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Old 2006-01-01, 06:20   #1
devarajkandadai
 
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Default Mersenne Prime Factors of v.large numbers

For the above refer to "Minimum Universal Exponent Generalisation of
Fermat's Theorem" on site:

www.crorepatibaniye.com/failurefunctions

A.K. Devaraj (dkandadai@yahoo.com)
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Old 2006-01-02, 17:03   #2
alpertron
 
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In your web site you wrote:

\lambda (m) is the minimum universal exponent of m (m belongs to N).

What do you mean by minimum universal exponent?

I suppose that by N you refer to N, the set of natural numbers.
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Old 2006-01-04, 03:54   #3
devarajkandadai
 
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Quote:
Originally Posted by alpertron
In your web site you wrote:

\lambda (m) is the minimum universal exponent of m (m belongs to N).

What do you mean by minimum universal exponent?

I suppose that by N you refer to N, the set of natural numbers.
Yes N means set of natural numbers.To give an example of min. u.e.:

5 is the minimum universal exponent (w.r.t base 2) i.e.

(2^5) - 1 =31 and 5 is the minimum exp such that 2^n - 1 is congruent to
zero (mod 31).

Devaraj
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Old 2006-01-04, 04:05   #4
devarajkandadai
 
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Quote:
Originally Posted by devarajkandadai
For the above refer to "Minimum Universal Exponent Generalisation of
Fermat's Theorem" on site:

www.crorepatibaniye.com/failurefunctions

A.K. Devaraj (dkandadai@yahoo.com)
Two numerical corralaries:a) (2^x) + 29: this function of x is a multiple of
31, a Mersenne Prime , for any x ending with 1 or 6.

b) 127, another Mersene prime, is an impossible factor of this function i.e.
it cannot be a factor of (2^x) + 29, no matter how large x is.
Devaraj
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Old 2006-01-04, 06:58   #5
cheesehead
 
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Quote:
Originally Posted by devarajkandadai
5 is the minimum universal exponent (w.r.t base 2) i.e.

(2^5) - 1 =31 and 5 is the minimum exp such that 2^n - 1 is congruent to
zero (mod 31).
But why is 5 the minimum universal exponent with respect to base 2, and not 3 or 2 or even 1, for example? After all, 3 or 2 (or 1) satisfies the same statement you give for 5:

(2^3) - 1 =7 and 3 is the minimum (positive) exponent such that 2^n - 1 is congruent to zero (mod 7).

(2^2) - 1 =3 and 2 is the minimum (positive) exponent such that 2^n - 1 is congruent to zero (mod 3).
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Old 2006-01-04, 09:36   #6
Numbers
 
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Quote:
Originally Posted by cheesehead
But why is 5 the minimum universal exponent
I think you will find that Mr Kanadadai meant that where 2^x - 1 = y, the minimum universal exponent is the lowest positive x that makes y congruent to 0(mod 31), not 0(mod y).

I think this is a consequence of the corollary he mentioned in his previous post
Quote:
Originally Posted by devarajkandadai
(2^x) + 29: this function of x is a multiple of 31...
I'm not saying I follow his argument, or even agree with him. I'm just offering to clarify what I think he said.
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Old 2006-01-04, 22:44   #7
cheesehead
 
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Okay. Thanks.
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