[QUOTE=blob100;204149]Ho thank you so much CRGreathouse,
You understood what I meant and now i'll use the sybols.

Tomer[/QUOTE]Woohoo, we finally got it. And it only took 121 messages (plus a few deleted also) to get there.

Okay so moving on ...

Death in Venice
 

[quote=__HRB__;204140]We were expecting a link to pornography with exiting sex, natural looks and oscar-wildesque dialog. Consider yourself in the doghouse for using bait-and-switch.[/quote]
Someone (and I thought it was you) said their next musical project
was to get into Mahler.

[URL]http://www.youtube.com/watch?v=FTP7XFVGnxQ[/URL]

This may be an entree and tick a few boxes.

I know :threadhijacked:again.

[quote=cmd;204146]anything you think "really" is true

you feel



( 5 ? )


[/quote]
I know that 5 is a factor of 430, I just wanted to show that n is 10 and 10 isn't odd.

So do my conjecture found before, trivially ture/false, or even false (not triviallic).

[quote=retina;204150]Woohoo, we finally got it. And it only took 121 messages (plus a few deleted also) to get there.

Okay so moving on ...[/quote]
Not deleted. Justed risen up to the soap box under the
guise of "Cunning Linguists".

Talking in tongues
 

[quote=__HRB__;204140]We were expecting a link to pornography with exiting sex, natural looks and oscar-wildesque dialog. Consider yourself in the doghouse for using bait-and-switch.[/quote]

[quote=cmd;204158]you know and accept ... [URL]http://www.coulee-de-serrant.com/:smile:[/URL][/quote]
Sorry, can't get that link here in civilization.

[QUOTE=blob100;204149][QUOTE=CRGreathouse;204148]Here's my guess as to what you mean. Is this right?

Denote by f(N) the smallest factor of N-1 such that 2^f(N) > N. Let s = f(p) where p is an odd prime. Conjecture: if s is odd, then [tex]2^s\equiv1\pmod p[/tex].[/QUOTE]
Ho thank you so much CRGreathouse,
You understood what I meant and now i'll use the sybols.

Tomer[/QUOTE]

Excellent, we have communication!

But it looks
[code]f(N)=fordiv(N-1,d,if(2^d>N,return(d)))
test(p,s)=if(s%2,if(Mod(2,p)^s==1,print(p),print("Conjecture fails at "p))))
forprime(p=3,97,test(p,f(p)))[/code]
like the conjecture fails at p = 11, s = 5, since 2^5 = 10 (mod 11).

[QUOTE=CRGreathouse;204148]Here's my guess as to what you mean. Is this right?

Denote by f(N) the smallest factor of N-1 such that 2^f(N) > N. Let s = f(p) where p is an odd prime. Conjecture: if s is odd, then [tex]2^s\equiv1\pmod p[/tex].[/QUOTE]

This is clearly false. Take (say) p = 1151. p-1 = 2 * 5^2 * 23
The smallest s dividing p-1 such that 2^s > 1151 is s = 23,
but 2^23 mod 1151 = 120, not 1.

Indeed. There are good reasons why the conjecture should not be
correct. In passing from Z/pZ* to the inequality 2^f(N) > N, one
is going from a local field (which is unordered!) to a global field, which
is very much ordered. The topologies are totally different.

Beat you to it, Bob. :smile:

[QUOTE=R.D. Silverman;204172]This is clearly false. Take (say) p = 1151. p-1 = 2 * 5^2 * 23
The smallest s dividing p-1 such that 2^s > 1151 is s = 23,
but 2^23 mod 1151 = 120, not 1.

Indeed. There are good reasons why the conjecture should not be
correct. In passing from Z/pZ* to the inequality 2^f(N) > N, one
is going from a local field (which is unordered!) to a global field, which
is very much ordered. The topologies are totally different.[/QUOTE]

Let me also point out that if 2 is a primitive root of the prime p, then
the smallest s such that 2^s = 1 mod p is s = p-1.

This conjecture had no chance to be correct.

[QUOTE=CRGreathouse;204173]Beat you to it, Bob. :smile:[/QUOTE]

How can we suggest to the OP (without his taking offense)
that before attempting any further such
"conjectures", that he actually learn some number theory?

I suggest that for the kind of conjecture that he stated, the book
by D. Shanks "Solved and Unsolved Problems in Number Theory" is
superb. It gives the best introduction to the connections between
arithmetic mod p and the underlying multiplicative group Z/pZ* that
I have ever seen.