My problem is really a conceptual one, rather than a specific one, but I'll provide an example question to illustrate where my difficulty lies. This is in an exercise set provided by my professor.
Let $f(x) = x^4 + 2x + 1$, and let $0<k$ be an integer. Find the number of elements $r$ in the following set: $\{0, 1, \ldots, 7^k - 1\}$ satisfying the congruence $f(r) \equiv 0$ mod $7^k$. Do not provide the solutions themselves, and instead provide the number of solutions.
I've used Hensel's lemma before to find specific integer solutions to a problem like this. In the problems I've done before, instead of working in something like mod $7^k$ where $k$ is some arbitrary unknown integer, I've worked with a concrete value (say, $7^2 = 49$). In specific cases like these, I can pretty comfortably use Hensel's lemma to find the integer solutions to the provided congruence.
My conceptual difficulty arises when generalizing to problems like these. In specific, "non-general" problems I've done, the initial step involves finding solutions to $f(x) \equiv 0$ mod $p^{k-1}$. But if $k$ is some arbitrary value, I'm not sure how we reliably find these solutions. I also don't understand how we can ascertain the number of elements that would satisfy the given congruence without actually figuring out what integers satisfy the congruence.
Though I want to emphasize my problem is conceptual and not particularly linked to the specific question I provided, I'm also wondering how in the given question, the $r$ we find to satisfy $f(r) \equiv 0$ mod $7^k$ is also the number of elements in the set that satisfy the congruence, which is also denoted $r$. This doesn't seem to be true in more specific cases of Hensel's lemma that I've done, so I'm wondering if this is something conceptual I'm not understanding or if the professor has just reused a variable here.
The tl;dr: Considering a congruence of the form $f(r) \equiv 0$ mod $n^k$, where $n$ and the polynomial $f$ are given, how can we find solutions using Hensel's lemma for some arbitrary $k > 0$? Additionally, how can we find the number of solutions to the congruence without actually calculating all solutions?