2. Find all three-digit numbers $\overline{abc}$ with distinct digits for which $a^{2}-b^{2}-c^{2}=a-b-c$.
Solution. If $b=0$, then $a^{2}-c^{2}=a-c$, i.e., $(a-c)(a+c)=a-c$. Since from the condition of the problem $a \neq c$, we get $a+c=1$, and from there $a=1$ and $c=0$, meaning the digits $b$ and $c$ are equal, which is a contradiction. Therefore, $b \neq 0$ and similarly it is proven that $c \neq 0$. Similarly, it is proven that $b \neq 1$ and $c \neq 1$. If $a=1$, then from the condition it follows that
$b^{2}+c^{2}=b+c$, i.e., $b(b-1)+c(c-1)=0$, and on the other hand, since $b, c \neq 0$ and $b, c \neq 1$ we have $b(b-1)+c(c-1)>0$. Therefore, $a \neq 1$. We will further consider the equivalent equation

$$
a(a-1)=b(b-1)+c(c-1)
$$

The product on the left side of this equation belongs to the set

$$
\{2,6,12,20,30,42,56,72\}
$$

and the sum of the products on the right side in the set

$$
\begin{aligned}
& \{8,14,22,32,44,58,74,18,26,36,48,62,78,42 \\
& 54,68,84,50,76,92,72,86,102,98,114,128\}
\end{aligned}
$$

Since 42 and 72 are in the intersection of the two sets, we have:
i) $42=12+30$, i.e., $7 \cdot 6=4 \cdot 3+6 \cdot 5$, so we get the numbers 746 and 764;

ii) $72=30+42$, i.e., $9 \cdot 8=6 \cdot 5+7 \cdot 6$, so we get the numbers 967 and 976. Therefore, the required numbers are 746, 764, 967, and 976.