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Replies: 73   Last Post: Jul 24, 2011 8:55 PM

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 quasi Posts: 12,067 Registered: 7/15/05
Posted: Jul 22, 2011 11:30 PM

On Fri, 22 Jul 2011 19:16:52 -0700 (PDT), Simon Roberts
<2assume@gmail.com> wrote:

>Say the set S = {1,2,3,...,(n-2),(n-1)}.
>Then if we squared each element x_i in S, we would get a new
>set S' (with (n-1)/2 elements) whose elements would be
>x_i)^2 mod n. Note we are assuming (n-1)/2 is odd, so
>n = 3 mod 4. If we square each element in S'
>((x_i)^2)^2 mod n we would end up with the same exact set S'
>and so on.
>
>Do you agree with this statement?

Yes, assuming also that n is a prime. Actually, it still holds
if n is a product of an odd number of primes all congruent to
3 (mod 4). By the way -- usually the letter p is used to
emphasize that a variable is prime.

One of my main complaints with your "statements" is that you
assume things not stated. Can't you make a complete statement?
Why do I always have to do it for you?

Here's what you're claiming now:

Proposition:

If p is prime with p = 3 (mod 4), then the sets

{x^2 (mod p) | x = 1,2,3 ... (p-1)}

{x^4 (mod p) | x = 1,2,3 ... (p-1)}

are equal.

Remark:

The above claim is true and is easily proved, but that does
_not_ yield a proof of your prior claim:

False Proposition:

If p is prime with p = 3 (mod 4), then for all integers x,y
not congruent to 0,1,-1 (mod p), the sets

{x^(2^i) (mod p) | i = 1,2,3 ...}

{y^(2^j) (mod p) | j = 1,2,3 ...}

are equal.

Counterexample: p = 19.

quasi

Date Subject Author
7/19/11 Simon Roberts
7/19/11 Simon Roberts
7/19/11 Simon Roberts
7/19/11 byron
7/19/11 Simon Roberts
7/19/11 quasi
7/19/11 quasi
7/19/11 Simon Roberts
7/19/11 quasi
7/19/11 Simon Roberts
7/19/11 quasi
7/19/11 Simon Roberts
7/19/11 quasi
7/19/11 Simon Roberts
7/19/11 Simon Roberts
7/19/11 Simon Roberts
7/19/11 Simon Roberts
7/19/11 Simon Roberts
7/19/11 Simon Roberts
7/20/11 quasi
7/20/11 quasi
7/20/11 Simon Roberts
7/20/11 quasi
7/20/11 Simon Roberts
7/20/11 tommyrjensen@gmail.com
7/20/11 Simon Roberts
7/20/11 tommyrjensen@gmail.com
7/20/11 Simon Roberts
7/20/11 tommyrjensen@gmail.com
7/20/11 Simon Roberts
7/20/11 Simon Roberts
7/20/11 tommyrjensen@gmail.com
7/20/11 Simon Roberts
7/20/11 Simon Roberts
7/20/11 Simon Roberts
7/20/11 tommyrjensen@gmail.com
7/20/11 Simon Roberts
7/20/11 Simon Roberts
7/20/11 Simon Roberts
7/20/11 Simon Roberts
7/20/11 Simon Roberts
7/20/11 Simon Roberts
7/20/11 Simon Roberts
7/20/11 Simon Roberts
7/22/11 Simon Roberts
7/22/11 quasi
7/22/11 Simon Roberts
7/22/11 Simon Roberts
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7/24/11 quasi
7/22/11 Simon Roberts
7/22/11 quasi
7/22/11 Simon Roberts
7/22/11 Simon Roberts
7/22/11 Simon Roberts
7/22/11 Simon Roberts
7/22/11 quasi
7/23/11 Simon Roberts
7/23/11 quasi
7/23/11 Simon Roberts
7/23/11 Simon Roberts
7/23/11 quasi
7/23/11 Simon Roberts
7/23/11 Simon Roberts
7/23/11 Simon Roberts
7/23/11 quasi
7/24/11 Bill Dubuque
7/24/11 Brian Q. Hutchings
7/23/11 quasi
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7/22/11 quasi