Further to the post here, I'm trying to find the $n \in \mathbb{Z}$ such that there is a solution to the equation
$$ x^3 +3x+y^3+3y=n$$
in $\mathbb{Z}_3$.
Now, I've been able to show that in the case that $3^2 | n$, then there is a solution by using the version of Hensel's Lemma from the link above:
- If we let $f(x)=x^3+3x -n$, then $|f(3)|_3=|9|_3|3+1-n/9|_3 \leq 3^{-2}$
- $f'(x)=3x^2+3$, so that $|f'(x)|_3=|3|_3|x^2+1|_3=3^{-1} \in (0,1)$ always
and
- $\dfrac{1}{2}f''(x)=3x$, so that for all $b \in \mathbb{Z}_3$, we have $|\dfrac{1}{2}f''(b)|_3 <1$.
In the case we don't have equality in 1., then we can use the usual Hensel's Lemma to find a solution. In the case that we have equality, by noting that 1., 2., and 3. hold and using the version of Hensel's Lemma in the link, we have that there is a solution.
Now things seem to get a bit trickier in the case that $n$ is not divisible by 9. Certainly if a solution for some $n$ exists with both $x,y \in 3\mathbb{Z}_3$, then necessarily $3^2|n$ and there is a solution in this case as we have seen.
If $3$ divides $n$, but $9$ does not, then writing out the $3$-adic expansion of a possible solution $x$ and $y$, putting them in and reducing modulo $3$ shows that $|x|_3=|y|_3=1$, and that if $a_0$ is the first term in the expansion of $x$, then $-a_0$ is the first term in the expansion of $y$.
Now I've been trying to use this to analyse the case when $3|n$, but it isn't working out for me. I suspect there may be no solution, but I am not sure. Any tips?
Note that if you change variables $x=u-v$, $y=v$, then you obtain $$(u-v)^3+3(u-v)+v^3+3v=n$$ which yields $$u^3-3u^2v+3uv^2+3u=n.$$ If there is a solution with $x,y\in\mathbb{Z}_3$ to the original equation, then there is a solution with $u=x+y$, and $v=y\in\mathbb{Z}_3$. Now, if $3|n$, then $3|u^3$ and therefore $3|u$. However, the equation can also be rewritten as $$u(u^2-3uv+3v^2+3)=n,$$ and so if $3|u$, then $9|n$. So this reduces the case of $n\equiv 0\bmod 3$ to the case when $n\equiv 0\bmod 9$.