All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Rational Floor Query (Posted on 2015-06-03) Difficulty: 3 of 5
(A) Prove that the equation:
floor(x)*(x2 + 1) = x3 has precisely one real solution in each interval between consecutive positive integers.

(B) Can any of the solutions inclusive of (A) be rational?
If so, provide an example.
If not, prove it.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Part (A) solution, Part (B) answer w/o solution. Comment 1 of 1
Let n<x<n+1 so that floor(x)=n
So we have
n(x2 + 1) = x3

f(x)=x^2 is a parabola increasing from f(0)=0 to f(infinity)=infinity.
g(x)=n/(x-n) is a hyperbola decreasing from f(n)=(+infinity)  (well, technically undefined) to f(infinity)=0.
f(x) is never negative.
If x<n, g(x) is negative.
So they have to cross precisely once.

f(n+1)=(n+1)^2 > g(n+1)=n for all real numbers
So they have to cross between n and n+1 

For part (B) you'd could try to solve x^3 - nx^2 - n =0
Wolfram alpha give the integer solution x=0, n=0.  But that is not consecutive positive integers.
The equation it gives for x (the cubic equation) does not look promising.
If you add "over the rationals" it just says False.

  Posted by Jer on 2015-06-04 10:15:42
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information