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) 
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=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 
[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 
[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). 
[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. 
Okay. Thanks.

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