I found this problem in a number theory course, I am assuming (but not sure) it is supposed to be an application of Hensel's lemma.
For every $n \in \mathbb{N}_0$, determine the number of solutions over $\mathbb{Z}/5^n\mathbb{Z}$ of the equation $y^2 = x^3 + x^2 + 5$.
In the base case $n = 1$, there are four solutions $(0, 0), (3, 1), (3, 4), (4, 0)$. The problem with Hensel's lemma in the multivariate case seems to be, if you lift your root to a higher prime power, you can no longer be sure this new root is unique. So instead I figured for every root, I'd fix each variable separately, and consider the univariate problem in the other respective variable. From this I derived the following results:
- $(0, 0)$ doesn't have higher power solutions by fixing either variable.
- $(3, 1)$ has a unique higher power solution by fixing either variable.
- $(3, 4)$ has a unique higher power solution by fixing either variable.
- $(4, 0)$ has a unique higher power solution by fixing the second variable.
However I have no clue how to further deduce the amount of solutions apart from these implicit ones. Just to get an idea of where I was headed I computed all solutions in the case $n = 2$. This way I did notice there are patterns in the solutions:
- $(0, 0)$ mod $5$, there are no corresponding solutions mod $25$.
- $(3, 1)$ mod $5$, the solutions are $(3 + 5k, 21 – 5k)$ mod $25$ for each $k$.
- $(3, 4)$ mod $5$, the solutions are $(3 + 5k, 4 + 5k)$ mod $25$ for each $k$.
- $(4, 0)$ mod $5$, the solutions are $(19, 5k)$ mod $25$ for each $k$.
I can sense there must be a connection between the results I got in the univariate case, and the results I have to get in the multivariate case. But I can't seem to figure out how to formally argue, based on the results of Hensel's lemma, why those patterns in the solutions of $n = 2$ turned out the way they are. Neither do I have a clue how to generalize my results towards an arbitrary power of $5$.
Best Answer
COMMENT.- In general, every solution $x_0$ of $f (x) = 0 \mod p$, except when $\color{red}{f '(x_0) \ne 0 \mod p}$, gives a solution of $f (x) = 0 \mod p ^ n$solving successively the equation modulo $p^2,p^3,\cdots p^n$.
Because of $f (x_0) = 0\mod p ^ n\Rightarrow f (x_0) = 0 \mod p$ putting $x=x_0+px_1$ we can solve $$f(x_0+px_1)=0\mod p^2$$ getting after simplifications$$f(x_0)+px_1f'(x_0)=0\mod p^2$$ which implies $$f(x_0)+px_1f'(x_0)=0\mod p$$ so we get $$x_1=\frac{-f(x_0)}{pf'(x_0)}\mod p$$ For example the given solution $(3,1)$ modulo $5$ of $y^2=x^3+x^2+5$ gives the univariable equation $$1=x^3+x^2+5$$ which is the equation $$f(x)=x^3+x^2+4=0$$ and with the exposed methode one get $x_1=\dfrac{-f(3)}{5f'(3)}=\dfrac{-(27+9+4)}{5(3\cdot3^2+2\cdot3)}=-\dfrac 33=4$.
Thus $(3,4)$ is a solution of $y^2=x^3+x^2+5$ modulo $25$.
In the proposed equation we have to go first to $3$ distinct equations because if $a$ is a square modulo $25$ we must have $f(x)=x^3+x^2+5-a=0\mod 5$ and because of $x^3+x^2\in\{1,0,2\}$ one has $$\begin{cases}f_1(x)=x^3+x^2+5=0\\f_2(x)=x^3+x^2+4=0\\f_3(x)=x^3+x^2+3=0\end{cases}$$ Some work gives then the following solutions modulo 25: $(3,\pm4),(8,\pm9),(13,\pm11),(18,\pm6),(19,0),(23,\pm1)$.