4. Let the function $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$ satisfy:
(1) If $x \leqslant y$, then $f(x) \leqslant f(y)$;
(2) When $x>0$,
$$
\begin{array}{l}
f\left(x^{4}\right)+f\left(x^{2}\right)+f(x)+f(1) \\
=x^{4}+x^{2}+x+1 .
\end{array}
$$

Find the function $f$.
4. Substituting $x=1$ into equation (1) yields $f(1)=1$.

Replacing $x$ with $x^{2}$ and substituting into equation (1) gives
$$
\begin{array}{l}
f\left(x^{8}\right)+f\left(x^{4}\right)+f\left(x^{2}\right)+1 \\
=x^{8}+x^{4}+x^{2}+1 .
\end{array}
$$

Subtracting equation (1) from this yields
$$
f\left(x^{8}\right)-f(x)=x^{8}-x \text {. }
$$

Define $C(x)=f(x)-x$.
Then $C(x)=C\left(x^{8}\right)$.
By induction, for $k \in \mathbf{Z}$, we have $C(x)=C\left(x^{8 k}\right)$.
First, consider $x>0$. Then take a sufficiently small negative integer $k$ such that
$$
\begin{array}{l}
C(x)+x^{8^{k}}>1 \\
\Rightarrow f\left(x^{8^{k}}\right)=C(x)+x^{8^{k}}>1=f(1) .
\end{array}
$$

But $0<x^{8^{k}}<1$, which contradicts the fact that $f(x)$ is increasing for $x>0$.
If $C(x)<0$, then $f(x)<x<f(1)$ for $0<x<1$, which also contradicts the fact that $f(x)$ is increasing for $x>0$.
If $C(x)>0$, then substituting $f(x)=x+C(x)$ into equation (1) gives
$$
\begin{array}{l}
x^{4}+C\left(x^{4}\right)+x^{2}+C\left(x^{2}\right)+x+C(x)+1 \\
=x^{4}+x^{2}+x+1 \\
\Rightarrow C\left(x^{4}\right)+C\left(x^{2}\right)+C(x)=0 \\
\Rightarrow C\left(x^{4}\right)+C\left(x^{2}\right)<0 \text{ (since } C(x)>0 \text{) }.
\end{array}
$$
This is a contradiction.
In summary, $f(x)=x$ for $x>0$ is the unique solution.