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-   -   Mersenne Prime Factors of v.large numbers (https://www.mersenneforum.org/showthread.php?t=5258)

devarajkandadai 2006-01-01 06:20

Mersenne Prime Factors of v.large numbers
 
For the above refer to "Minimum Universal Exponent Generalisation of
Fermat's Theorem" on site:

[url]www.crorepatibaniye.com/failurefunctions[/url]

A.K. Devaraj (dkandadai@yahoo.com)

alpertron 2006-01-02 17:03

In your web site you wrote:

[tex]\lambda (m)[/tex] is the minimum universal exponent of m (m belongs to N).

What do you mean by [I]minimum universal exponent[/I]?

I suppose that by N you refer to [b]N[/b], the set of natural numbers.

devarajkandadai 2006-01-04 03:54

[QUOTE=alpertron]In your web site you wrote:

[tex]\lambda (m)[/tex] is the minimum universal exponent of m (m belongs to N).

What do you mean by [I]minimum universal exponent[/I]?

I suppose that by N you refer to [b]N[/b], the set of natural numbers.[/QUOTE]
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

devarajkandadai 2006-01-04 04:05

[QUOTE=devarajkandadai]For the above refer to "Minimum Universal Exponent Generalisation of
Fermat's Theorem" on site:

[url]www.crorepatibaniye.com/failurefunctions[/url]

A.K. Devaraj (dkandadai@yahoo.com)[/QUOTE]

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

cheesehead 2006-01-04 06:58

[quote=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).[/quote]
But why is [U]5[/U] 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).

Numbers 2006-01-04 09:36

[quote=cheesehead]But why is 5 the minimum universal exponent[/quote]

I [b]think[/b] 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 [b]think[/b] this is a consequence of the corollary he mentioned in his previous post
[quote=devarajkandadai](2^x) + 29: this function of x is a multiple of 31...[/quote]

I'm not saying I follow his argument, or even agree with him. I'm just offering to clarify what I think he said.

cheesehead 2006-01-04 22:44

Okay. Thanks.


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