Find all distinct $a,b,c$ such that $a +1/b = b + 1/c = c+1/a$ Could anyone advise me how to find all values of $a,b, c$ such that $a + \dfrac{1}{b} = b + \dfrac{1}{c} = c + \dfrac{1}{a} \ , $ where $a,b,c $ are distinct real numbers?
I have found that $a, b, c$ can take on either $1,-2, $ or $-\dfrac{1}{2} $, so there are at least 6 sets of solutions. Thank you.
 A: Your solution can be generalized to 
$a={1\over n-1}$
$b=-{n-1\over n}$
$c=-n$
where $a+{1\over b}=b+{1\over c}=c+{1\over a}=-1$ so at least there are infinitely many solutions.
Actually it can be further generalized to the following:
$a={m\over n-m}$
$b=-{n-m\over n}$
$c=-{n\over m}$
Inspired by comments by Oleg567, here is a way to find all solutions:
Let $a+{1\over b}=b+{1\over c}=c+{1\over a}=r$
Then $c=r-{1\over a}$ and $b={1\over r-a}$.
Hence ${1\over r-a} + {a\over ar-1} = r$
${2ar-a^2-1\over (r-a)(ar-1)}=r$
$2ar-a^2-1=ar^3-a^2r^2-r^2+ar$
$(r^2-1)(a^2-ra+1)=0$
For $r=\pm1$ there are infinitely many solutions given by my solution above section and Oleg567's solution.
For $|r|<1$ or $1<|r|<2$ there is no real solution.
For $|r|>2$ there exist a pair of real solutions for each r: $a={r\pm \sqrt{r^2-4}\over2}$
However this implies $b$ and $c$ must equal to one of the pair as well by symmetry, so at least two of $a,b,c$ must be equal. And hence $r=\pm1$ is the only possible solution.
A: Easy parameterization:
$$
(a,b,c) = \left(a, \frac{1}{1-a}, 1-\frac{1}{a}\right),
$$
where $a$ is any real, $a\ne 0, a\ne \frac{1}{2}, a\ne 1$.
Then $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}=1$.
