Prime conjecture

[alpertron]

Based on the previous post: let n = (2^2p -2^p+1)/3 where p is a prime number, is it true that 2^n-1 = 1 (mod n) ?

[R. Gerbicz]

Originally Posted by alpertron

Based on the previous post: let n = (2^2p -2^p+1)/3 where p is a prime number, is it true that 2^n-1 = 1 (mod n) ?

Nice problem. Proof: let p>2 prime number

Observe that: (2^(2*p)-2^p+1)/3 | 2^(6*p)-1
so it is enough to prove that for p>2: 2^(n-1)==1 mod 2^(6*p)-1
it is equivalent to n-1==0 mod (6*p)
so (2^(2*p)-2^p-2)/3==0 mod (6*p)
(18*p)|2^p*(2^p-1)-2

Let p>3 then it is true mod p (use Fermat's little theorem), 2 also divides, this is trivial since 2^p is even. For 9 use: p=6k+-1 then 2^p=={2,5} mod 9, for the two cases: 2^p*(2^p-1)-2=={0,18}==0 mod 9.
Since 2,9,p are relative prime numbers for p>3 it means that the divisibility is also true for 2*9*p=18*p.
For p=3: 2^(n-1)==1 mod n is also true.

[alpertron]

Thanks.

[Mounir]

Hi everyone


I know I am a bit late.... BUT I have the answer to your problem:


take N = 2^524287 - 1.



Notice that 524287 = 2^19 - 1, this is a clue for the proof ;)


Indeed, you just need to consider the sequence X(n+1) = 2^X(n) - 1 with X(0) = 19 and proove that all terms of the sequence verify X(n)*X(n+1) divide 2^X(n) - 2.


Have a nice day!

[kijinSeija]

Hi

I would like to know if a^(2^n) mod (2^n-1) = a² when a is between 0 and sqrt(2^n-1) and a is an integer when 2^n-1 is prime and only if it's prime ?

I tried this with some Mersenne exposant and it apparently works.

It can't be used to eliminate non-Mersenne exponant quickly ?

Thanks :D

[MattcAnderson]

In My Humble Opinion (IMHO) the even numbers are easier to keep track of

[LaurV]

Originally Posted by kijinSeija

Hi

I would like to know if a^(2^n) mod (2^n-1) = a² when a is between 0 and sqrt(2^n-1) and a is an integer when 2^n-1 is prime and only if it's prime ?

I tried this with some Mersenne exposant and it apparently works.

It can't be used to eliminate non-Mersenne exponant quickly ?

Thanks :D

That is a PRP test in disguise. From Fermat theorem, a^p=a (mod p) if p is prime. Amplify with a, you get a^(p+1)=a^2 (mod p). Replace p with Mp and you have your test. Replace a with 3 and that's what P95 and gpuOwl are doing.

Yes, it can be used to eliminate composite Mp's, but not "fast". Well, not faster than we are doing already.

[kijinSeija]

Originally Posted by LaurV

That is a PRP test in disguise. From Fermat theorem, a^p=a (mod p) if p is prime. Amplify with a, you get a^(p+1)=a^2 (mod p). Replace p with Mp and you have your test. Replace a with 3 and that's what P95 and gpuOwl are doing.

Yes, it can be used to eliminate composite Mp's, but not "fast". Well, not faster than we are doing already.

Hi and thanks for your reply.

And by the way I noticed something interesting : if a+b+c = 2^n-1 you can found number with a²+b²+c² = (4^n-1)/3 and it seems it works with Mersenne number (it seems it doesn't work with composite Mersenne number like 2^11-1 or 2^23-1)

For example 4+2+1 = 7 (2^3-1) and 4²+2²+1² = 21 (4^3-1)/3
another example : 14+9+8 = 31 (2^5-1) and 14²+9²+8² = 341 (4^5-1)/3

There is a way to explain this ? Thanks :)

[slandrum]

I'm not sure what you are trying to say here...

4+3+0 = 7 (2^3-1), but 4^2 + 3^2 + 0^2 = 25 which is not (4^3-1)/3

13 + 10 + 8 = 31 (2^5-1) but 13^2 + 10^2 + 8^2 = 333 which is not (4^5-1)/3

So clearly there's something I'm missing.

[kijinSeija]

Originally Posted by slandrum

I'm not sure what you are trying to say here...

4+3+0 = 7 (2^3-1), but 4^2 + 3^2 + 0^2 = 25 which is not (4^3-1)/3

13 + 10 + 8 = 31 (2^5-1) but 13^2 + 10^2 + 8^2 = 333 which is not (4^5-1)/3

So clearly there's something I'm missing.

I mean you can find three random numbers (a, b, c) that respect that :

a+b+c = 2^n-1
and a²+b²+c²=(4^n-1)/3

If a+b+c is a prime Mersenne number.

you can't find any numbers for example for a+b+c = 2^11-1 and a²+b²+c² for (4^11-1)/3 it seems it doesn't exist.