$8 \cdot 21$ Find the integer $k$ such that $36+k, 300+k, 596+k$ are the squares of three consecutive terms in an arithmetic sequence.
[Solution] From the conditions satisfied by $k$, there exist real numbers $n$ and $d$ such that
$$
\left\{\begin{array}{l}
(n-d)^{2}=36+k, \\
n^{2}=300+k, \\
(n+d)^{2}=596+k .
\end{array}\right.
$$
(3) - (1) gives $n d=140$.
(2) - (1) - (3) gives $d^{2}=16$.
From this, we know $n^{2}=35^{2}=1225$, so
$$
k=n^{2}-300=925 .
$$