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 2012-09-19, 16:06 #34 alpertron     Aug 2002 Buenos Aires, Argentina 22×5×71 Posts Based on the previous post: let n = (22p -2p+1)/3 where p is a prime number, is it true that 2n-1 = 1 (mod n) ?
2012-09-19, 16:55   #35
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

2×32×5×17 Posts

Quote:
 Originally Posted by alpertron Based on the previous post: let n = (22p -2p+1)/3 where p is a prime number, is it true that 2n-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.

 2012-09-19, 17:18 #36 alpertron     Aug 2002 Buenos Aires, Argentina 22×5×71 Posts Thanks.
 2021-03-20, 08:03 #37 Mounir   Mar 2021 1 Posts 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!
 2021-03-21, 16:14 #38 kijinSeija   Mar 2021 France 2×32 Posts 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
 2021-03-21, 22:43 #39 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 2×503 Posts In My Humble Opinion (IMHO) the even numbers are easier to keep track of
2021-03-22, 03:03   #40
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

268316 Posts

Quote:
 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.

2021-05-23, 15:32   #41
kijinSeija

Mar 2021
France

2×32 Posts

Quote:
 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.

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 :)

 2021-05-23, 16:55 #42 slandrum   Jan 2021 California 13×23 Posts 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.
2021-05-23, 17:09   #43
kijinSeija

Mar 2021
France

2×32 Posts

Quote:
 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.

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