For prime $p \ge 5$ there exists an $n$ with $2 \le n \lt p -1$ with $[n]$ a primitive root of unity of $(\mathbb{Z}/{p^2}\mathbb{Z})^\times$.

Question

Let $p$ be a prime satisfying $p \ge 5$.

Is the following true?

There exists an integer $n$ satisfying

$\quad 2 \le n \lt p -1$
$\quad \text{The residue class } $[n]$ \text{ generates the multiplicative group } (\mathbb{Z}/{p^2}\mathbb{Z})^\times$
$\quad$(i.e. $[n]$ is a primitive root of unity)

If the statement is true there is a follow-up question,

Is there a prime number that can be chosen for $n$?

My work

I've been 'playing around' in number theory to the point that this is now an intuitive 'sure thing', but it can all be blown apart with a counter example. Since, if true, the answer might be involved, I added the reference request tag. I also added the conjecture tag, but I'll delete that if it becomes untenable from the feedback I get.

Answer 1

Okay I figured out the general case. I'll still leave up my other answer though.

Recall that $\mathbb{Z}/p^2\mathbb{Z}^\times \cong C_{p(p-1)} \cong C_p \times C_{p-1}$.

In particular each primitive root $\alpha$ mod $p$ has exactly one lift $\hat{\alpha}$ mod $p^2$ which is not primitive, and it corresponds to the one that lives in the $\{e\} \times C_{p-1}$ subgroup in the above isomorphism. We can see from this that if $\hat{\alpha}$ is primitive mod $p$ but not mod $p^2$ than its multiplicative inverse mod $p^2$ (which is $\hat{\alpha}^{p-2}$ in this case) is also primitive mod $p$ but not mod $p^2$.

Okay now suppose $\alpha < p$ is a primitive root mod $p$ but not $p^2$. Consider the unique number $\beta < p$ such that $\alpha \beta \equiv 1$ mod $p$. I claim that $\beta$ must be a primitive root mod $p^2$. Suppose not, then $\beta$ must be the inverse of $\alpha$ mod $p^2$ since there is a unique non-primitive element congruent to $\beta$ mod $p$, and we know the inverse of $\alpha$ is one. However since $\alpha < p $ and $\beta < p$ we have that $\alpha \beta < p^2$, so they can't possibly be inverses.

Answer 2

Here's a proof for when $p \equiv 1 \ (\text{mod } 4)$:

First note that if $p \equiv 1 \mod 4$ then $\alpha$ is a primitive root mod $p$ iff $-\alpha$ is. Suppose $(-\alpha)^b \equiv 1$ for some $b < p-1$. If $b$ were even then we'd have $\alpha^b \equiv 1$, which is a contradiction as $\alpha$ is primitive. If $b$ were odd then $\alpha^b \equiv -1$, which only happens when $b = \frac{p-1}{2}$ but thats not odd since $p \equiv 1 \ (\text{mod } 4)$.

Okay so now lets look mod $p^2$. I claim that if $\alpha < p$ is a primitive root mod $p$ then at least one of $\alpha$ or $p-\alpha$ is primitive mod $p^2$.

Since $\alpha$ and $p-\alpha$ are primitive mod $p$, then mod $p^2$ they are either primitive or they have order $p-1$. Suppose we have that both $\alpha^{p-1}$ and $(p-\alpha)^{p-1}$ are congruent to $1$ mod $p^2$. Expanding this with the binomial theorem we get:

$$1 \equiv (p-\alpha)^{p-1} \equiv -\binom{p-1}{1}pa + \alpha^{p-1} \equiv -(p-1)p\alpha +1$$

Which means $(p-1)p\alpha$ is divisible by $p^2$, but that is a contradiction since $p$ is prime and $\alpha < p$.