I know I've been beat to it mathematically, but just to also show it experimentally:
I wrote a Java program to check the conjecture. Searching all primes below 10000, I found 658 counterexamples and 570 agreeing examples. If anyone's interested, I've attached the full output of the program.
Prior to doing this, I randomly picked 3 primes, and all of them turned out to be agreeing examples. What a coincidence...

[quote=CRGreathouse;204171]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]
Yes I saw it by myself.
I found that there are p numbers which according to mersenne and some for 2[sup]n[/sup]+1.
11=p is one of the numbers that are according to 2[sup]n[/sup]+1.
We can say that 2^s\equiv+/-1\(mod p) which isn't perfect but more logical.

But now the problem is finding on which p numbers we get a s that is according to mersenne or to 2[SUP]n[/SUP]+1.

[quote=blob100;204189]We can say that 2^s\equiv+/-1\(mod p) which isn't perfect but more logical.[/quote]
All this does is make 8 of the 658 counterexamples where p < 10000 into agreeing examples.

[QUOTE=R.D. Silverman;204175]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.[/QUOTE]Good suggestion, but...

Do you believe it suitable for someone with the mathematical sophistication likely to be possessed by a 14-year old? Note I state "sophistication" and make no estimation about intelligence.

What would you suggest for someone smart and of that age?

Paul

Mini geek,
Can you give me examples for numbers that are not agreeing with my conjecture?
Never mind, I saw the whole program.

Ouch I guess I can't find sturcture from now..
Just made a garbage and wasted the time of the whole group of peaple that answered my question.

[QUOTE=blob100;204189]
But now the problem is finding on which p numbers we get a s that is according to mersenne or to 2[SUP]n[/SUP]+1.[/QUOTE]


This is no problem. I will give some hints: look up "quadratic residue".
Look up "Euler's Criterion". Ask the following questions:

What are the modular requirements on a prime p in order for 2 to be
a quadratic residue (respectively non-residue)?

What are the requirements related to quadratic residues that determine
when a prime p divides 2^q + 1, rather than 2^q -1??

[QUOTE=blob100;204203]Ouch I guess I can't find sturcture from now..
Just made a garbage and wasted the time of the whole group of peaple that answered my question.[/QUOTE]You have shown far more respect for the help that you have gotten than many others have in the past. I think, that other than the off topic items, this has been one of the sanest and most well behaved thread of this sort. I think that you have paid attention to the help of others.

I also think that Bob has done well.:bow: Often times those that are unlearned cause him to react a bit strongly.

[QUOTE=xilman;204193]Good suggestion, but...

Do you believe it suitable for someone with the mathematical sophistication likely to be possessed by a 14-year old? Note I state "sophistication" and make no estimation about intelligence.

What would you suggest for someone smart and of that age?

Paul[/QUOTE]

Shank's book should be readable by a teenager possessing a reasonable
amount of mathematical maturity.

[QUOTE=blob100;204189][QUOTE=CRGreathouse;204171]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]
Yes I saw it by myself.
I found that there are p numbers which according to mersenne and some for 2[sup]n[/sup]+1.
11=p is one of the numbers that are according to 2[sup]n[/sup]+1.
We can say that 2^s\equiv+/-1\(mod p) which isn't perfect but more logical.

But now the problem is finding on which p numbers we get a s that is according to mersenne or to 2[SUP]n[/SUP]+1.[/QUOTE]

You are not paying attention to the examples or you would have noticed the superfluous ")" in the second line of the [URL="http://pari.math.u-bordeaux.fr/"]PARI/GP[/URL] script. CRGreathouse took the time and had the smarts to figure out what you meant, but to thank him for it you don't even bother to recreate the results?

[QUOTE=CRGreathouse;204171]
like the conjecture fails at p = 11, s = 5, since 2^5 = 10 (mod 11).[/QUOTE]

Yes, [TEX]p=11 \ s=5[/TEX] was the counterexample I mentioned on my re-written version of the conjeture at post #110
So we can conclude the conjeture was false. [TEX]\qed[/TEX]

Another work well done.