Difference between revisions of "User:Herefishyfishy1"

(New page: <math>i^n=cis(\frac{\pi n}{2})</math> <math>\forall a,b,c,n\in \mathbb{N}, n>2\Longrightarrow a^n+b^n\not=c^n</math>)
(No difference)

Revision as of 10:47, 13 July 2008

$i^n=cis(\frac{\pi n}{2})$

$\forall a,b,c,n\in \mathbb{N}, n>2\Longrightarrow a^n+b^n\not=c^n$