Peter s,
The following "translation" of your primitive root link may be easier to follow: Let p be a prime number and let b be any number greater than 1 but less than p. Divide each of the following p powers of b by p: 1 (=b^0), b, b^2, b^3, ..., b^(p-1). If there are p-1 distinct non-zero remainders, then b is a primitive root of b. Otherwise, it is not a primitive root of p.
For p=7 and b=3, the 7 powers of 3 are 1, 3, 9, 27, 81, 243, 729 and the corresponding remainders are 1, 3, 2, 6, 4, 5 ,1. Since 6 (=7-1) of these remainders are distinct, then 3 is a primitive root of 7. However, if b=2, the 7 powers of 2 are 1, 2, 4, 8, 16, 32, 64 and the corresponding remainders are 1, 2, 4, 1, 2, 4, 1. Since there are only 3 distinct remainders, then 2 is not a primitive root of 7.
John
The following "translation" of your primitive root link may be easier to follow: Let p be a prime number and let b be any number greater than 1 but less than p. Divide each of the following p powers of b by p: 1 (=b^0), b, b^2, b^3, ..., b^(p-1). If there are p-1 distinct non-zero remainders, then b is a primitive root of b. Otherwise, it is not a primitive root of p.
For p=7 and b=3, the 7 powers of 3 are 1, 3, 9, 27, 81, 243, 729 and the corresponding remainders are 1, 3, 2, 6, 4, 5 ,1. Since 6 (=7-1) of these remainders are distinct, then 3 is a primitive root of 7. However, if b=2, the 7 powers of 2 are 1, 2, 4, 8, 16, 32, 64 and the corresponding remainders are 1, 2, 4, 1, 2, 4, 1. Since there are only 3 distinct remainders, then 2 is not a primitive root of 7.
John