AUT_ABooklet_2020

Answer. The smallest possible integer with that property is $n=13$.

We note that we have $xyz\mid (x+y+z{)}^n$ if and only if for each prime $p$ the inequality $v_p(xyz)\le v_p((x+y+z{)}^n)$ holds, where as usual $v_p(m)$ denotes the exponent of $p$ in the prime factorization of $m$. Let $x$, $y$ and $z$ be positive integers with $x\mid y^3$, $y\mid z^3$ and $z\mid x^3$. Let $p$ be an arbitrary prime, and w.l.o.g. let the multiplicity of $p$ be lowest in $z$, that is, $v_p(z)=\min \{v_p(x),v_p(y),v_p(z)\}$. Then we have $v_p(x+y+z)\ge v_p(z)$, and from the divisibility constraints we get $v_p(x)\le 3v_p(y)\le 9v_p(z)$. It follows that $$\begin{array}{rl}{v}_p(xyz)& =v_p(x)+v_p(y)+v_p(z)\\ & \le 9v_p(z)+3v_p(z)+v_p(z)=13v_p(z)\\ & \le 13v_p(x+y+z)=v_p((x+y+z{)}^{13}),\end{array}$$ which proves that for $n=13$ the desired property is satisfied. It remains to show that this is indeed the smallest possible integer with this property. For doing so, let $n$ now be a number that has the desired property. By setting $(x,y,z)=(p^9,p^3,p^1)$ with an arbitrary prime $p$ (in order to achieve that both inequalities in the previous calculation become equalities), we get $$\begin{array}{rl}13& =v_p(p^{13})=v_p(p^9\cdot p^3\cdot p^1)=v_p(xyz)\\ & \le v_p((x+y+z{)}^n)=v_p((p^9+p^3+p^1{)}^n)=n\cdot v_p(p(p^8+p^2+1))=n,\end{array}$$ which yields $n\ge 13$.