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.
 A: 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 $x\ne 1,x>0$. 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 $n\equiv k\text{ (mod} \;3)$, $x>1$ 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 $x>1$. The case where $0<x<1$ 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 $x>0$.
A: First we show that $\lim_{x \to 1} f(x) = f(1) = 1$. Note that the limit of an increasing function from below or from above always exists. Let the limit from below be $a$ and the limit from above be $b$. If we take the limit from below of $f(x) + f(x^2) + f(x^4) = x + x^2 + x^4$, we get $a + a + a = 3$, so $a = 1$ (since as $x$ approaches $1$, $x^2$ and $x^4$ also approach $1$). Similarly from above, $b + b + b = 3$, so $b = 1$. Finally, we already know $f(1) = 1$. So
$$
\lim_{x \to 1} f(1) = 1.
$$
(In fact, we could show $f$ is continuous for all $x$, using a similar argument.)
To finish, consider your equation $f(x) + f(x^{8^k}) = x + x^{8^{k}}$. Take the limit as $k \to -\infty$. Regardless of $x$, $x^{8^k}$ approaches $x^{0} = 1$. So, we get
\begin{align*}
f(x) + 1 = x + 1 \\
\implies & \boxed{f(x) = x},
\end{align*}
which holds for all $x$. So $f(x) = x$ is the only solution.
A: If we are interested in functions smooth enough to admit some power series expansion (around $x=0$):
$$f(x) = \sum_{k=0}^\infty c_kx^k$$
Since $$f(x^n) = \sum_{k=0}^\infty c_kx^{kn}$$
It will force relations between the coefficients in a systematic way which you can then investigate.
A: 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 > 0, f[x^4] + f[x^2] + f[x] - x^4 - x^2 - x // FullSimplify]

