Example 15. (IMO21-6)
Find all real numbers $a$ such that there exist non-negative real numbers $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$, $x_{5}$ satisfying the following relations:
$$
\sum_{\mathrm{k}=1}^{\mathrm{s}} k x_{k}=a, \sum_{\mathrm{k}=1}^{5} k^{3} x_{k}=a^{2}
$$

and $\quad \sum_{k=1}^{5} k^{5} x_{k}=a^{3}$.
Proof: Let there be non-negative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, then we have
$$
\begin{aligned}
\mathfrak{a}^{4} & =\left(\sum_{k=1}^{5} k^{3} x_{k}\right)^{2} \\
& =\left[\sum_{k=1}^{5}\left(k^{\frac{1}{2}} \sqrt{x_{k}}\right)\left(k^{5} \sqrt{x_{k}}\right)\right)^{2} \\
& \leqslant\left(\sum_{k=1}^{5} k x_{k}\right)\left(\sum_{k=1}^{5} k^{3} x_{k}\right)=a^{4}
\end{aligned}
$$

Thus, the inequality in the above equation can only be an equality, so
$$
\frac{k^{\frac{5}{2}} \sqrt{x_{k}}}{k^{2} \sqrt{x_{k}}}=k^{2} \frac{\sqrt{x_{k}}}{\sqrt{x_{k}}}=\text { constant. }(k=1,2,3,4,5)
$$
If $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ have two or more non-zero values, the above equation cannot hold. Therefore, there can only be the following two scenarios:
1) $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0$, in this case $a=0$,
2) Some $x_{k}=c \neq 0$, and the rest $x_{i}=0(i \neq k)$, in this case from the given conditions we have
$$
k c=a, k^{3} c=a^{2}, k^{5} c=a^{3}.
$$

From this, we get $k^{2}=a$ and $c=k .(k=1,2,3,4,5)$
In summary, when $a=0,1,4,9,16,25$, there exist non-negative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ that satisfy the given system of equations.
(Author's affiliation: China University of Science and Technology)