[GAP Forum] A problem simplified generalization

[Bulutoglu Dursun A Civ AFIT/ENC]

	Dear GAP Forum;
	I am trying to generalize the result I stated before as follows:

Let Z_p={0,1,2,...p-1} be the finite field with p elements and x
be a primitive element. Let r be some positive integer.
 Also let A={1,2,3,...,(p-1)/2} and 
B={x^r,x^(r+1),x^(r+2),...,x^(r+(p-3)/2)} be subsets of Z_p^* then A can
not be equal to B for p>7.

Using the equations Sum(A)=Sum(B) and Sum(A^3)=Sum(B^3) I was able to
show that x has to satisfy 11x^2+10x+11=0 in Z_p
Using the equation Sum(A^5)=Sum(B^5) with the above two equations I got
(Assuming that I did not make an algebra mistake) the equation:

 (3*2^12+1)+(3*2^12-4)x+(3*2^12+6)x^2+(3*2^12-4)x^3+(3*2^12+1)x^4=0
in Z_p. 

The proof should come from equalities of the form Sum(A^k)=Sum(B^k) as 
the values of Sum(A^k),Sum(B^k) for k=1,2,3...p-2 completely determine
the distribution of the elements of A and B respectively. (For k=2 and 4
I get Sum(A^k)=Sum(B^k)=0 which are not useful). 
I was wondering what your ideas are on this.
Thanks in advance, your ideas have been very helpful.

Dursun.



-----Original Message-----
From: forum-bounces at gap-system.org [mailto:forum-bounces at gap-system.org]
On Behalf Of Mowsey
Sent: Tuesday, January 11, 2005 6:53 PM
To: Bulutoglu Dursun A Civ AFIT/ENC; forum at gap-system.org
Subject: Re: [GAP Forum] A problem simplified

I looked over John D. Dixon's solution and Marco Costantini's
code, and noticed that the solution suggested p=7 should not work
while the code indicated it did. There was a tiny error in the
solution, but it can be easily fixed:

For A={1,2,3,...,(p-1)/2}, Sum(A) = ( (1 + (p-1)/2)*(p-1)/2 ) / 2
= (p^2-1)/8. Previously the result was reported as (p^2-1)/4. When
the sum is taken mod p, one gets Sum(A) = -1/8.

The argument continues then as Sum(B) = (x^( (p-3)/2 + 1) - 1) / (x-1)
= ( x^( (p-1)/2 ) - 1 ) / ( x - 1 ) and since x has order p-1, this
is = (-1 - 1)/(x-1) = -2/(x-1).

If A=B, then Sum(A)=Sum(B) and -1/8 = -2/(x-1), so x-1=16 and x=17.
Previosuly the conclusion had been x=9 which was a contradiction,
but no longer.

The method can be continued however to finish. We consider
Sum(A^3) = 1^3 + 2^3 + ... + ((p-1)/2)^3 = (p^2-1)^2/64 = 1/64 mod p.
Sum(B^3) = 1^3 + x^3 + x^6 + ... x^(((p-3)/2)^3)
= ( (x^3)^((p-1)/2) - 1 ) / (x^3 - 1) = ((-1)^3 - 1)/(x^3 - 1)
= -2/(x^3-1).
Equating Sum(A^3)=Sum(B^3), we get 1/64 = -2/(x^3-1) and
x^3 - 1 = -128 and x^3 = -127.

Hence if p is such that A = B, then x = 17 mod p and x^3 = -127 mod p.
This just says 17^3 = -127 mod p, so p divides 5040 = 2^4*3^2*5*7,
and p can only be 2, 3, 5, or 7.

> I reduced my problem to the following simpler problem:
> Let Z_p={0,1,2,...p-1} be the finite field with p elements and x
> be a primitive element.
> Also let A={1,2,3,...,(p-1)/2} and B={1,x,x^2,...,x^(p-3)/2} be
> subsets of Z_p^* then A can not be equal to B for p>7.
> I was wondering if you have any ideas how this could be solved.

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