2003 AIME II Problems/Problem 10

Problem

Two positive integers differ by $60$ . 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 $b$ and $b+60$ , so we have $\sqrt{b}+\sqrt{b+60}=\sqrt{c}$ . Square both sides to get $2b+60+2\sqrt{b^2+60b}=c$ . Thus, $b^2+60b$ must be a square, so we have $b^2+60b=n^2$ , and $(b+n+30)(b-n+30)=900$ . The sum of these two factors is $2b+60$ , so they must both be even. To maximize $b$ , we want to maximixe $b+n+30$ , so we let it equal $450$ and the other factor $2$ , but solving gives $b=196$ , 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 $b=48$ . This checks, so the solution is $48+108=\boxed{156}$ .

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
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2003 AIME II Problems/Problem 10

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