Congruence equation - Find all possible $k$ Let $x,\ y,\ z > 1 \in \mathbb{Z_+}$ and $\gcd(x,y,z)=1$.
Find all possible $k$ such :
$$ \sum_{\mathrm{cyc}}{x}\equiv\sum_{\mathrm{cyc}}{x^2y}\equiv\sum_{\mathrm{cyc}}{xy^2}\equiv 3xyz\equiv 0\pmod{k} $$
PS: It's a part of bigger task (I reduce it by myself).
 A: First, if $p$ is a prime not equal to 3 that divides $k$, then we have $xyz \equiv 0 \pmod{p}$, so WLOG $x \equiv 0 \pmod{p}$. Then, since $ x + y + z \equiv 0 \pmod{p}$, hence $y = - z \pmod{p}$. Substituting this into the other equation, we get that $ y^3 \equiv 0 \pmod{p}$, so $y \equiv 0 \pmod{p}$. But this gives $p \mid \gcd(x,y,z)$.
Hence, $ k = 3^n$ for some $n$. Consider $n \geq 2$. From $3xyz \equiv 0 \pmod{3^n}$, we know that one of the terms must be a multiple of 3. If 2 terms are a multiple of 3, then from $x+y+z \equiv 0 \pmod{3^n}$, hence all 3 terms are, contradicting $\gcd(x,y,z)=1$. WLOG, let $x$ be the term that is a multiple of $3^{n-1}$. If $ x \equiv 0 \pmod{3^n}$, then we likewise have $ y \equiv -z \pmod{3^n} \Rightarrow y \equiv 0 \pmod{3^n}$, so there are no solutions.
WLOG, we may assume that $ x \equiv 3^{n-1} \pmod{3}$ (otherwise, multiply each term by 2). From $x + y + z \equiv 0 \pmod{3^n}$, we can set $ y \equiv 3^{n-1} +k, z \equiv 3^{n-1} -k \pmod{3^n}$. From the other equality, we get that $k^3 \equiv 0 \pmod{3^n}$, which implies that $k$ is a multiple of 3. This contradicts $\gcd(x,y,z)=1$.
Hence, we must have $k=3$ as the only possibility. A quick check shows that this works, e.g. $(x,y,z)= (4,7,10).$
