Thread: Some arithmetic... View Single Post
2014-07-15, 01:23   #4
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by Zeta-Flux $2^{(p-1)/2} \equiv \pm 1\pmod{p^2}$.
Sorry for quoting this twice. One thing that just came to me is that this is equivalent of saying $2^n \eq \pm 1\pmod{4n^2+4n+1}$ and this equals $2^n-1 \eq 0 \pmod{4n^2+4n+1}$ or $2^n+1 \eq 0 \pmod{4n^2+4n+1}$ which when you consider that if 2m+1 divides 2k+1, $k \eq m \pmod{2m+1}$ we can bring this down to $2^{n-1}-1 \eq 2n^2+2n \pmod{4n^2+4n+1}$ or $2^{n-1} \eq 2n^2+2n\pmod{4n^2+4n+1}$ I'm I getting better or just making it worse ?

Last fiddled with by science_man_88 on 2014-07-15 at 01:29