Let n<x<n+1 so that floor(x)=n
So we have
n(x^{2} + 1) = x^{3}
^{or}
x^{2}=n/(xn)
Informally,
f(x)=x^2 is a parabola increasing from f(0)=0 to f(infinity)=infinity.
g(x)=n/(xn) 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 20150604 10:15:42 