Is the equation $x^3-y^3\equiv k \mod n$ always solveable, when neither $7$ nor $9$ divides $n$? Here
A Diophantine equation involving factorial
solutions of $x^3-y^3=z!$ are wanted. To rule out some $z$, I wanted to find modules $n$, such that $$x^3-y^3\equiv k\mod n$$ is not solveable for some $k$.
For $n=7$, we have no solution , if $k$ is $3$ or $4$. And if $n=9$, we have no solution if $k$ is $3,4,5$ or $6$.
But the verification of the numbers $n=2$ to $500$ showed that always a solution exists if neither $7$ nor $9$ divides $n$.

Is the following conjecture true ?
If $n>1$ and neither $7$ nor $9$ divides $n$, then the equation $$x^3-y^3\equiv k\mod n$$ is solveable for all $k$ ?

If the above conjecture is actually true does this imply that $x^3-y^3\equiv z!\mod n$ is always solveable for $z\ge 7$ and $n>1$ ?
 A: For $n=3$ the equation is clearly solvable for all $k$. To show that there is a solution for every $k$ whenever $7$ and $9$ do not divide $n$, by the Chinese remainder theorem it suffices to show that solutions exists for all prime powers $p^m$ with $p\neq3,7$. By Hensel's lemma it suffces to show that solutions exist for all primes $p\neq3,7$.
If $p\equiv2\pmod{3}$ then $3$ is coprime to $\varphi(p^m)=p^{m-1}(p-1)$, the order of the group $(\Bbb{Z}/p^m\Bbb{Z})^{\times}$. Hence every element of $\Bbb{Z}/p^m\Bbb{Z}$ that is not a multiple of $p$ is a cube, so the equation has solutions for every $k$.
If $p\equiv1\pmod{3}$ then the set of cubes $C$ is a subgroup of index $3$ in $(\Bbb{Z}/p\Bbb{Z})^{\times}$. 

Fact: If $p>7$ then there exists $c\in C$ with $c\neq\pm1$ and $c-1\notin C$.

The cosets of $C$ in $(\Bbb{Z}/p\Bbb{Z})^{\times}$ are then $C$, $(c-1)C$ and $(c-1)^2C$. For every $k\in C$ there is clearly a solution, and the fact that $c-1\in C+C$ shows that $(c-1)C\subset C+C$ so for every $k\in(c-1)C$ there is also a solution. It remains to show that $(c-1)^2C\subset C+C$.
Suppose toward a contradiction that $(c-1)^2C\not\subset C+C$. Then $(c-1)^2C\cap(C+C)=\varnothing$ and so
$$1+c\notin(c-1)^2C\qquad\text{ and }\qquad(c+1)(c-1)=c^2-1\notin(c-1)^2C.$$
By cancelling $c-1$ we get $c+1\notin(c-1)C$, so $c+1\in C$. 
Similarly $2=1+1\notin(c-1)^2C$, and $2\notin(c-1)C$ because otherwise
$$c+1=(c-1)+2\in(c-1)C+(c-1)C,$$
contradicting $C\cap((c-1)C+(c-1)C)=\varnothing$, hence also $2\in C$ and so $-2c\in C$.
Now $c^2+1\in C+C$ and hence $c^2+1\notin(c-1)^2C$ by assumption. If $c^2+1\in(c-1)C$ then
$$c^4-1=(c^2+1)(c+1)(c-1)\in(c-1)^2C,$$
but $c^4-1\in C+C$, a contradiction. Hence $c^2+1\in C$, but then
$$(c-1)^2=(c^2+1)+(-2c)\in C+C,$$
again contradicting $(c-1)^2C\cap(C+C)=\varnothing$. This shows that also $(c-1)^2C\subset C+C$ and therefore $C+C=\Bbb{Z}/p\Bbb{Z}$. So if $p\equiv1\pmod{3}$ and $p\neq7$ then there is also a solution for every $k$.
