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

Elementary, Watson... (yawn)
 

[QUOTE=jnml;393516][TEX]\forall \ n \in \mathbb{N}, \ z \in \mathbb{Z}:\ 2^n-z | 2z^2-1 \ \leftrightarrow \ 2^n-z | 2^{2n+1}-1.[/TEX][/QUOTE]
Let x = 2^n-z.
Is [TEX]2^{2n+1}-1 = 0\ (mod x) \ \leftrightarrow \ 2z^2-1 = 0\ (mod x)[/TEX]?
Bloody trivially true, because
[TEX]2^{2n+1}-1 = 2z^2-1\ (mod x)[/TEX], because (after rearrangement)
[TEX]2*(2^{2n} - z^2) = 2*(2^n - z)*(2^n + z) = 0\ (mod x)[/TEX]