Math Forum Discussions

Felicis@gmail.com

Posts: 48
Registered: 7/19/06

Re: help required -- ring, units, mod... etc
Posted: Mar 17, 2007 4:02 PM

Plain Text

Dear Maria-
try taking a look at integers modulo p, where p is prime.

Z_2, Z_3, Z_5, etc. What do you notice?

More specifically- which elements are units? Which are not?

Now- which are square roots of 1?

As to a method of proof- if I look at the nonzero elements of these
sets, along with the operation of multiplication, what do I get?

Let's start with that - what do you find out?

cheers-
Eric

On Mar 16, 1:57 pm, "Maria Bertouli" <a...@spam.net> wrote:
> "Hanford Carr @lmcinvestments.com>" <"hwcarr<nospam> wrote in messagenews:45FABDDE.7E43E9C6@lmcinvestments.com...
>
>
>
> > Maria Bertouli wrote:
>
> > > "Rupert" <rswarbr...@googlemail.com> wrote in message
> > >news:1173980327.840236.35150@l75g2000hse.googlegroups.com...
> > > > On Mar 15, 1:43 pm, "Maria Bertouli" <a...@spam.net> wrote:
> > > > > I would like to find all the positive integers n such that every
> unit u
> > > in
> > > > > the Integers mod n satisfies u^2=1.
>
> > > > > How can I find these?
>
> > > > > Thanks.
>
> > > > Firstly, you probably should go about finding which units in integers
> > > > mod 5, say, have a square of one. Then try a non-prime, say 6? It
> > > > should become clearer after trying some examples, I think.
>
> > > i think i've figured out it should be [u mod n]^2=1 that we are
> satisfying
> > > and not u^2 =1. but still wouldn't mind confirmation on this.
>
> > > so now, i know the integers mod 2, 3, 4, 6, 8, work. (i've only worked
> up to
> > > the integers mod 10)
>
> > > i would be interested to know if it could be proved for how many it
> would
> > > work for? i don't want an answer of course but i would like an idea of
> how
> > > to approach a method, if there is one.
>
> > > thanks.
> > snip
>
> > Use the Carmichael lambda function.
> >http://mathworld.wolfram.com/CarmichaelFunction.html
> > Find all numbers for which lambda (n) = 2.
>
> that looks beyond me. thanks for the link though. is there a simpler
> explaination of the same thing anywhere? thanks in advance.
>
> > Regards Hanford