2.14 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 numbers $(17,1967, 1983)$ are obtained. Could the three numbers initially written on the blackboard have been:
(1) $(2,2,2)$;
(2) $(3,3,3)$.
(17th All-Soviet Union Mathematical Olympiad, 1983)
［Solution］We perform a backward deduction:
$$
\begin{array}{l}
1983=1967+17-1, \\
1967=1951+17-1,
\end{array}
$$

Thus, $\quad(17,1967,1951) \rightarrow(17,1967,1983)$.
Also,
$$
1951=1935+17-1,
$$

which means
$$
(17,1935,1951) \rightarrow(17,1967,1951) .
$$

From this, we can deduce
$$
\begin{aligned}
1983 & =1967+17-1 \\
& =1951+2 \cdot 17-2 \\
& =1935+3 \cdot 17-3 .
\end{aligned}
$$

Continuing this backward deduction, since
$$
\begin{array}{l}
1983=31+122 \cdot 17-122, \\
1967=15+122 \cdot 17-122 .
\end{array}
$$

After 123 transformations, we have
$$
(17,1967,1983) \leftarrow(17,15,31) .
$$

Continuing this transformation, we get
$$
\begin{array}{l}
(17,15,31) \leftarrow(17,15,3) \leftarrow(13,15,3) \leftarrow(13,11,3) \leftarrow(9,11,3) \\
\leftarrow(9,7,3) \leftarrow(7,5,3) \leftarrow(5,3,3) \leftarrow(3,3,3) .
\end{array}
$$

Therefore, at the beginning, the numbers on the blackboard could have been $(3,3,3)$ but not $(2,2,2)$.