4. 231 Find all real numbers $a$ such that there exist non-negative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfying
$$\sum_{k=1}^{5} k x_{k}=a, \sum_{k=1}^{5} k^{3} \cdot x_{k}=a^{2}, \sum_{k=1}^{5} k^{5} x_{k}=a^{3} .$$
[Solution] Let there be non-negative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ that satisfy the conditions. Then, by the Cauchy-Schwarz inequality,
$$\begin{array}{l}
\quad a^{4}=\left(\sum_{k=1}^{5} k^{3} x_{k}\right)^{2}=\left[\sum_{k=1}^{5}\left(k^{\frac{1}{2}} x_{k}^{\frac{1}{2}}\right)\left(k^{\frac{5}{2}} x_{k}^{\frac{1}{2}}\right)\right]^{2} \\
\leqslant\left(\sum_{k=1}^{5} k x_{k}\right)\left(\sum_{k=1}^{5} k^{5} x_{k}\right)=a \cdot a^{3}=a^{4}, \\
\text { thus } \quad \frac{k^{\frac{5}{2}} \cdot x_{k}^{\frac{1}{2}}}{k^{\frac{1}{2}} \cdot x_{k}^{\frac{1}{2}}}=\frac{k^{2} \cdot x_{k}^{\frac{1}{2}}}{x_{k}^{\frac{1}{2}}}=\text { constant }
\end{array}$$

If $x_{1}=x_{2}=\cdots=x_{5}=0$, then $a=0$. If $x_{1}, x_{2}, \cdots, x_{5}$ have two or more non-zero values, then $\circledast$ cannot hold. Therefore, $x_{1}, x_{2}, \cdots, x_{5}$ can have only one non-zero value. Suppose $x_{j} \neq 0$, then,
$$j x_{j}=a, j^{3} x_{j}=a^{2}, j^{5} x_{j}=a^{3},$$

From the first two equations, we get $a=j^{2}$, and $x_{j}=j$, with the rest being 0. Since $j$ can be $1,2,3,4,5$, the only values for $a$ that satisfy the conditions are $a=0,1,4,9,16,25$.