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