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 20120611 12:31:26 