We know that $2^{28} \equiv 1 \bmod 29$, i.e. $16^7 \equiv 1 \bmod 29$. However we can also conclude from this that:
$16^{14},16^{21},16^{28},16^{35},16^{42},16^{49} \equiv 1 \bmod 29$.
So $16^2, 16^3, 16^4, 16^5, 16^6, 16^7$ are the other $6$ solutions.
Alternatively: we know every $x\in\mathbb{Z}/29\mathbb{Z}$ can be written as $x=2^k$ for some $k$.
To satisfy $x^7 \equiv 1 \bmod 29$ we require $2^{7k} \equiv 1 \bmod 29$. But since $2$ is a primitive root mod $29$ this means that $7k \equiv 0 \bmod 28$, i.e. $k \equiv 0 \bmod 4$ so we may take $k=0,4,8,12,16,20,24$ (after this the powers of $2$ repeat).
Let $m\geq 3$ and $g$ is a primitive root modulo $m$. Then $\mathbb Z^*_m$ can be fully represented by $\{g_0,\dots,g^{\varphi(m)-1}\}$. Hence, any $x\in \mathbb Z^*_m$ can be written as $x=g^k$ for some $0\leq k<\varphi (m)$ and working in $\mathbb Z_m$:
$$x^2=1 \Leftrightarrow (g^k)^2= 1 \Leftrightarrow g^{2k}= 1 \Leftrightarrow \varphi(m) \mid 2k$$
which is only possible for at most two values of $k$, namely $0$ and $\frac{\varphi(m)}{2}$. As $\pm 1$ are always solutions and distinct, those are exactly the two solutions.
The other direction is most likely more difficult, I don't have a proof yet, but you can of course deduce this from Euler's theorem:
If there is no primitive root modulo $m$, then by Euler's theorem $m$ can not be of the form
$$2, 4, p^k, 2p^k$$
where $p$ is an odd prime and $k$ is a positive integer. Check carefully, that this implies that $m$ is either a power of two or can be written as product of two coprime numbers which are both at least $3$. In both cases we find more than two solutions:
- $m=2^k$. Define $x=2^{k-1}-1$. Then $x\not\equiv \pm 1\pmod{2^k}$, but
$$x^2=(2^{k-1}-1)^2=2^k-2\cdot 2^{k-1} +1 \equiv 1\pmod{2^k}$$
- $m=ab, (a,b)=1, a,b\geq 3$: Then $x^2\equiv 1\pmod a$ and $x^2\equiv 1\pmod b$ have both at least two distinct solutions ($\pm 1$) and the Chinese remainder theorem tells that $x^2\equiv 1\pmod{ab}$ has at least four solutions.
Best Answer
Here is an indirect solution - a solution that doesn't use any specific primitive root. I think a better solution exists, but I would still like to post it because it gives a different perspective.
The solutions $1$ and $m-1$ are obviously always solutions. So the difficulty is proving there aren't any more.
By the primitive root theorem, the possibilities for $m$ are $m=2$, $m=4$, $m=p^k$, or $m=2p^k$ where $p$ is an odd prime and $k\ge 1$.
For $m=2$ we actually have $1=m-1$, and there is a unique square root of $1$.
For $m=4$ you may check directly that the claim holds, just by calculating.
For $m=p^k$, we prove by induction on $k$. For $k=1$ there are no zero divisors, so $(x-1)(x+1)=0$ implies $x-1=0$ or $x+1=0$. For $k>1$, assuming there are exactly two solutions modulo $p^{k-1}$, we use Hensel's lemma to claim that these lift to exactly two solutions modulo $p^k$.
For $m=2p^k$ we use the Chinese remainder theorem to combine the two solutions modulo $p^k$ with the unique solution modulo $2$ to get two solutions modulo $m$.