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


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 nonprime, 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



