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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 6:32 PM

On Fri, 22 Jul 2011 17:50:35 -0500, quasi <quasi@null.set> wrote:

>On Fri, 22 Jul 2011 16:11:22 -0500, quasi <quasi@null.set> wrote:
>

>>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) (mod p), i = 1,2,3, ...}
>>
>> {y^(2^j) (mod p), j = 1,2,3, ...}
>>
>>have the same number of elements.

>
>It's true -- I can prove it.
>
>But for now, I'll leave it as a challenge.

To Simon Roberts:

The above result, while elementary, is (I think) interesting
worthwhile.

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