Example 2. Given an integer $n$, if $n$ plus 38 is a perfect square, prove that $n$ plus 38 is a perfect square, find $n$.

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Example 2. Given an integer $n$, if $n$ plus 38 is a perfect square, prove that $n$ plus 38 is a perfect square, and find $n$.
Let $n-51=x^{2},(x, y \in Z)$
$$
n+38=y^{2} .
$$

Subtracting the equations yields
$$
\begin{array}{l}
(y-x)(y+x)=89=1 \cdot 89 \text { ? } \\
=(-1)(-89) .
\end{array}
$$

Thus, $\left\{\begin{array}{l}y+x=1,89,-1,-89, \\ y-x=89,1,-89,-1 .\end{array}\right.$
Therefore, $\left\{\begin{array}{l}x=-44,41,41, \quad 41, \\ y=45,45,-45,-45,\end{array}\right.$ so $n=1987$.