AIME 2003 II · 第 10 题

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