Question is to prove that
For any prime $p$ and any positive integer $m$ not divisible by $p$, there exists a primitive $m$-th root of unity if and only if $m$ divides $p-1$.
Let $\zeta=a_0+a_1p+a_2p^2+\cdots$ be a primitive $m$ th root of unity. We then have
$$\sqrt[m]{1}=a_0+a_1p+a_2p^2+\cdots$$
Considering m th power and going modulo $p$ we see that $a_0^m\equiv 1\mod p$.
As $a^{p-1}\equiv 1\mod p$ for all non zero $a$, we can find $t<p-1$ such that
$a_0^t\equiv 1\mod p$.
Suppose $(t,p-1)=1$ then we have $a_0\equiv 1\mod p$.
I have two questions to solve.
Does $a_0\equiv 1\mod p$ imply $\zeta=1$?
Does $a_0^t\equiv 1\mod p$ imply $\zeta^t=1$?
Positive answer for second question implies positive answer for first question.
If it is the case then i am done as i have $\zeta^t=1$, contradicting the condition that $\zeta$ is a primitive $m$ th root of unity.
The condition $a_0^t\equiv 1 \mod p$ with Hensel's lemma implies $\alpha^t=1$ for some $\alpha=a_0+b_1p+b_2p^2+\cdots$. This does not immediately say that $\zeta^t=1$.
Any hints are welcome.
In some sense, i am trying to prove/disprove that
Given $f(X)=X^n+c_1X^{n-1}+\cdots+c_n$ and $\alpha=a_0+a_1p+a_2p^2+\cdots$
and $\beta=a_0+b_1p+b_2p^2+\cdots$ we have $f(\alpha)=0$ iff $f(\beta)=0$
Best Answer
First, recall Hensel's lemma:
Theorem: Let $a\in\Bbb Z_p$. If $f(X)\in\Bbb Z_p[X]$, $f(a) = 0\pmod{p}$ and $f'(a)\not\equiv 0\pmod{p}$, then there exists a unique $\alpha\in\Bbb Z_p$ such that $f(\alpha) = 0$ and $\alpha\equiv a\pmod{p}$.
This might not be quite the version you're used to, but it's essentially the same, except we're starting with a $p$-adic number and polynomial with $p$-adic coefficients. Existence and uniqueness will both be important for us in the following.
Remark: This version of Hensel's lemma answers both of your questions in the affirmative. If you have $\zeta = a_0 + a_1 p + \dots$ with $\zeta^m = 1$ in $\Bbb Z_p$ ($p\not\mid m$), and $a_0 = 1$, then $\zeta = 1$, since $1$ is already an element of $\Bbb Z_p$ satisfying $X^m - 1 = 0$, and thus uniqueness implies $\zeta = 1$. The answer to your second question is again yes by uniqueness, because $\zeta^t$ still satisfies $X^m - 1 = 0$ (and so does $1$) and $\zeta^t\equiv 1\pmod{p}$.
Here is the way I'd probably make the argument.
Lemma: There exists a primitive $p-1$st root of unity $\zeta\in\Bbb Z_p^\times$.
Note that this proves the existence portion of your question, because if $m>0$ divides $p - 1$, then an appropriate power of $\zeta$ will be a primitive $m$th root of unity.
Proof: Consider the polynomial $X^{p - 1} - 1\in\Bbb Z_p[X]$, and the elements $1,2,\dots, p - 1\in\Bbb Z_p$. Each of these elements satisfies $a^{p - 1} - 1\equiv 0\pmod{p}$. Moreover, $-a^{p - 2}\not\equiv 0\pmod{p}$, so for each of the elements we considered we obtain an $\alpha\equiv a\pmod{p}$ with $\alpha^{p - 1} - 1 = 0$. So we have found $p - 1$ $p - 1$st roots of unity. To see that one is primitive, recall that a finite subgroup of the multiplicative group of a field is cyclic, and the elements in $\Bbb Q_p^\times$ satisfying $X^{p - 1} - 1 = 0$ form a subgroup. Those elements are precisely the $p - 1$ $\alpha$'s we found, because a degree $n$ polynomial over a field can have at most $n$ roots.
Now, we must show that if $m$ does not divide $p - 1$, there is not a primitive $m$th root of unity.
Lemma: Let $m > 0$ be a positive integer such that $p\not\mid m$. Suppose that $\alpha\in\Bbb Z_p^\times$ satisfies $\alpha^m = 1$. Then $\alpha$ satisfies $\alpha^n = 1$ for some $n\mid p - 1$.
Proof: (adapted from the notes in my above comment) Suppose $\zeta\in\Bbb Z_p^\times$ satisfies $\zeta^m = 1$. Then $\zeta\not\equiv 0\pmod{p}$, so there exists some $p - 1$st root of unity $\alpha$ with $\alpha\equiv\zeta\pmod{p}$ by our argument for the last lemma. Then $\alpha$ and $\zeta$ both satisfy $f(X) = X^{m(p - 1)} - 1 = 0$. (Note that $f'(\alpha),f'(\zeta)\not\equiv 0\pmod{p}$.) But now we have two elements of $\Bbb Z_p$ satisfying the same polynomial $f\in\Bbb Z_p[X]$ which are equivalent modulo $p$, so by uniqueness in Hensel's lemma, $\zeta = \alpha$. Thus, $\zeta$ is a $p-1$st root of unity, so it is a primitive $n$th root only for some $n$ dividing $p - 1$.
Note that we did not need to assume $\alpha$ is a primitive $m$th root for the last argument. The lemma holds for any $m$th root, and will thus hold for a primitive $m$th root.