User:Herefishyfishy1

Revision as of 09:47, 13 July 2008 by Herefishyfishy1 (talk | contribs) (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>)
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$i^n=cis(\frac{\pi n}{2})$

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