Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Perfect square and Integers.
Find the sum of all positive integers n for which \(n^{2}-19n+99\) is a perfect square.
Perfect Square
Integers
Inequalities
Answer: is 38.
AIME I, 1999, Question 3
Elementary Number Theory by David Burton
\((n-10)^{2}\) \(\lt\) \(n^{2}-19n+99\) \(\lt\) \((n-8)^{2}\) and \(n^{2}-19n+99\) is perfect square then \(n^{2}-19n+99\)=\((n-9)^{2}\) that is n=18
and \(n^{2}-19n+99\) also perfect square for n=1,9,10
then adding 1+9+10+18=38.
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Very simple problem.