Competition problem — Number theory

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Number theory Difficulty 5.3 AIME, harder Find the answer

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

The text seems to have some typographical errors or unclear phrasing. Here is a corrected version for clarity:

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 $n$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

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 $n=1987$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.