Mersenne Prime Factors of v.large numbers

[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)

[alpertron]

In your web site you wrote:

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.

[devarajkandadai]

Originally Posted by alpertron

In your web site you wrote:

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

[devarajkandadai]

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

[cheesehead]

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).

[Numbers]

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

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.

[cheesehead]

Okay. Thanks.