> From [EMAIL PROTECTED] Tue Mar 13 16:06:33 2001
> I wonder if there is possible to find primitive roots in GF(q^2), say a
> + ib, with the condition
>
> a^2 + b^2 = 1 mod q
>
> If this were possible, then
>
> (a + ib)*(a - ib) = 1 + 0i (mod q)
>
> i, e. the inverse of the root would be the conjugate, and the
> transformed of a real integer signal should have the same symmetries
> than in trigonometric complex case. Could it resolve the problem for a
> non-power-of-two FFT length ?. If the answer is yes: any trick to find
> the root other than brute force?
>
> Regards.
>
> Guillermo.
No. Given alpha = a + ib in GF(q^2), where q == 3 (mod 4), observe that
alpha^q = (a + ib)^q = a^q + (ib)^q = a^q + i^q b^q = a - ib
is the conjugate of alpha. Therefore
alpha^(q+1) = alpha * alpha^q = (a + ib)*(a - ib) = a^2 + b^2.
If alpha is a primitive root in GF(q^2) (order q^2 - 1),
then alpha^(q + 1) is a primitive root in GF(q) (order q - 1).
In particular, a^2 + b^2 <> 1.
(I deleted names other than [EMAIL PROTECTED] from the To: and Cc:
lines, so we don't receive duplicates).
Peter
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