Example 7. Find all real numbers $a$ such that there exist non-negative real numbers $x_{\mathrm{k}}, k=1,2,3,4,5$, satisfying the relations
$$\begin{array}{c}
\sum_{k=1}^{5} k x_{\mathrm{k}}=a, \quad \sum_{k=1}^{5} k^{3} x_{k}=a^{2} \\
\sum_{k=1}^{5} k^{5} x_{k}=a^{3}
\end{array}$$
If there is such a real number $a$ that makes the above three equations hold for the corresponding $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, then we have

On the other hand, by the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(\sum_{k=1}^{5} k^{3} x_{\mathrm{k}}\right)^{2}=\left(\sum_{\mathrm{k}=1}^{5} k^{\frac{1}{2}} \sqrt{x_{\mathrm{k}}} \cdot \frac{5}{k^{2}} \sqrt{x_{\mathrm{k}}}\right)^{2} \\
\leqslant\left(\sum_{\mathrm{k}=1}^{5} k x_{\mathrm{k}}\right)\left(\sum_{\mathrm{k}=1}^{5} k^{5} x_{\mathrm{k}}\right)
\end{array}$$
Equation (1) precisely indicates that the inequality (2) must hold with equality, which is true if and only if
$$k^{2} \sqrt{x_{\mathrm{k}}} / \sqrt{x_{\mathrm{xk}}}=\lambda, k=1,2,3,4,5 .$$

If at least two of $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are non-zero, then equation (3) cannot hold. Therefore, at most one of $x_{1}, x_{2}, x_{3}, x_{4}^{2}, x_{5}$ can be non-zero.

If $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0$, then $a=0$; if $x_{\mathrm{r}}=c \neq 0, x_{1}=0, j \neq \gamma$, then we have
$\gamma c=a, \gamma^{3} c=a^{2}, \gamma^{5} c=a^{3}$. A deeper understanding reveals: $a=\gamma^{2}, \gamma=c_{0}$. But $1 \leqslant \gamma \leqslant 5$, hence we get $a=1,4,9,16, 25$. Therefore, all $a$ that satisfy the requirements of the problem are $0,1,4,9,16$ and 25. The condition for equality is satisfied.