Problem:
Let P(x)=x2−3x−9. A real number x is chosen at random from the interval 5≤x≤15. The probability that ⌊P(x)​⌋=P(⌊x⌋)​ is equal to ea​+b​+c​−d​, where a,b,c,d, and e are positive integers, and none of a,b, or c is divisible by the square of a prime. Find a+b+c+d+e.
Solution:
Suppose x satisfies the given equation and n≤x<n+1 for some integer 0≤n≤14. Because ⌊P(x)​⌋=P(⌊x⌋)​=m is an integer, it follows that n2−3n−9=m2, that is (2n−3+2m)(2n−3−2m)=4n2−12n+9−4m2=45=32⋅5. Suppose 2n−3+2m=k is a divisor of 45. Then 2n−3−2m=k45​, and adding the last two equations yields n=41​(6+k+k45​). Because n≥0, the possible factors of 45 are k=1,3,5,9,15, and 45. The integer solutions for n are 5, 6, and 13; the corresponding values of m are 1, 3, and 11. Thus the values x that satisfy the required identity are restricted to one of the intervals [5,6),[6,7), and [13,14). The polynomial P(x) is increasing for x>23​, and thus in this range P(⌊x⌋)​≤P(x)​ and m=P(⌊x⌋)​≤P(x)​. Therefore the given equation holds if and only if P(x)​<m+1, that is, x2−3x−9<m2+2m+1. Solving for x yields
x<21​(3+45+4(m+1)2​)
The pair (m,n)=(1,5) gives x<21​(3+61​), so 5≤x<21​(3+61​) determines the set of solutions in [5,6). Similarly, (m,n)=(3,6) gives x<21​(3+109​), so 6≤x<21​(3+109​) determines the set of solutions in [6,7), and (m,n)=(11,13) gives x<21​(3+621​), so 13≤x<21​(3+621​) determines the set of solutions in [13,14). The requested probability equals