Let $p$ be a prime number and $k\geq 1$ be an integer. I am just wondering if there is any reference stating that (or reference that proves) if $gcd(a,p^k)=1$ then no integer in the set $S=\{p^0,p^1,\ldots,p^{k-1}\}$ solves the congruence
$$ax\equiv 1(mod\ p^k).$$
Note that the solution to the above congruence is given by $x\equiv a^{-1}(mod\ p^k)$ and there are many known algorithms that solves this congruence. In order to avoid those algorithms, I am trying to look for other possible ways to prove the claim above.
What I had proved so far is for the case where $x=a$. Meaning the statement above holds true for the congruence equation
$$x^2\equiv 1(mod\ p^k)$$.
Thank you in advance for your help.