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

[QUOTE=devarajkandadai;321158]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.
[/QUOTE]

I didn't fully understand this. I think its the part "a,n & c belong to N". What is "N"? Sorry for asking basic questions.

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.

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...

[QUOTE=devarajkandadai;321158]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.[/QUOTE]

If p | Mp, then this would be trivially true. But since p | Mp-1, this is unlikely to be true. And LaurV has already given a counterexample.

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.

[QUOTE=axn;321173]If p | Mp, then this would be trivially true. But since p | Mp-1, this is unlikely to be true. And LaurV has already given a counterexample.

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.[/QUOTE]
Indeed, take p=7, then Mp=127. There is no f(n)=7^n+1 to be divisible by 7, all are 1 (mod 7), but 7^63+1 is divisible by 127.

Two points
 

[QUOTE=LaurV;321167]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.[/QUOTE]

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.

[QUOTE=devarajkandadai;321252]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.[/QUOTE]
Well, "[I]any theory is good if, to confirm it, we need to scrap no more then 50% of the experimental findings[/I]." (from the Murphy's collection).
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.

Two points
 

[QUOTE=Batalov;321170]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...[/QUOTE]

Example: 2^n+7 is divisible by 5 for some values of n. It is not divisible by 31, the relevant M_p, for any value of n. This shows that the converse is not true.

[QUOTE=devarajkandadai;321254]Example: 2^n+7 is divisible by 5 for some values of n. It is not divisible by 31, the relevant M_p, for any value of n. This shows that the converse is not true.[/QUOTE]
Man, that is not the converse.
An anyhow, what you are trying to do is trivial, if even me can understand it :smile:
The order of 2 in any prime p is a factor of p-1. The order of 2 in any Mp is p. As p and p-1 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 2-4-3-1-...-2-4-3-1 (mod 5). The order of 2 in 2^5-1=31 is 5, the string is 2-4-8-16-1. Now if you chose convenient c, say c=4, the strings become 2+4, 4+4, 3+4, 1+4, i.e 1-3-2-0 (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 2-4-3-1 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.

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.