\} + \{\frac 1x\} = 1$ means nothing more or less than $x + \frac 1x$ is an integer.

....

So $x + \frac 1x = n$ where $n$ is an integer can be solved by

$x^2 + 1 = nx$ and $x \ne 0$ (but as $x = 0\implies x^2 + 1 = 1\ne 0 =nx$ we will not have to worry about that condition.)

$x^2 -nx + 1 = 0$

So $x = \frac {n \pm \sqrt{n^2 - 4}}{2}$.

And those real numbers will exist for any integer $|n| \ge 2$.

There are clearly infinitely many such $x = \frac {n \pm \sqrt{n^2 - 4}}{2}; |n| \ge 2$.

NEW: However we must omit the $x = \frac {n \pm \sqrt{n^2 - 4}}{2}; |n| \ge 2=\pm 1$

i.e

$n \pm 2= \pm\sqrt{n^2 - 4}$

$n^2 \pm 4n + 4 = n^2 - 4$

$n = \pm 2$.

So for the infinite number of integers $n; |n| > 2$ we will have so an $x$.

======

If we want to convince ourselves.

Let $x = \frac {-17 +\sqrt{17^2 -4}}2$ then:

$17^2 - 4 = 285$ And $15^2 = 225 < 285 < 289 = 17^2$ so

$15 <\sqrt{285} < 17$

$-2 <-17+\sqrt{285} < 0$

$-1 < x = \frac {-17 +\sqrt{285}}2 < 0$ so

$\{x\} = \frac {-17 +\sqrt{285}}2 + 1$.

And $\frac 1x = \frac 2{-17+\sqrt{17^2 -4}}=$

$\frac {2(-17 - \sqrt{285})}{(-17 + \sqrt{285})(-17-\sqrt 285)}=$

$\frac {2(-17 - \sqrt{285})}{289-285}$

$\frac {-17 - \sqrt{285}}2$

So $\frac {-17 - 17}2 < \frac 1x < \frac {-17 - 15}2$

So $-17 < \frac 1x < -16$ so

$\{\frac 1x\} = \frac 1x + 17 = \frac {-17 - \sqrt{285}}2 + 17$.

So $\{x\} + \{\frac 1x\} = \frac {-17 +\sqrt{285}}2 + 1 + \frac {-17 - \sqrt{285}}2 + 17=1$.

Ta-da... I guess that is worth doing at least once in a lifetime......