1999 AIME Problems/Problem 3
Problem
Find the sum of all positive integers for which is a perfect square.
Solution 1
If for some positive integer , then rearranging we get . Now from the quadratic formula,
Because is an integer, this means for some nonnegative integer . Rearranging gives . Thus or , giving or . This gives or , and the sum is .
Solution 2
Suppose there is some such that . Completing the square, we have that , that is, . Multiplying both sides by 4 and rearranging, we see that . Thus, . We then proceed as we did in the previous solution.
Solution 3
When , we have
So if and is a perfect square, then
or .
For , it is easy to check that is a perfect square when and ( using the identity
We conclude that the answer is
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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