Fri, 02 May 2008 23:33:59 -0600 from Virgil <Virgil@[EMAIL PROTECTED]
>:
> In article <030520080037307285%plsperry@[EMAIL PROTECTED]
>,
> Paul Sperry <plsperry@[EMAIL PROTECTED]
> wrote:
>
> > --
> > Paul Sperry
> > Columbia, SC (USA)
>
> 4 == -1 (mod 5) so 4^n == (-1)^n (mod 5).
I never really studied exponents in modular arithmetic, so thinking
out loud I'll try to apply this to 3^n mod 5:
(-2)^n mod 5 could be anything but 0.
Hmm: 3, 9, 27, 81, 243, 729, ...
Seems to work.
I'm guessing that this is a shorthand for binomial expansion. I could
write out
3^n = (5-2)^n
= 5^2 + n*(-2)*5^(n-1) + ... + n*(-2)^(n-1)*5 + (-2)^n
Taking that modulo 5, every term drops out except the last. So
3^n mod 5 = (-2)^n mod 5.
What can the values of (-2)^n mod 5 be? -2 becomes 3, 4, -8 becomes
2, 16 becomes 1, -32 becomes 3, 64 becomes 4, -128 becomes 2, 256
becomes 1, -512 becomes 3, anything but 0.
It seems that 3^n mod 5 or (-2)^n mod 5 cycles 3,4,2,1 endlessly. I
guess an inductive proof would be necessary to show it rigorously.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
****kata ga nai...
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