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2021-09-21, 13:34   #6
Dr Sardonicus

Feb 2017
Nowhere

2×2,687 Posts

Quote:
 Originally Posted by tetramur 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).
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 Fq 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 q2 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 Fq x Fq of two copies of Fq.

An example of the latter case where (2^p + 1)/3 is composite is p = 53, for which q = 107 == 9 (mod 14).