2003 AIME II Problems/Problem 10
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
and the other factor
, 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
and the other
, which gives
. This checks, so the solution is
.
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 |
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