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 $48+108=\boxed{156}$.

See also

2003 AIME II (ProblemsAnswer KeyResources)
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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