15. We note that $\sum_{i=1}^{5} i\left(a-i^{2}\right)^{2} x_{i}=a^{2} \sum_{i=1}^{5} i x_{i}-2 a \sum_{i=1}^{5} i^{3} x_{i}+\sum_{i=1}^{5} i^{5} x_{i}=a^{2} \cdot a-2 a \cdot a^{2}+a^{3}=0$. Since the terms in the sum on the left are all nonnegative, it follows that all the terms have to be 0. Thus, either $x_{i}=0$ for all $i$, in which case $a=0$, or $a=j^{2}$ for some $j$ and $x_{i}=0$ for $i \neq j$. In this case, $x_{j}=a / j=j$. Hence, the only possible values of $a$ are $\{0,1,4,9,16,25\}$.