20190426, 09:22  #1 
Feb 2012
Prague, Czech Republ
244_{8} Posts 
Constraints on k in 2kp+1
If divides a Mersenne number , prime , then can be written as .
IIUC, it can be shown that . Additionally I think I've seen somewhere that when . Q: Are there other known constraints? Or any other interesting properties of ? 
20190426, 12:56  #2 
Einyen
Dec 2003
Denmark
3×5×199 Posts 
k=0 (mod p) is ok but only 1 known example p=93077.

20190426, 13:15  #3 
"Brian"
Jul 2007
The Netherlands
CC3_{16} Posts 

20190426, 13:27  #4 
Feb 2017
Nowhere
2·3·11·59 Posts 
For p > 2,
so 2 is a quadratic residue of every prime divisor q, whence q == 1 or 1 (mod 8). Consequently, if q = 2*k*p + 1, then k == 0 (mod 4) (q == 1 (mod 8)) or k == p (mod 4) (q == 1 (mod 8)). 
20190426, 14:05  #5  
Mar 2018
3·43 Posts 
Quote:
11686604129694847 = 1 + 2 · 3^4 · 11 · 757 · 93077^2 Edit: Which is to say, k is not divisible by p^2 in this example, factor minus one is divisible by p^2. Last fiddled with by DukeBG on 20190426 at 14:08 

20190426, 14:19  #6 
Mar 2018
3×43 Posts 
In case it'll be of interest, a little intuition about why 2kp+1, where it comes from... For any prime number f, we'll have M(f1) being divisible by f because Fermat's little theorem. Also you know (very easy to show) that M(x)  M(x * y). So the number f might be divisible in M(f1) because it's a factor of some M(n) where n  f1. Thus for M(p), their factors have p  f1. I.e. f = 1 mod p. I.e. f=k'*p+1. However, f is obviously odd, so k' should be even, so we can write it as 2k or f=2kp+1.
Last fiddled with by DukeBG on 20190426 at 14:19 
20190426, 14:20  #7  
"Brian"
Jul 2007
The Netherlands
110011000011_{2} Posts 
Quote:
I am asking if it is known that if k=0 (mod p) then k=0 (mod p^2) (for k<>0). 

20190426, 14:22  #8 
Mar 2018
3×43 Posts 
What do you think k is in this example? 3^4 · 11 · 757 · 93077 is not divisible by 93077^2.

20190426, 14:30  #9  
"Brian"
Jul 2007
The Netherlands
6303_{8} Posts 
Quote:
p=93077 According to ATH p=93077 is the only known example where pk. However I also observed that p^2  k. Is that just a monstrous coincidence? Or is it necessary? 

20190426, 14:59  #10 
"Brian"
Jul 2007
The Netherlands
3267_{10} Posts 
Apologies!
I've totally misunderstood what k was and that led to a breakdown of communications. Ignore my posts please. 
20190426, 15:03  #11 
Einyen
Dec 2003
Denmark
3·5·199 Posts 
There is no other GIMPS factor p<1B with this property and I also searched a bit beyond GIMPS limits without finding any other examples.
Last fiddled with by ATH on 20190426 at 15:04 
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