A27 (21-5, Israel) 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 equations
$$
\sum_{k=1}^{5} k x_{k}=a, \sum_{k=1}^{5} k^{3} x_{k}=a^{2}, \sum_{k=1}^{5} k^{5} x_{k}=a^{3} .
$$
Solution 1 The required relation is
$$
a^{2} \sum_{k=1}^{5} k x_{k}+\sum_{k=1}^{5} k^{5} x_{k}=2 a \sum_{k=1}^{5} k^{3} x_{k} \text {. }
$$

Or
$$
\sum_{k=1}^{5} k\left(a-k^{2}\right)^{2} x_{k}=0 .
$$

Thus, these five non-negative terms are all zero, i.e.,
$$
k\left(a-k^{2}\right)^{2} x_{k}=0, k=1,2,3,4,5 .
$$

If $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0$, then $a=0$; otherwise, for some $j, x_{j} \neq 0$, then $a=j^{2}$.
Thus, the 6 possible values of $a$ are $0,1,4,9,16,25$.
Solution 2 By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
\left(a^{2}\right)^{2} & =\left(\sum_{k=1}^{5} k^{3} x_{k}\right)^{2} \\
& \leqslant\left(\sum_{k=1}^{5} k x_{k}\right)\left(\sum_{k=1}^{5} k^{5} x_{k}\right) \\
& =a^{4} .
\end{aligned}
$$

The inequality is actually an equality. By the condition for the Cauchy-Schwarz inequality to become an equality, there exists a real number $\lambda$ such that

or
$$
\begin{aligned}
\sqrt{k^{5} x_{k}} & =\lambda \sqrt{k x_{k}}, \\
k^{4} x_{k} & =\lambda^{2} x_{k},
\end{aligned}
$$

i.e.,
$$
\left(k^{4}-\lambda^{2}\right) x_{k}=0, \quad k=1,2,3,4,5 .
$$

If for some $j, x_{j} \neq 0$, then $\lambda=j^{2}$. The other $x_{i}$ are all zero, substituting into the original equation, we get
$$
\begin{aligned}
j x_{j} & =a, \\
j^{3} x_{j} & =a^{2} .
\end{aligned}
$$

Thus, $a=j^{2}=\lambda$, and it also satisfies $j^{5} x_{j}=a^{3}$.
Another case is $x_{k}=0, k=1,2,3,4,5$. In this case, $a=0$.
Thus, the possible values of $a$ are $0,1,4,9,16,25$.