How to find $\#\{1\le x\le 5^k:5^k|(x^4-1)\}$? Find $\#\{1\le x\le 5^k:5^k|(x^4-1)\}$. I am not so sure how it is done, nor am I completely sure if it is about any specific $k$ or all of them together. What I did arrive at, not being really sure, is: 
If $k=0$, the answer is $1$, I guess, so suppose $k>1$.
I need to have $x^4\equiv 1 \pmod{5^k}$ satisfied. I know it is for $x\equiv 1 \pmod{5^k}$. Since $5^k=5^{k-1}\cdot 4+5^{k-1}=4\cdot5^{k-1}+4\cdot5^{k-2}+5^{k-3}=4(5^{k-1}+...+5^2+5)+5\equiv 1 \pmod4$, then by a theorem taught in class, $x\equiv -1$ is a primitive root modulo $5^k$. That is, $x=5^k-1\in \{1\le x\le 5^k:5^k|(x^4-1)\}$. So $\#\{1\le x\le 5^k:5^k|(x^4-1)\}\ge 2$. I am stuck at the moment. I don't know how to find more $x$ solutions such that $x^4\equiv 1 \pmod{5^k}$. How have my argues been so far? I could really use your guiding. 
 A: $G=\left(\mathbb{Z}_{/(5^k\mathbb{Z})}\right)^*$ is a cyclic group having order $o(G)=4\cdot 5^{k-1}$ and we are just looking for the number of $g\in G$ such that $o(g)$ is $1$,$2$ or $4$. Assuming that $G=\langle h \rangle$, such elements are $h^{o(G)},h^{o(G)/2},h^{o(G)/4}$ and $h^{3 o(G)/4}$, so their number is:
$$ \varphi(1)+\varphi(2)+\varphi(4) = \sum_{d\mid 4}\varphi(d) = 4$$
no matter what $k$ is.
A: First note that if $x$ works for $k+1$ then $x$ (reduced modulo $5^k$) also works for $k$, so every solution for $k+1$ comes from a solution for $k$. 
Now suppose we have a solution for $k$, that is, $a^4-1\equiv0\bmod{5^k}$. Let $b=a+5^kt$; then there is exactly one value of $t$ in $\{\,0,1,2,3,4\,\}$ such that $b$ is a solution for $k+1$. We see this as follows. 
$b^4-1=a^4+4a^35^kt+5^{2k}c-1$ for some $c$, so dividing $b^4-1\equiv0\bmod{5^{k+1}}$ through by $5^k$ we get $${a^4-1\over5^k}+4a^3t\equiv0\bmod5$$ Since $a\not\equiv0\bmod5$, this equation has a unique solution $t$, and we are done. 
All I have done here is I have gone through the standard proof of Hensel's Lemma for this particular congruence. 
