Find the Ф(28) = 12 primitive roots modulo 29...
Only had a very brief introduction to primitive roots and a bit lost with this problem.
Based on my limited understanding so far I have:
Ф(28) = 12
Ф(7) * Ф(4) = 12
Ф(7) * Ф(2) * Ф(2) = 12
Ф(7) * 1 * 1 = 12
Not sure if I am on the right track, any help is greatly appreciated.
A primitive root modulo $29$ is a generator of the cyclic group $C_{28}$. Hence if $a$ is a primitive root modulo $29$, then $a^k$ is also a primitive root modulo $29$ if and only if $gcd(k,\phi(29))=1$. Hence the primitive roots for $29$ are: $2$, $2^3 = 8$, $2^5 \equiv 3$, $2^9 \equiv 19$, $2^{11}\equiv 18$,$2^{13}\equiv 14$,$2^{15}\equiv 27$, $2^{17}\equiv 21$, $2^{19}\equiv 26$, $2^{23}\equiv 10$, $2^{25}\equiv 11$,and $2^{27}\equiv 15 \bmod 29$.