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jnml · 2015-01-25, 18:31

$\forall \ n \in \mathbb{N}, \ z \in \mathbb{Z}:\ 2^n-z | 2z^2-1 \ \leftrightarrow \ 2^n-z | 2^{2n+1}-1.$

Batalov · 2015-01-25, 19:06

Quote:

Originally Posted by jnml

$\forall \ n \in \mathbb{N}, \ z \in \mathbb{Z}:\ 2^n-z | 2z^2-1 \ \leftrightarrow \ 2^n-z | 2^{2n+1}-1.$

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)
$2*(2^{2n} - z^2) = 2*(2^n - z)*(2^n + z) = 0\ (mod x)$

2015-01-25, 18:31	#1
jnml Feb 2012 Prague, Czech Republ 10110010₂ Posts	. $\forall \ n \in \mathbb{N}, \ z \in \mathbb{Z}:\ 2^n-z \| 2z^2-1 \ \leftrightarrow \ 2^n-z \| 2^{2n+1}-1.$