2011 AIME II Problem 15

For $x \in [n, n+1)$ the right side is $\sqrt{P(n)},$ which must be an integer: find the $n$ with $P(n)$ a perfect square

Only $n = 5,$ $6,$ and $13$ work; on each such interval solve $P(x) \lt \left(\sqrt{P(n)} + 1\right)^2$ for the valid subinterval

For $x \in [n, n + 1)$ the right-hand side is $\sqrt{P(n)},$ which must be an integer, so $P(n) = n^2 - 3n - 9$ must be a perfect square. For $n = 5,$ $6,$ $\ldots,$ $14$ the values are $1,$ $9,$ $19,$ $31,$ $45,$ $61,$ $79,$ $99,$ $121,$ $145:$ only $n = 5,$ $6,$ and $13$ give squares, with $\sqrt{P(n)} = 1,$ $3,$ and $11,$ respectively.

$P$ is increasing on $[5, 15],$ so for $x \in [n, n + 1)$ we automatically have $\sqrt{P(x)} \ge \sqrt{P(n)},$ and $\left\lfloor \sqrt{P(x)} \right\rfloor = \sqrt{P(n)} = m$ holds exactly when $P(x) \lt (m + 1)^2,$ i.e. $x \lt \frac{3 + \sqrt{45 + 4(m+1)^2}}{2}.$ For $m = 1,$ $3,$ and $11$ the cutoffs are $\frac{3 + \sqrt{61}}{2},$ $\frac{3 + \sqrt{109}}{2},$ $\frac{3 + \sqrt{621}}{2},$ each lying inside the corresponding unit interval, so the successful subintervals have lengths $\frac{\sqrt{61} - 7}{2},$ $\frac{\sqrt{109} - 9}{2},$ $\frac{\sqrt{621} - 23}{2}.$

The interval $[5, 15]$ has length $10,$ so the probability is $$\begin{aligned} &\frac{1}{10} \\ &\quad {}\cdot \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{2} \\ &= \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20}, \end{aligned}$$ giving $a + b + c + d + e =$ $61 + 109 + 621 + 39 + 20 =$ $850.$