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Old 2012-12-10, 05:50   #1
devarajkandadai
 
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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
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Old 2012-12-10, 06:47   #2
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Quote:
Originally Posted by devarajkandadai View Post
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.
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.
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Old 2012-12-10, 07:20   #3
LaurV
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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.

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Old 2012-12-10, 07:36   #4
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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...
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Old 2012-12-10, 08:28   #5
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Quote:
Originally Posted by devarajkandadai View Post
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.
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.
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Old 2012-12-10, 10:13   #6
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Originally Posted by axn View Post
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.
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.

Last fiddled with by LaurV on 2012-12-10 at 10:19
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Old 2012-12-11, 03:51   #7
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Originally Posted by LaurV View Post
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.
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.
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Old 2012-12-11, 04:10   #8
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Quote:
Originally Posted by devarajkandadai View Post
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.
Well, "any theory is good if, to confirm it, we need to scrap no more then 50% of the experimental findings." (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.

Last fiddled with by LaurV on 2012-12-11 at 04:15
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Old 2012-12-11, 04:15   #9
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Originally Posted by Batalov View Post
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...
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.
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Old 2012-12-11, 04:29   #10
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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.
Man, that is not the converse.
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 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.

Last fiddled with by LaurV on 2012-12-11 at 05:20
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Old 2012-12-12, 13:50   #11
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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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