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Old 2015-01-25, 19:06   #2
Batalov
 
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Mar 2008
Phi(4,2^7658614+1)/2

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Talking Elementary, Watson... (yawn)

Quote:
Originally Posted by jnml View Post
\forall \ n \in \mathbb{N}, \ z \in \mathbb{Z}:\ 2^n-z | 2z^2-1 \ \leftrightarrow \ 2^n-z | 2^{2n+1}-1.
Let x = 2^n-z.
Is 2^{2n+1}-1 = 0\ (mod x) \ \leftrightarrow \ 2z^2-1 = 0\ (mod x)?
Bloody trivially true, because
2^{2n+1}-1 = 2z^2-1\ (mod x), because (after rearrangement)
2*(2^{2n} - z^2) = 2*(2^n - z)*(2^n + z) = 0\ (mod x)
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