Sat, 03 May 2008 13:54:09 -0700 from The World Wide Wade
<aderamey.addw@[EMAIL PROTECTED]
>:
> In article <MPG.22865099b5c7cbd298b61d@[EMAIL PROTECTED]
>,
> Stan Brown <the_stan_brown@[EMAIL PROTECTED]
> wrote:
> > I never really studied exponents in modular arithmetic, so thinking
> > out loud I'll try to apply this to 3^n mod 5:
>
> The exponent property is based on a result I'm sure you know: If a = b
> (mod m) and c = d (mod m), then ac = bd (mod m). Proving a = b (mod
> m) => a^n = b^n (mod m) for n = 1, 2, ... is then a two-line induction
> argument.
Thanks, Wade. I didn't know that property, but once it's stated I
can see how to prove it easily enough.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
****kata ga nai...
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