37th Iranian Mathematical Olympiad

Assume that $(x_1,x_2,\dots ,x_n)$ is a solution to the equation. Without loss of generality, let $x_n\ge x_{n-1}\ge \cdots \ge x_1$. Since $(x_1{x}_2\dots x_n{)}^2$ is divisible by $x_n^2$, we have $$x_n^2\mid x_1^3+x_2^3+\cdots +x_{n-1}^3⟹x_n^2\le x_1^3+x_2^3+\cdots +x_{n-1}^3.$$ On the other hand, $$(x_1{x}_2\dots x_n{)}^2=x_1^3+x_2^3+\cdots +x_n^3$$ So, $$n{x}_n^3\ge (x_1{x}_2\dots x_n{)}^2⟹x_n\ge \frac{(x_1{x}_2\dots x_{n-1}{)}^2}{n}.$$ Combining these two facts, we get $$\begin{array}{rl}{x}_1^3+x_2^3+\cdots +x_{n-1}^3& \ge \frac{(x_1{x}_2\dots x_{n-1}{)}^4}{n^2}\\ implies(n-1)x_{n-1}^3& \ge \frac{(x_1{x}_2\dots x_{n-1}{)}^4}{n^2}\\ implies{n}^2(n-1)& \ge x_1^4{x}_2^4\dots x_{n-2}^4{x}_{n-1}.\end{array}$$ Since for every index $i$, we have $x_i\ge 2$, we'll get $$n^2(n-1)\ge 2^{4n-7}.$$ Which is obviously wrong for $n\ge 2$ and we're done.