Example 2. Write three integers on the blackboard, then erase one of them and replace it with the difference of 1 and the sum of the other two numbers. After repeating this process several times, the result is
(17, 1967, 1983).

Could the three numbers initially written on the blackboard have been
(1.) $(2,2,2)_{3}$
(2) (3, 3, 3).
(17th All-Union Mathematical Competition, 1983)
Given the problem, if in a certain operation the set of numbers is
$$
(a, b, c), 0<a<b<c \text{, }
$$

then, $c=a+b-1$.
After removing $c$, we have
$$
b=a+d-1 \text{. }
$$
Substituting (2) into (1) gives
$$
c=d+2a-2=d+2(a-1) \text{. }
$$

After some operations, it must be that
$$
\begin{array}{l} 
c=p+k(a-1) . \quad(p<a) \\
\text{Since } 1983=31+122 \times 16 \\
=31+122 \times 17-122, \\
1967=15+122 \times 16 \\
=15+122 \times 17-122,
\end{array}
$$

Thus, after 123 transformations, 1983 $\rightarrow 31$, $1967 \rightarrow 15$.
After 123 transformations,
hence $(17,1967,1983) \rightarrow (17,15,31)$. Since 15 is the smallest number, the subsequent operations are
$$
\begin{array}{r}
(17,15,31) \leftarrow(3,15,17) \leftarrow(3,13,15) \\
\leftarrow(3,11,13) \leftarrow(3,9,11) \leftarrow(3,7,9) \leftarrow(3,7,
\end{array}
$$
$$
5) \leftarrow(3,3,5) \leftarrow(3,3,3) \text{. }
$$

Therefore, the numbers initially written on the blackboard are $(3,3,3)$, not $(2,2,2)$.