, 2021

Solution:
We have
$$8 a^{a} b^{b} = 27 a^{b} b^{a} \Longleftrightarrow \frac{a^{a} b^{b}}{a^{b} b^{a}} = \frac{27}{8} \Longleftrightarrow \frac{a^{a-b}}{b^{a-b}} = \frac{27}{8} \Longleftrightarrow \left(\frac{a}{b}\right)^{a-b} = \frac{27}{8}.$$
Since $27 = 3^{3}$ and $8 = 2^{3}$, there are only four possibilities:
- $a / b = 3 / 2$ and $a-b = 3$, which yields $a = 9$ and $b = 6$;
- $a / b = 27 / 8$ and $a-b = 1$, which yields no solutions;
- $a / b = 2 / 3$ and $a-b = -3$, which yields $a = 6$ and $b = 9$;
- $a / b = 8 / 27$ and $a-b = -1$, which yields no solutions.
Therefore $a^{2} + b^{2}$ must equal $6^{2} + 9^{2} = 117$.