Determine, with proof, all possible irrational numbers n such that each of n³ - 6n and n⁴ - 8n² is a rational number.

I haven't considered higher order roots but here's what I have so far:

For two rational numbers a, b with √b irrational let
n = a + √b
n² = a² + b + 2a√b
n³ = a³ + 3ab + (3a² + b)√b
n^4 = a^4 +6a²b + b² + (4a³ + 4ab)√b

n³ - 6n = a³ + 3ab -6a  + (3a² + b - 6)√b
n^4 - 8n = a^4 + 6a²b + b² - 8a² - b + (4a³ + 4ab - 16a)√b

For these to be rational the terms in parentheses must both equal zero.  This makes a system:
(3a² + b - 6) = 0
(4a³ + 4ab - 16a) = 0
Solve each for b
b = 6 - 3a²
b = 4 - a²

So 4 - a² = 6 - 3a²
a = ±1
and b = 3

The only values for  n with the type sought are n = ±1 + √3

The rational expressions each come out to 4 or -4

Posted by Jer on 2012-06-11 12:31:26