Two points

[LaurV]

Originally Posted by devarajkandadai

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.

So, you say: "given p a prime, for which Mp is a prime too, there exists some c such as 2^n+c is not divisible by p, neither by Mp", this is true, and is trivial. Without the blue part, this is false, and I just proved to you. I gave you p=5, Mp=31, c=15, for which 2^n+c is never divisible by 5, but every 2^(5k+4)+15 is divisible by 31. What "counter" you need?

[MattcAnderson]

In reply to post number 2, N is the set of natural numbers.
N = {1, 2, 3, ...}

[cheesehead]

Unicode codes for double-struck letters ( ℝ ℂ ℕ ℙ ℚ ℤ ) denoting commonly-referenced sets:

http://symbolcodes.tlt.psu.edu/bylan...chart.html#let

[henryzz]

Don't forget - quaternions

[cheesehead]

Originally Posted by henryzz

Don't forget - quaternions

(TEX expression rather than Unicode)

[henryzz]

Originally Posted by cheesehead

(TEX expression rather than Unicode)

http://en.wikibooks.org/wiki/Unicode...ence/2000-2FFF
210D

edit:
http://en.wikipedia.org/wiki/Blackboard_bold

[science_man_88]

Originally Posted by devarajkandadai

I presume I should have said let p be an odd prime. Secondly the base, a, can only be 2 since Mersenne primes have a meaning only when base is 2.

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.

what's the fixed value of c ? if not fixed:

2^n = 1 mod 5 for n=0 mod 4 ; 2^n+4 = 0 mod 5 for these n
2^n = 2 mod 5 for n=1 mod 4 ; 2^n+3 = 0 mod 5 for these n
2^n = 4 mod 5 for n=2 mod 4 ; 2^n+1 = 0 mod 5 for these n
2^n = 3 mod 5 for n=3 mod 4 ; 2^n+2 = 0 mod 5 for these n

2^n = 1 mod 31 for n=0 mod 5 ; 2^n+30 = 0 mod 31 for these n
2^n = 2 mod 31 for n=1 mod 5 ; 2^n+29 = 0 mod 31 for these n
2^n = 4 mod 31 for n=2 mod 5 ; 2^n+28 = 0 mod 31 for these n
2^n = 8 mod 31 for n=3 mod 5 ; 2^n+23 = 0 mod 31 for these n
2^n = 16 mod 31 for n=4 mod 5 ; 2^n+15 = 0 mod 31 for these n

29-4 =25 a multiple of 5 n=0 mod 4 && n=1 mod 5 minimum n=16 to fit both so for the n that fit both they both divide. can you give me an example from with this data ? also if you extend what I said to the next one that divides 31 you see it doesn't divide by 5.

[LaurV]

Originally Posted by science_man_88

2^n = 4 mod 31 for n=2 mod 5 ; 2^n+[COLOR=Red]28[/COLOR]27 = 0 mod 31 for these n

Essentially you put in bullets what I said in plain text few posts above, that is why the typo (28->27) "stroke" me.

[science_man_88]

Originally Posted by LaurV

Essentially you put in bullets what I said in plain text few posts above, that is why the typo (28->27) "stroke" me.

sorry for the error, my point was that if you find one example were 2^n+c is divisible by both 5 and 31 c<-c+31 the amount added to c now doesn't allow c to divide by 5 but it keeps it divisible by 31 for a consistent c across the 2 calculations.