Taiwan IMO Selection Camp, 2018

Suppose there exist coprime positive integers $x, y$ such that $x^n + y^n$ is a multiple of $(x+y)^3$. Let $s = x+y$, then $s > 2$.

(1) If $n$ is even, since $x^n + y^n = x^n + (s-x)^n = 2x^n \pmod s$. But at the same time, $x^n + y^n$ is a multiple of $(x+y)^3$, so $x^n + y^n = 0 \pmod s$, hence $2x^n = 0 \pmod s$. But since $x$ and $y$ are coprime, $x$ and $s$ are coprime, therefore we must have $2 = 0 \pmod s$, that is, $x = y = 1$, a contradiction!

(2) If $n$ is odd, since $x^n + y^n = x^n + (s-x)^n = C_2^n s^2 (-x)^{n-2} + C_1^n s(-x)^{n-1} \pmod{s^3}$. But at the same time, $x^n + y^n$ is a multiple of $(x+y)^3 = s^3$, so $x^n + y^n = 0 \pmod{s^3}$, hence $C_2^n s^2 (-x)^{n-2} + C_1^n s(-x)^{n-1} = 0 \pmod{s^3}$. This means there exists an integer $k$ such that
$$-\frac{1}{2}n(n-1)sx^{n-2} + nx^{n-1} = ks^2. \quad (1)$$
From this we know $s|nx^{n-1}$, and since $(x,s) = 1$, we get $s|nx^{n-2}$, hence $s^2|\frac{1}{2}n(n-1)sx^{n-2}$. Substituting back into Eq. (1), we obtain $s^2|nx^{n-1}$, and again since $(x,s) = 1$ we know $s^2|n$, but this contradicts the fact that $n$ has no square factor, a contradiction!