perplexus.info :: Just Math : Rational Floor Query

Rational Floor Query (Posted on 2015-06-03)

(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.

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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

^or

x²=n/(x-n)

Informally,

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

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