[GAP Forum] A problem simplified

[Mowsey]

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.