Quote:
Originally Posted by tetramur
<snip>
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).
<snip>

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 nonresidue (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).