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
n� = a� + b + 2a√b
n� = a� + 3ab + (3a� + b)√b
n^4 = a^4 +6a�b + b� + (4a� + 4ab)√b
n� - 6n = a� + 3ab -6a + (3a� + b - 6)√b
n^4 - 8n = a^4 + 6a�b + b� - 8a� - b + (4a� + 4ab - 16a)√b
For these to be rational the terms in parentheses must both equal zero. This makes a system:
(3a� + b - 6) = 0
(4a� + 4ab - 16a) = 0
Solve each for b
b = 6 - 3a�
b = 4 - a�
So 4 - a� = 6 - 3a�
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
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Posted by Jer
on 2012-06-11 12:31:26 |