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
Algebraic Thoughts
