IMO Team Selection Contest

For $n = 1$ one of the solutions is $x = y = z = 3$ and for $n = 3$ one of the solutions is $x = y = z = 1$.

If $n$ is even and there exists an integer solution, then the right-hand side of the equation is even. This is possible only if at least one of the numbers $x, y, z$ is even. Then the right-hand side is divisible by $4$. Since the remainders modulo $4$ of the squares of integers can only be $0$ or $1$, all numbers $x, y, z$ must be even. Let $x = 2a$, $y = 2b$, $z = 2c$. Then $a^2 + b^2 + c^2 = 2nabc$, hence $(a, b, c)$ satisfy a similar equation with doubled $n$, so they must be even. Continuing this process reveals that $x, y, z$ must be divisible by arbitrarily large powers of $2$ which is impossible. Consequently, there are no solutions for even $n$.

Suppose that for some odd $n > 3$ the equation has an integer solution $(x, y, z)$. The given equation is equivalent to $z^2 - nxy \cdot z + (x^2 + y^2) = 0$; let $z'$ be the other root of this quadratic equation. Then $z' > 0$ since positive $nxy$ and $x^2 + y^2$ enable only positive solutions of the quadratic equation. By Viéte's formulae, $z' = nxy - z$. On the other hand, assume without loss of generality that $z = \max(x, y, z)$; then $x^2 \le xz \le xyz$ and $y^2 \le yz \le xyz$, whence $z^2 \ge (n-2)xyz$ and $z \ge (n-2)xy$. Therefore $z' \le 2xy < (n-2)xy \le z$, implying $x + y + z' < x + y + z$. Thus we can infinitely reduce the sum of the components of the solution, which is impossible.