Find all the solutions of this equation:$x+3y=4y^3 \ , y+3z=4z^3\ , z+3x=4x^3 $ in reals.

Find all the solutions of this equation

$x+3y=4y^3 \ , y+3z=4z^3\ , z+3x=4x^3 $

My attempt:In the hint of the question it was written that show $x,y,z \in [-1,1]$ then add all equations and conclude $(x,y,z)=(-1,-1,-1),(0,0,0),(1,1,1)$ are the solutions.I can show that $x,y,z \in [-1,1]$like below:

By symmetry consider $x \ge y \ge z$ then if $x>1$ we have:

$4x^3-4x>0 \Rightarrow 4x^3-3x>x \Rightarrow z>x$

which is wrong.Now consider $x<-1$ we have:

$x<-1 \Rightarrow y<-1 \Rightarrow 4y^3-4y<0 \Rightarrow 4y^3-3y<y \Rightarrow x<y$

By the same way we can get $x,y,z \in [-1,1]$.By adding the equations we can get:


But what should I do now?

Solutions Collecting From Web of "Find all the solutions of this equation:$x+3y=4y^3 \ , y+3z=4z^3\ , z+3x=4x^3 $ in reals."

Let $f(u) = 4u^3-3u$, the system of equations at hand can be rewritten as

$$x = f(y),\;y = f(z),\;z = f(x)$$
This implies $x$ (and so do $y$ and $z$) are roots of following polynomial of degree $27$:

f(f(f(u))) – u =
& \;\;67108864\,{u}^{27}-452984832\,{u}^{25}+1358954496\,{u}^{23}\\
& -2387607552\,{u}^{21}+2724986880\,{u}^{19}-2118057984\,{u}^{17}\\
& +1143078912\,{u}^{15}-428654592\,{u}^{13}+109983744\,{u}^{11}\\
& -18670080\,{u}^{9}+1976832\,{u}^{7}-117936\,{u}^{5}+3276\,{u}^{3}-28\,u
This implies there are at most $27$ distinct solutions for $(x,y,z)$.

Let’s consider the case $|x| \le 1$. Pick a $t \in [0,\pi]$ such that $u = \cos(t)$. Notice
$$f(\cos(t)) = 4\cos(t)^3 – 3\cos(t) = \cos(3t)$$
We have
$$f(f(f(u))) – u = 0 \iff \cos(27t) – \cos(t) = 0
\iff \sin(14t)\sin(13t) = 0$$

When $\sin(14t) = 0$, we have $15$ choices of $t = \frac{\pi}{14}k$ for $0 \le k \le 14$.
When $\sin(13t) = 0$, we have $14$ choices of $t = \frac{\pi}{13}\ell$ for $0 \le \ell \le 13$.

Among these two list of solution, the $t = 0$ and $t = \pi$ are duplicated.
These means there are $27$ solutions of $x$ within the interval $[-1,1]$.

$$\pm 1,\quad \cos\left(\frac{\pi}{14}k\right), 1 \le k \le 13\quad\text{ and }\quad
\cos\left(\frac{\pi}{13}\ell\right), 1 \le \ell \le 12 $$

Since there are at most $27$ solutions, these are all possible solutions:

$$(x,y,z) =
(1,1,1) \\
(-1,-1,-1) \\
\cos\left(\frac{3\pi}{14}k\right)\right),& 1 \le k \le 13\\
\cos\left(\frac{3\pi}{13}\ell\right)\right),& 1 \le \ell \le 12\\
When one limit to integer solutions, the $3^{rd}$ case give another integer solution at $k = 7$.
= \left(\cos\frac{\pi}{2},
\cos\frac{3\pi}{2}\right) = (0,0,0)$$
This means there are totally $3$ integer solutions for the system of equations.

$$(x,y,z) = (1,1,1), (0,0,0)\;\text{ and }\;(-1,-1,-1)$$

Preliminary lemma: knowing the classical relationship

$\tag{0}\cos(3t)=4\cos^3(t)-3\cos(t)$ (recalled by @Achille Hui)

and knowing the property/definition of Chebyshev polynomials:

$T_n(cos(t))=cos(nt)$, we have :


Proof : $cos(27t)=cos(3(3(3t))).$

The system of 3 equations can be written under the form

$$x=f(y), y=f(z), z=f(x)$$

with $f(t):=t^3-3t$.

Thus, using the lemma: $f(f(f(t))=T_{27}(t)$.

It is clear that $x,y,z$ are solutions of fixed point relationships :


knowing that the curve of any $T_n$ maps $[0,1]$ onto [0,1], and in this case, that $T_{27}(1)=1.$

Thus, as shown on the graphical representation below, the abscissas of intersection points with the straight line with equation $y=x$ are confined in $[-1,1]$.

enter image description here

Remark: there are 27 intersection points. Thus we await 27 solutions, knowing that, once we have chosen $x$, there is no more choice for $y=f(x)$ and for $z=f(f(x))$ among these roots.

Let us find an explicit expression for these systems $(x,y,z)$ of roots, under a trigonometric form.

These roots, being in $[-1,1]$, can be represented resp. as

$$x=\cos(a), y=\cos(b), z=\cos(c)$$

for certain values of $a,b,c$. Using (0), we can rewrite the initial system into the equivalent form:

$$\cos(a)=\cos(3b), \ \ \cos(b)=\cos(3c), \ \ \cos(c)=\cos(3a)$$

which is equivalent mod $2 \pi$ to:

$$\tag{1} a=s_1 \ 3b, \ \ b=s_2 \ 3c, \ \ c=s_3 \ 3a$$

(where $s_k=\pm1$). Taking the product of these 3 congruences, the only possible solutions are

$$abc=27abc \ \text{mod} \ 2\pi \ \ \text{or} \ \ abc=-27abc \ \text{mod} \ 2\pi$$

giving $26 abc=k 2 \pi$ or $28 abc=k 2 \pi$ for a certain integer $k$, i.e., $abc=\dfrac{k 2\pi}{26}$ or $abc=\dfrac{k 2 \pi}{28}.$

Now, due to (1), $abc=s 27 a$ (where $s=\pm1$); whence two categories of cases :

$$\tag{2}\begin{cases}a&=&s\dfrac{27}{26} k 2 \pi=s\dfrac{1}{26} k 2 \pi \\
a&=&s\dfrac{27}{28} k 2 \pi=s \dfrac{1}{28} k 2 \pi\end{cases}$$

(where $s=\pm1$). It means that taking in (2), successively, $k=0,1, \cdots $, we will obtain the different solutions for $a$. Due to (2), these results for $a$ will generate all the solutions under the form:

$$\tag{3}(x,y,z)=(\cos(a),\cos(9a),\cos(3a)) \ \text{with either} \ a=\dfrac{k \pi}{13} \ or \ a=\dfrac{k \pi}{14},$$

with $k \in \mathbb{Z}$. (note that the signs have been dropped because $\cos(-t)=\cos(t)$).

In conclusion, one can verify that relationships (3) do not generate $26+28$ solutions, but in fact $27$, due to a certain number of “coincidences”.

Among the solutions, one finds $(x,y,z) = (0,0,0), (1,1,1)$ and $(-1,-1,-1)$.

Brute force. If we use $y=4z^3-3z$ and $z=4x^3-3x$ in $x=4y^3-3y$, we obtain a huge polynomial of $27$th degree in $x$ which can be factorized (by some not-human algebraic manipulator) as

If we are interested in the integers solutions then we have only $x=0$, $x=1$ and $x=-1$ because the other factors do not have integer roots (check each constant term).

Now $x\in\{-1,0,1\}$ implies that $z=4x^3-3x=x$, $y=4z^3-3z=x$ and therefore the integer solutions of the system are: $(0,0,0)$, $(1,1,1)$ and $(-1,-1,-1)$.

I am assuming that you want integer solutions. First note that if any of $x$, $y$, and $z$ is zero, then all of them are. Suppose now that they are all nonzero. Then, you have relations $y\mid x$, $z\mid y$, and $x\mid z$, whence $|x|=|y|=|z|=:t$. Then, $x+3y=4y^3$ implies that $$4t\geq |x+3y|=\left|4y^3\right|=4t^3\,,$$
making $t=1$ the only possibility. The rest is straightforward, and the only solutions are $(x,y,z)=(\alpha,\alpha,\alpha)$ with $\alpha\in\{-1,0,+1\}$.