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Old 2014-07-15, 01:23   #4
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"Forget I exist"
Jul 2009

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Originally Posted by Zeta-Flux View Post
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
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