20121210, 05:50  #1 
May 2004
2^{2}×7×11 Posts 
Two points
This is with reference to my number theory video presentation on You tube:
1) Let p be a prime and M_p be the corresponding Mersenneprime .Consider f(n) = a^n + c where a,n & c belong to N, a & c are fixed. Then if p is not a factor of f(n) for any value of n then M_p also cannot be a factor of f(n) for any value of n, however large. 2) The converse is not true. A.K. Devaraj 
20121210, 06:47  #2  
"Åke Tilander"
Apr 2011
Sandviken, Sweden
566_{10} Posts 
Quote:


20121210, 07:20  #3 
Romulan Interpreter
Jun 2011
Thailand
2^{3}·11·97 Posts 
Let p=2. Then Mp=3. Let f(n)=2^n+1. All f(n) are odd, so they can't be divisible by 2. Don't tell me that 2^3+1 (and generally 2^(2k+1)+1) is not divisible by 3....
edit: @Ake, we assume N is the set "0,1,2..." of natural numbers. Last fiddled with by LaurV on 20121210 at 07:23 
20121210, 07:36  #4 
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3
3·31·97 Posts 
2) What was meant to be the converse? if reversing the roles of p and M_p, then take p=2, a=3, c=1...

20121210, 08:28  #5  
Jun 2003
10757_{8} Posts 
Quote:
In fact, if you construct f(n) = p^n+1, the first part (p not being a factor) is trivially satisfied. If the order of p (mod Mp) is even, then there'll be an f(n) which is divisible by Mp. 

20121210, 10:13  #6  
Romulan Interpreter
Jun 2011
Thailand
2^{3}×11×97 Posts 
Quote:
Last fiddled with by LaurV on 20121210 at 10:19 

20121211, 03:51  #7  
May 2004
308_{10} Posts 
Two points
Quote:
Example: 2^n+5 is not divisible by 5 for any value of n. 2^n + 5 is not divisible by 31, the relevant M_p, for any value of n. 

20121211, 04:10  #8  
Romulan Interpreter
Jun 2011
Thailand
2^{3}×11×97 Posts 
Quote:
How about p=5, Mp=31, a=2, c=15, does this satisfy your hypothesis? Or you will now start imposing supplementary conditions to the "c" parameter too? Because if 2^n+5 is not divisible by 5 (and indeed is not, as 2^n is not, and 5 it is), I just added 10 to it and get 2^n+15 again, not divisible by 5 for any n. But the trick is that I found out 2^4+5 was 21, so by adding 10 we get 2^4+15 is divisible to 31... And generally, because of the periodicity of 2^n to 31, we get 2^(5k+4)+15 is always divisible to 31, for any k. Last fiddled with by LaurV on 20121211 at 04:15 

20121211, 04:15  #9 
May 2004
2^{2}×7×11 Posts 
Two points

20121211, 04:29  #10  
Romulan Interpreter
Jun 2011
Thailand
2^{3}·11·97 Posts 
Quote:
An anyhow, what you are trying to do is trivial, if even me can understand it The order of 2 in any prime p is a factor of p1. The order of 2 in any Mp is p. As p and p1 are always coprime, you can always find values which satisfy all 4 combinations (x,y), (x,not y), (not x, y) and (not x, not y). For example, the order of 2 in 5 is 4, because 2^1=2, 2^2=4, 2^3=3, 2^4=1, then they repeat 2431...2431 (mod 5). The order of 2 in 2^51=31 is 5, the string is 248161. Now if you chose convenient c, say c=4, the strings become 2+4, 4+4, 3+4, 1+4, i.e 1320 (mod 5) respective 6,8,12,20,5 (mod 31). You are now in the case (x, not y), i.e. sometimes 5 divides 2^n+4, but 31 never divides 2^n+4. Take another c and you find other case. If c=0 (mod 5) then the first string stays always 2431 and there will be no n for which 2^n+5k is divisible by 5. But now if the same c is 29,27,23,15 or 30 mod 31, then sometime 2^n+c will divide even to 31. If you take c=5,10,20,25,35,40,45,50,55,65,etc (i.e. i skipped 15, 30 and 60, to be 0 mod 5 and not 15,30,29 mod 31) then your "theorem" is true. If only. Last fiddled with by LaurV on 20121211 at 05:20 

20121212, 13:50  #11 
May 2004
2^{2}×7×11 Posts 
Two points
The fact remains that no counter has been furnished to my statement that if p is a given odd prime such that its M_p the relevant Mersenne prime exists and if f(n)= 2^n + c( where c is fixed ) is not divisible by p then certainly 2^n + c is not divisible by M_p, however large n may be.

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