The incongruent primitive roots modulo 17 are 3, 5, 6, 7, 10, 11, 12, and 14.
Determine the order of:
(a) $-3 \pmod {17}$
(b) $-5 \pmod {17}$
(c) $-6 \pmod {17}$
(d) $-7 \pmod {17}$
(e) $-10 \pmod {17}$
(f) $-11 \pmod {17}$
(g) $-12 \pmod {17}$
(h) $-14 \pmod {17}$
We note that the all the values are negations of known primitive roots. So we'll first prove a general lemma that.
Lemma. For prime $p$ where $\frac{p-1}{2}$ is even, if $a$ is a primitive root of $p$, then the order of $-a$ is $p-1$.
If $k$ is the order of $(-a)$, we have
$$ (-a)^k \equiv 1 \pmod p $$
We consider two cases for $k$, even and odd.
If $k$ is even, we have
$$ (-a)^k \equiv (a)^k \equiv 1 $$
Because $a$ is a primitive root, the smallest positive $k$ is $k=p-1$.
If $k$ is odd, we have
$$ ((-a)^k)^2 \equiv a^{2k}\equiv 1 \pmod p $$
Since $a$ is a primitive root, it must be that $p - 1 \mid 2k$, or equivalently $\frac{p-1}{2} \mid k$. But $\frac{p-1}{2}$ is even, and $k$ is odd. An even number cannot divide an odd number, and so we have a contradiction. This means $k$ even is not possible.
The only possibly case is $k$ even, and that leads to $k=p-1$, and so the order of $-a$ is $p-1$.
(a) Since 3 is a primitive root of prime 17, and since $\frac{17-1}{2}=8$ is even, then by the Lemma, the order of $-3$ is 16.
(b) Since 5 is a primitive root of prime 17, and since $\frac{17-1}{2}=8$ is even, then by the Lemma, the order of $-5$ is 16.
(c) Since 6 is a primitive root of prime 17, and since $\frac{17-1}{2}=8$ is even, then by the Lemma, the order of $-6$ is 16.
(d) Since 7 is a primitive root of prime 17, and since $\frac{17-1}{2}=8$ is even, then by the Lemma, the order of $-7$ is 16.
(e) Since 10 is a primitive root of prime 17, and since $\frac{17-1}{2}=8$ is even, then by the Lemma, the order of $-10$ is 16.
(f) Since 11 is a primitive root of prime 17, and since $\frac{17-1}{2}=8$ is even, then by the Lemma, the order of $-11$ is 16.
(g) Since 12 is a primitive root of prime 17, and since $\frac{17-1}{2}=8$ is even, then by the Lemma, the order of $-12$ is 16.
(h) Since 14 is a primitive root of prime 17, and since $\frac{17-1}{2}=8$ is even, then by the Lemma, the order of $-14$ is 16.