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


But try the problem first…

Answer: is 38.

Source
Suggested Reading

AIME I, 1999, Question 3

Elementary Number Theory by David Burton

Try with Hints


First hint

\((n-10)^{2}\) \(\lt\) \(n^{2}-19n+99\) \(\lt\) \((n-8)^{2}\) and \(n^{2}-19n+99\) is perfect square then \(n^{2}-19n+99\)=\((n-9)^{2}\) that is n=18

Second Hint

and \(n^{2}-19n+99\) also perfect square for n=1,9,10

Final Step

then adding 1+9+10+18=38.

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