***Oh well, I got gibberish too but I ain't fixing it because it isn't anything new***
Each of the numbers in the cube roots must be the cubes of relatively simple numbers with ¡î2 in them. If one has a positive ¡î2 term and the other its opposite they will cancel out on adding.
Amazing to me was that I found them on my first try:
(1+¡î2)©ø = 7+5¡î2
(1-¡î2)©ø = 7-5¡î2
(1+¡î2) + (1-¡î2) = 2
a=2 which is an integer
-Jer
Edited on October 13, 2004, 4:07 pm
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Posted by Jer
on 2004-10-13 16:03:57 |
I way I see it
