quasi
Posts:
9,903
Registered:
7/15/05


Re: a contradiction worth reading (unless a silly mistake was made)
Posted:
Jul 24, 2011 12:02 AM


On Fri, 22 Jul 2011 18:24:04 0500, quasi <quasi@null.set> wrote:
>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 (p1)/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 >and not entirely trivial, so your ideas did lead to something >worthwhile.
I take it back  I've now found a simple, straightforward proof which makes it clear that proving the above claim is just a routine exercise.
quasi

