Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Perfect square and Integers.

## Perfect square Problem – AIME I, 1999

Find the sum of all positive integers n for which \(n^{2}-19n+99\) is a perfect square.

- is 107
- is 38
- is 840
- cannot be determined from the given information

**Key Concepts**

Perfect Square

Integers

Inequalities

## Check the Answer

Answer: is 38.

AIME I, 1999, Question 3

Elementary Number Theory by David Burton

## Try with Hints

\((n-10)^{2}\) \(\lt\) \(n^{2}-19n+99\) \(\lt\) \((n-8)^{2}\) 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.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

## One reply on “Perfect square Problem | AIME I, 1999 | Question 3”

Very simple problem.