Quote:
Originally Posted by tetramur
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So, I think it is almost proven, but there is one issue.
Conjecture 1. Let p be prime p > 3, q be the smallest divisor of Wp = (2^p+1)/3 and both a, b be rationals mod q, then the order of the element w = 3+√-7/4 in the field X of {a+b √-7} is equal to 2^(2*p).
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There is no such thing as "the rationals mod q." The rational number 1/q is not defined "mod q."
The residue ring Z/qZ of the
integers mod q
is a field, the finite field F
q of q elements. In the field K = Q(sqrt(-7)), the ring of algebraic integers is R = Z[(1 + sqrt(-7))/2].
If -7 is a quadratic non-residue (mod q) [that is, if q == 3, 5, or 13 (mod 14)], then the residue ring R/qR is the finite field of q
2 elements.
If -7 is a quadratic residue (mod q) [that is, if q == 1, 9, or 11 (mod 14)] the residue ring R/qR is
not a field, but is the direct product F
q x F
q of two copies of F
q.
An example of the latter case where (2^p + 1)/3 is composite is p = 53, for which q = 107 == 9 (mod 14).