2003 AIME II Problems/Problem 10
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Problem
Two positive integers differ by 
The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?
Solution
Call the two integers 
and 
, so we have 
. Square both sides to get 
. Thus, 
must be a square, so we have 
, and 
. The sum of these two factors is 
, so they must both be even. To maximize , we want to maximixe 
, so we let it equal 450 and the other factor 2, but solving gives 
, which is already a perfect square, so we have to keep going. In order to keep both factors even, we let the larger one equal 150 and the other 6, which gives 
. This checks, so the solution is 48+108=156.
See also
| 2003 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
