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

Home > Just Math
Irrational to Rational (Posted on 2012-06-09) Difficulty: 3 of 5
Determine, with proof, all possible irrational numbers n such that each of n3 - 6n and n4 - 8n2 is a rational number.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Algebraic Thoughts Comment 2 of 2 |
Let n^4-8n^2 = X and n^3-6n = Y, with X and Y being rational values as stated.

n must be an algebraic number to satisfy a polynomial with rational coefficients.  Furthermore the degree of n must be at most 3 for the cubic.  Then the quartic must be factorable into a linear times the cubic:
n^4 - 8n^2 - X = (n^3 - 6n - Y) * (n+R)

The right side multiplies to become:
n^4 + Rn^3 - 6n^2 - (6R + Y)n - R*Y.

Then the quartic equality simplifies to:
0 = n^3 + (2/R)n^2 - (6 + Y/R)n + (X/R - Y)

n must be a root of this equation and the original cubic n^3-6n-Y = 0.  The simplified difference of these equations is 2n^2 - Yn + X = 0.

Therefore n must be an algebraic number of at most degree 2.

Jer's values of n = +/-1 + sqrt(3), X=-4, Y=4 satisfy this quadratic equation.

In retrospect, it looks like all I did was apply the Euclidean GCD algorithm to n^4-8n^2-X and n^3-6n-Y to get 2n^2-Yn+X, but the conclusion still stands.

Edited on July 4, 2016, 10:07 pm
  Posted by Brian Smith on 2016-07-04 22:06:23

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 (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information