On Tue, February 7, 2017 10:21 am, Surinder Kaur wrote:
> If F is a field with q = p^m -1 elements where m is even and p is an odd
> prime,
Since the existence of a field F with q elements implies that q is a prime power
and since for an odd prime p we have q = p^m - 1 even, we can conclude that
q is in fact a power of 2, and that F has characteristic 2.
> A the set of all quadratic residues and B = {4 b - 1 | b is
> primitive element of F}.
Since F has characteristic 2, we have B = {1}.
> Then my aim is to prove A intersection B is non
> empty i.e. for some primitive element b the element 4b-1 is a square of
> some element.
Just take any primitive element b. -- Then we have 4b-1 = 1 (since F has
characteristic 2), and 1 is the square of itself. qed.
-- Does this help you?
Best regards,
Stefan Kohl
P.S.: The GAP Forum is intended for questions on GAP.
-- General questions on mathematics are more on-topic on Math.SE.
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