Quote:
Originally Posted by carpetpool
When working with finite fields F(q) of order q (q is an odd prime), F(q) has exactly q elements [0, 1, 2, 3......q3, q2, q1]. Addition, subtraction, multiplication, and division are performed on these elements. If e is an element in F(q), then e^(q1) = 1 by Fermat's Little Theorem.
For the construction of F(q^2), choose a quadratic non residue r mod q. Then let s be a symbol such that s^2 = r in the same sense as i is a symbol such that i^2 = 1.
Elements are of the form e = a*s+b in F(q^2) where a and b are reduced integers mod q. Addition, subtraction, multiplication, and division are defined in F(q^2). Like in F(q), there are exactly q^2 elements in F(q^2).
 Show that for any element e in F(q^2), e^(q^21) = 1.
 Show that there are exactly phi(q^21) primitive elements e such that q^21 is the smallest integer m such that e^m = 1.
Can someone please provide further information on this? Thank you.

first point follows from the factorization of q^21 as (q1)*(q+1) does it not ??
as to the second point I'm not sure.