How do you find the real solutions to these simultaneous equations?

Question

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 $a = \tan(x)$.

Answer 1

$b = \frac {2a}{1-a^2}$

$a = \tan t$

$b = \tan 2t\\ c = \tan 4t\\ a = \tan 8t$

$tan t = tan 8t\\ t + n\pi = 8t\\ 7t = n\pi\\ t = \frac {n}{7} \pi$

$a,b,c = \tan \frac {n\pi}{7}, \tan \frac {2n\pi}{7},\tan \frac {4n\pi}{7}$

Answer 2

Prove and use

$$\frac{2\tan(x)}{1-\tan^2(x)} = \tan(2x)$$

Answer 3

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}

$$ \begin{pmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 1 & 0 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}= \begin{pmatrix} p\pi \\ q\pi \\ r\pi \end{pmatrix} $$

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix}= \begin{pmatrix} -\frac{4}{7} & -\frac{2}{7} & -\frac{1}{7} \\ -\frac{1}{7} & -\frac{4}{7} & -\frac{2}{7} \\ -\frac{2}{7} & -\frac{1}{7} & -\frac{4}{7} \end{pmatrix} \begin{pmatrix} p\pi \\ q\pi \\ r\pi \end{pmatrix}$$

Answer 4

Since it's an algebra problem, let me give a pure algebraic solution without using any trigonometric functions, so that it's accessible to those who haven't learnt them.

Make a variable the subject of each equation like this.

\begin{cases} b =& \dfrac{2a}{1-a^2} \\ c =& \dfrac{2b}{1-b^2} \\ a =& \dfrac{2c}{1-c^2} \tag{S} \label{S} \end{cases}

Note that each variable can't be equal to $\pm1$. (We have $2a + a^2 b = b$, so if $a^2 = 1$, $2a = 0$, which is equivalent to $a = 0$, which is a contradiction.) Thus, writing each variable as a fraction in \eqref{S} is allowed.

Multiply the three equations in \eqref{S}.

\begin{align} abc &= \frac{8abc}{(1-a^2)(1-b^2)(1-c^2)} \\ abc = 0 \quad &\text{or}\quad (1-a^2)(1-b^2)(1-c^2) = 8 \end{align}

If we have $abc = 0$, meaning that one of them must be zero. No matter which one we start with (say $a = 0$), we can always use \eqref{S} to deduce that the other two variables are also equal to zero. (Since $b \stackrel{\eqref{S}}= \dfrac{2a}{1-a^2}$, $b = 0$, and similarly, $c = 0$.)

In the second case, it suffices to find $a$ and $b$.

\begin{align} 1 - \frac{8}{(1-a^2)(1-b^2)} &= c^2 \\ 1-b^2 - \frac{8}{1-a^2} &= \frac{4b^2}{1-b^2} \\ 1-\frac{4a^2}{(1-a^2)^2} - \frac{8}{1-a^2} &= \frac{16a^2}{(1-a^2)^2-4a^2} \\ \frac{(1-a^2)^2-8(1-a^2)-4a^2}{a^4-2a^2+1} &= \frac{16a^2}{a^4-6a^2+1} \\ \frac{a^4+2a^2-7}{a^4-2a^2+1} &= \frac{16a^2}{a^4-6a^2+1} \\ a^8 - 4a^6 - 18a^4 + 44a^2 - 7 &= 16a^6 - 32a^4 + 16a^2 \\ a^8 - 20a^6 + 14a^4 + 28a^2 - 7 &= 0 \\ (a^2 + 1)(a^6 - 21a^4 + 35a^2 - 7) &= 0 \\ a^6 - 21a^4 + 35a^2 - 7 &= 0 \\ \end{align}

By Eisenstein's criterion with $p = 7$, this polynomial has no rational roots. We note temporarily $a = 1$, $b = -21$, $c = 35$ and $d = -7$ for solving this cubic equation

$$ x^3 - 21x^2 + 35x - 7 = 0. \tag{#}\label{cubic} $$

\begin{align} \Delta_0 &= b^2-3ac = 336 \\ \Delta_1 &= 2b^3 - 9abc + 27a^d = -12096 \\ \Delta &= -\frac{\Delta_1^2-4\Delta_0^3}{27a^2} = 200704 > 0, \quad \text{so we have 3 distinct real roots.} \\ C &= \sqrt[3]{\frac12 \left(\sqrt{\Delta_1^2-4\Delta_0^3}+\Delta_1\right)} = \sqrt[3]{672\left(-9+i\sqrt{3}\right)} \\ x &= -\frac{1}{3a}\left(b+\omega^n C+\frac{\Delta_0}{\omega^n C}\right), \end{align}

where $\omega = -\dfrac12+\dfrac{\sqrt3}{2}$ and $n \in \{0,1,2\}$.

Denote the roots of the cubic equation \eqref{cubic} by $\alpha$, $\beta$ and $\gamma$.

Possible values of $a$ is $\sqrt\alpha$,$\sqrt\beta$, and $\sqrt\gamma$. By observing the symmetry under cyclic permutation of $a$, $b$ and $c$, we see that the permutations of the three roots of \eqref{cubic} are possible solutions for $a^2$, $b^2$ and $c^2$.

In \eqref{S}, we observe that we can change the sign of each variable $a$, $b$ or $c$.

To conclude, apart from $(a,b,c) = (0,0,0)$, we have six solutions to \eqref{S}, by permuting the roots and signs of $\{\pm\sqrt\alpha,\pm\sqrt\beta,\pm\sqrt\gamma\}$.

Answer 5

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.