A formula related to multiplicative order

[praneeth]

I want to submit a paper on proof of this formula, so can some one please tell whether this already exists or not?

Let S’(A) be the set of elements in GF(p) such that S’(A) = {x/ O(x,p) = A}. Here A should be
the factor of (p-1) and A>2, where p is prime, then
∑x = μ(A) + ½* T(A)*p;
where the summation is over all the elements of set S’(A) and
O(x,p) : Order of x with respect to p, (by order it is meant to be multiplicative order).
μ(A) : Mobius function of A.
T(A) : Euler’s-Totient function of A.

 

[praneeth]

no replies which means I can write a paper on proof of this theorem. Is that it??

 

[matt grime]

It means no one has looked at it, or thought about it, not that it is new or novel. Not that this is necessarily the correct place to even ask such a question.

The most obvious problem with what you wrote is that it does not and cannot make sense. You have an equality. The LHS is a sum of elements in a finite field, the right hand side is an integer.

 

[praneeth]

1) If you look at it as a mathematical expression, is that correct?? addition of elements taken from a finite field in the integer domain (changing the domain).
(or)
2) (∑x - μ(A))mod p = 0
is that correct? I just want to know if it has got any significance?

 

[matt grime]

You cannot arbitrarily coerce elements from GF(p) into Z, as you did. There is no canonical choice - should I pick -1 or p-1, for instance?

You can of course, take an integer like u(A), and reduce it mod p, if you wish.

Of course, since you're in GF(p), the multiplicative group is cyclic, so one has a nice description of what the elements are with a given order, and how many of them there are etc. Try Le Veque for more information.

 

[praneeth]

Mr. Grime, So, the 1st equation is wrong. I don't know "Le Veque". If its new what should I do with it. Can you suggest anyone journal to write to about this? because every journal I saw won't accept new proofs to theorems as its articles.

Also I want to mention that
{∑(x^r)-μ(A)} mod p=0; for any r co-prime to A. This result might be useful in cryptography.