Functional equation involving $f(x^4)+f(x^2)+f(x)$ Find all increasing functions $f$ from positive reals to positive reals satisfying $f(x^4) + f(x^2) + f(x) = x^4 + x^2 + x$.


It's easy to show that $f(1)=1$, and I was also able to show that
$$f(x)-x = f(x^{(8^k)}) - x^{(8^k)}$$
for all integers $k$.

But where to go from here? Any hints would be much appreciated.
Let $g(x) = f(x)-x$. We obtain $g(x^4)+g(x^2)+g(x)=0$. Define $a_n = g(x^{2^n})$ for every $n\in\Bbb Z$ where $x\ne 1,x&gt;0$. It follows that
$$
a_{n+2}+a_{n+1}+a_n=0,
$$ and from this 2nd order linear equation we get for some $A,B$,$$
a_n =\begin{cases}A,\quad n\equiv 0 \;\text{(mod }3)\\ B,\quad n\equiv 1 \;\text{(mod }3)\\-A-B,\quad n\equiv 2 \;\text{(mod }3)\end{cases}.
$$ 

It says that $(a_n)$ is a $3$-periodic sequence. Now we claim that $a_n = 0$ for every $n$. Since $f(x)=x+g(x)$ is increasing, it holds for each $k=0,1,2$ and all $n\equiv k\text{ (mod} \;3)$, $x&gt;1$ that 
$$
x^{2^n}+a_k \le x^{2^{n+1}}+a_{k+1}\le x^{2^{n+2}}+a_{k+2}.
$$ Let $n\equiv k\text{ (mod} \;3)$ tend to $-\infty$ to obtain
$$
a_k\le a_{k+1}\le a_{k+2}
$$ for each $k=0,1,2$. This implies $a_0=a_1=a_2=0$ and thus $a_n = 0,\ \forall n$ follows for $x&gt;1$. The case where $0&lt;x&lt;1$ can be treated similarly by noting that
$$
x^{2^n}+a_k \ge x^{2^{n+1}}+a_{k+1}\ge x^{2^{n+2}}+a_{k+2}.
$$
This proves $f(x)=x+g(x)=x$ for all $x&gt;0$.
Assuming $f(x)$ as a real function, calling $g(x) = f(x)-x$ we have

$$
g(x^4)+g(x^2)+g(x) = 0
$$

now calling $e^y = x$ we have

$$
g(e^{4y})+g(e^{2y})+g(e^y) = 0
$$

or

$$
G(4y)+G(2y)+G(y) = 0
$$

This recurrence equation has as solution

$$
G(y) = C_1 y^{-\frac{2 i \pi }{3 \ln 2}}+C_2 y^{\frac{2 i \pi }{3 \ln 2}}
$$

because

$$
\mathcal{G}(\log_2 (4y))+\mathcal{G}(\log_2(2y))+\mathcal{G}(\log_2 y)=0\\
\mathcal{G}(\log_2 y+2)+\mathcal{G}(\log_2 y +1)+\mathcal{G}(\log_2 y)=0
$$

and making $z = \log_2 y$ we get

$$
\mathcal{G}(z+2)+\mathcal{G}(z+1)+\mathcal{G}(z)=0
$$

which can be characterized as second order linear difference equation.

Then follows

$$
G(y) = C'_1 \cos \left(\frac{2 \pi}{3}  \log_2 y\right)+C'_2\sin \left(\frac{2 \pi}{3}  \log_2 y\right) = g(e^y) = f(e^y)-e^y
$$

and finally

$$
f(x) = C'_1\cos\left(\frac{2\pi}{3}\log_2(\ln x)\right)+C'_2\sin\left(\frac{2\pi}{3}\log_2(\ln x)\right)+x
$$

In case of $f(x)$ increasing then the solution is $f(x) = x$ with $C'_1=C'_2=0$

NOTE

Use this MATHEMATICA script to verify the recurrence equation as well as the final solution.

Clear[G]
G[y_] := Cos[(2 \[Pi] Log[2, y])/3] c1 + Sin[(2 \[Pi] Log[2, y])/3 ] c2
G[4 y] + G[2 y] + G[y] // FullSimplify

f[x_] := Cos[(2 \[Pi] Log[2, Log[x]])/3] c1 + Sin[(2 \[Pi] Log[2, Log[x]])/3 ] c2 + x
Assuming[x &gt; 0, f[x^4] + f[x^2] + f[x] - x^4 - x^2 - x // FullSimplify]