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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 4:13 PM

On Fri, 22 Jul 2011 15:50:24 -0500, quasi <quasi@null.set> wrote:

>On Fri, 22 Jul 2011 10:06:10 -0700 (PDT), Simon Roberts
><2assume@gmail.com> wrote:
>

>>Please correct me if this is not well written but for each
>>i>=1 and every x x^(2^i) mod n forms a set that is identical
>>to the set x^(2^j) mod n j>=1 j!=i.

>
>It's poorly written:
>
>(1) You never specified your intended restrictions on n.
>
>(2) You didn't specify any restrictions on x. What about
>x = 0 (mod p)? What about x = 1 or -1 (mod p)?
>
>Would it have hurt you to at least give a careful and
>complete statement of the proposition you're claiming is
>true? It shouldn't be necessary for people to piece
>together your claim from prior messages, many of which
>may have already been deleted from their news interface
>
>I'll state what it appears you're claiming ...
>
>Proposition:
>
>if p is prime with p = 3 (mod 4), then for all integers x,y
>with x,y not congruent to 0,1,-1 (mod p),
>
> {x^(2^i), i = 1,2,3, ...} = {y^(2^j), j = 1,2,3, ...}

I meant:

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

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

>>I don't know if this is a well known result and if it has
>>or can be proved more simply.

>
>It's not well known, and it hasn't and can't be proved since
>it's false. For example, try p = 19.
>
>In fact, for primes p = 3 (mod 4), the claim of the above
>proposition is always false if (p-1)/2 is composite, and it's
>even false in many cases where (p-1)/2 is prime.
>
>The primes p = 3 (mod 4) with p < 1000 and (p-1)/2 also prime
>for which the conclusion of the above proposition holds are:
>
> 7,11,23,59,107,167,263,347,359,587,839,887,983
>
>Question:
>
>For which primes p = 3 (mod 4) with (p-1)/2 also prime does
>the conclusion of the above proposition hold?
>
>
>Remark:
>
>However, a quick scan of the data suggests the following
>tentative claim ...
>
>Conjecture:
>
>If p is prime with p = 3 (mod 4) and (p-1)/2 also prime, then
>for all integers x,y with x,y not congruent to 0,1,-1 (mod p),
>the sets
>
> {x^(2^i), i = 1,2,3, ...}
>
> {y^(2^j), j = 1,2,3, ...}
>
>have the same number of elements.

I meant:

the sets

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

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

have the same number of elements.

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
7/22/11 quasi
7/22/11 quasi
7/22/11 quasi
7/22/11 quasi
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
7/24/11 quasi
7/22/11 quasi