Let's determine those two-digit numbers $\overline{a b}$ in the decimal system for which the greatest common divisor of $\overline{a b}$ and $\overline{b a}$ is $a^{2}-b^{2}$.
Since the greatest common divisor, $a^{2}-b^{2}$, is positive, it follows that $a>b$. Furthermore, $0 \leq a \leq 9, 0 \leq b \leq 9$, but $a$ and $b$ cannot both be 0. According to the condition,

$$
\left(a^{2}-b^{2}\right) \mid (10 a+b), \quad \text { and } \quad\left(a^{2}-b^{2}\right) \mid (10 b+a)
$$

Since the divisor of two numbers is also the divisor of their sum and difference:

$$
\left(a^{2}-b^{2}\right) \mid (11 a+11 b), \quad \text { and } \quad\left(a^{2}-b^{2}\right) \mid (9 a-9 b)
$$

Using the relationship $a^{2}-b^{2}=(a-b)(a+b)$ and the properties of divisibility:
(1) $a-b \mid 11$,
(2) $a+b \mid 9$.

Since $a-b>0$ and $a+b \leq 9$, it follows from (1) that $a-b=1$. From (2), since $a+b>0$, $a+b=1$, $a+b=3$, or $a+b=9$ can be possible. Examining these three cases:

$$
\begin{aligned}
& a+b=1) a=1 \\
& a+b=3(a=2 \\
& a-b=1\} b=0 \\
& a+b=9\} a=5 \\
& a-b=1\} b=1 \\
& a-b=1\} b=4
\end{aligned}
$$

It is easy to verify that the resulting pairs of numbers $(10,1),(21,12),(54,45)$ satisfy the conditions of the problem.