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.

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.
