Thread: Some arithmetic... View Single Post
2014-07-28, 22:04   #10
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by R.D. Silverman Would a moderator please move this (and succeeding) gibberish to the crank math sub-forum?
I decided I better show you how I got that:
• the fact that if 2n+1|2k+1 (| meaning divides), $k \eq n \pmod{2n+1}$ can be shown by showing 2n+1 divides when k=n and that 2n+1 divides one in every 2n+1 odd numbers.
• based on 2n+1=p we get: $2^{p-1}\eq 1 \pmod {p^2} == 2^{2n+1-1}\eq 1 \pmod {{2n+1}^2} == 2^{2n+1-1}\eq 1 \pmod{4n^2+4n+1} == 2^{2n}\eq 1 \pmod{4n^2+4n+1} == 2^{2n}-1\eq 0 \pmod{4n^2+4n+1$
• based on these other two we can show that if $2^{2n}-1\eq 0 \pmod{4n^2+4n+1}$ then $2^{2n-1}-1 \eq 2n^2+2n \pmod{4n^2+4n+1}$

therefore the original congruence plus the basics of division can show that what I said is true. edit: this last result can be rewritten as $2^{p-2}-1 \eq {\frac{p-1}{2}(p+1)} \pmod {p^2}$

Last fiddled with by science_man_88 on 2014-07-28 at 22:13