$12 \cdot 34$ Find all square numbers $s_{1}, s_{2}$ that satisfy $s_{1}-s_{2}=1989$.
(30th International Mathematical Olympiad Shortlist, 1989)
［Solution］Let $s_{1}=a^{2}, s_{2}=b^{2}, a>b>0$, then
$$
(a+b)(a-b)=1989=3^{2} \cdot 13 \cdot 17 .
$$

Thus,
$$
\left\{\begin{array}{l}
a+b=m, \\
a-b=n .
\end{array}\right.
$$

where $m n=1989$, and $m>n$.
From the above system of equations, we get
$$
\left\{\begin{array}{l}
a=\frac{m+n}{2}, \\
b=\frac{m-n}{2} .
\end{array}\right.
$$

Let $(m, n)=(1989,1),(663,3),(221,9)$, $(153,13),(117,17),(51,39)$.
$$
\begin{aligned}
(a, b)= & (995,994), \quad(333,330), \quad(115,106), \\
& (83,70), \quad(67,50), \quad(45,6) \\
\left(s_{1}, s_{2}\right)= & \left(995^{2}, 994^{2}\right), \quad\left(333^{2}, 330^{2}\right), \quad\left(115^{2}, 106^{2}\right), \\
& \left(83^{2}, 70^{2}\right), \quad\left(67^{2}, 50^{2}\right), \quad\left(45^{2}, 6^{2}\right)
\end{aligned}
$$