Finding all no-congruent primitive roots $\pmod{29}$

Question

Finding all no-congruent primitive roots $\pmod{29}$.

I have found that $2$ is a primitve root $\pmod{29}$ Then I found that is it 12 no-congruent roots, since $\varphi(\varphi(29)) = 12$ Then I found that:

$r_1=2^1=2\bmod (29)\\r_2=2^3=8\bmod (29)\\r_3=2^5=3\bmod (29)\\r_4=2^{11}=18\bmod (29)\\r_ 5=2^{13}=18\bmod (29)\\r_6=2^{17}=21\bmod (29)\\r_7=2^{19}=21\bmod (29)\\r_8=2^{23}=10\bmod (29)\\r_9=2^{27}=15\bmod (29)\\r_{10}=2^{29}=2\bmod (29)$

Is $10$ of these roots $12$ roots. Took the power of the primes from $1-29$ (not the primefactors of $\varphi,\ 2$ and $7$), but I am missing $2$ roots, and I don't understand how to find them. I have used all prime powers.

Answer 1

You should use all powers of $2$ that are relatively prime to $28$.

The two roots you are missing are $2^9$ and $2^{15}.$

$9$ and $15$ are not prime (they are multiples of $3$), but they share no factors with $28$.

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(I also note that you have wrong values for $2^{13}$ and $2^{19}\bmod29$;

they aren't the same as $2^{11}$ and $2^{17}$, respectively.)

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Also, you are missing $2^{25}$; you have $2^{29}$, which is the same as $2^{1}\bmod28$, instead.

Answer 2

Like my answer here Order of elements modulo p,

We can say $2^r$ is a primitive root of $29$ iff $$(r,\phi(29))=1$$

Now $$\phi(\phi(29))=\phi(7)\cdot\phi(4)=?$$