How do you find the real solutions to these simultaneous equations? I am looking for all real $(a,b,c)$ that satisfy the following
\begin{equation}
\left\{
\begin{array}{l}2a + a^2b = b\\ 
2b + b^2c = c\\
2c + c^2a = a\\
\end{array}
\right.
\end{equation}
I know that $a=b=c = 0$ is the only real solution to the problem I know of but I don't know how to prove it. 
I was also given the hint, substitute $a = \tan(x)$. 
 A: Rearrange
\begin{equation}
  \left\{
    \begin{array}{cccc}
      \tan y &=& b &=& \dfrac{2a}{1-a^2} \\ 
      \tan z &=& c &=& \dfrac{2b}{1-b^2} \\ 
      \tan x &=& a &=& \dfrac{2c}{1-c^2} \\ 
    \end{array}
  \right.
\end{equation}
\begin{equation}
  \left\{
    \begin{array}{cccc}
      \tan y &=& \tan 2x  \\ 
      \tan z &=& \tan 2y \\ 
      \tan x &=& \tan 2z \\ 
    \end{array}
  \right.
\end{equation}
\begin{equation}
  \left\{
    \begin{array}{ccc}
      y &=& p\pi+2x \\ 
      z &=& q\pi+2y \\ 
      x &=& r\pi+2z \\ 
    \end{array}
  \right.
\end{equation}
$$
\begin{pmatrix}
  -2 & 1 & 0 \\
  0 & -2 & 1 \\
  1 & 0 & -2
\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix}=
\begin{pmatrix} p\pi \\ q\pi \\ r\pi \end{pmatrix}
$$
$$\begin{pmatrix} x \\ y \\ z \end{pmatrix}=
\begin{pmatrix}
  -\frac{4}{7} & -\frac{2}{7} & -\frac{1}{7} \\
  -\frac{1}{7} & -\frac{4}{7} & -\frac{2}{7} \\
  -\frac{2}{7} & -\frac{1}{7} & -\frac{4}{7}
\end{pmatrix}
\begin{pmatrix} p\pi \\ q\pi \\ r\pi \end{pmatrix}$$
A: $b = \frac {2a}{1-a^2}$
$a = \tan t$
$b = \tan 2t\\
c = \tan 4t\\
a = \tan 8t$
$tan t = tan 8t\\
t + n\pi = 8t\\
7t = n\pi\\
t = \frac {n}{7} \pi$
$a,b,c = \tan \frac {n\pi}{7}, \tan \frac {2n\pi}{7},\tan \frac {4n\pi}{7}$
A: Prove and use
$$\frac{2\tan(x)}{1-\tan^2(x)} = \tan(2x)$$
A: The first equation implies c>a and c>b.
But the third equation implies b>c and b>a.
Therefore, the given simultaneous equations can only be satisfied by a=0, b=0, and c=0.
