Show that there is no pair of coprime positive integers $(x,y)$ such that $(x+y)^3 \mid (x^n+y^n)$ 
Let $n$ be a square-free integer. Show that there is no pair of coprime positive integers $(x,y)$ such that $$(x+y)^3 \Big| (x^n+y^n)$$

The problem can be apparently solved by LTE, but I don't know what are the cases to break it up to.
If I first take $n$ to be odd and $p >2$, then I need to assume that $p \mid x-y$ and now by LTE we get that $$v_p((x+y)^3)=3v_p(x+y) \le v_p(x^n+y^n)=v_p(x+y)+v_p(n) = v(x+y)+1$$ as $v_p(n)=1$ since $n$ was square-free. This implies that $2v_p(x+y) \le 1 \iff v_p(x+y) \le \frac12$ which is a contradiction.
I think I need to consider some other cases now such as $n$ being even and $p=2$ at least? How can I approach this one?
 A: When $n$ is even, any odd prime $p$ dividing $x+y$ does not divide $x^n+y^n=x^n-y^n+2y^n$ because $p$ divides $x^n-y^n$ and $y$ and $p$ are coprime.
So remaining case is $y=2^k-x$ where $x$ is odd. When $n$ is odd, you can use LTE. When $n$ is even, you can show that $x^n+y^n$ can be divided by $2$ only once because $x,y$ are odd.
A: 
$(x+y)^3 | (x^n+y^n), (x, y)=1.$

\begin{align}
\text{Case 1. } \; & n=1: \\
&(x+y)^2|1, x+y=\pm1. \Rightarrow \text{cannot be positive.} \\
\ \\
\text{Case 2. } \; & n=2: \\
&(x+y)^3 | x^2+y^2. \\
&\text{We can easily find that $(x+y)^3$ will be absolutely bigger than $x^2+y^2$ when $x, y$ increases.} \\
& \therefore (x, y)=(1, 1). (\because (1+2)^3>1^2+2^2.) \\
& \text{But it doesn't satisfies.} \\
\ \\
\text{Case 3. } \; & n\geq 3: \\
\ \\
\text{(i) } \; & n=\text{odd}, \exists p \text{ s.t. }p|(x+y), p \neq 2(p: \text{prime}): \\
& \text{Using LTE: } \\
& \text{Contradiction, as you did.} \\
\ \\
\text{(ii) } \; & n=\text{odd}, \exists! p \text{ s.t. } p|(x+y), p:\text{prime}(\text{which is } 2): \\
& 3v_2(x+y) \leq v_2(x^n+y^n)=v_2(x+y). \\
& \therefore v_2(x+y)\leq0, \text{Which is contradiction.} \\
\ \\
\text{(iii) } \; & n=\text{even}: \\
&\text{let } p: \forall \text{prime number s.t. } p|x+y. \\
\ \\
& \text{if }p=2: \\
& x^n=(x^2)^k \equiv 0, 1, 5 (\mod 8)(\because k \geq 2.) \\
& \therefore \text{if } 8|x^n+y^n, x^n \equiv y^n \equiv 0 (\mod 8.) \Rightarrow \text{Contradiction.} \\
\ \\
& \text{if } p\neq2: \\
& x \equiv -y(\mod p) \Rightarrow x^n-y^n \equiv x^n-(-x)^n \equiv 0 (\mod p) (\because n: \text{even.}) \\
& \text{if } p|y, x \equiv 0 (\mod p), \text{Contradiction.} \\
&\therefore p \not|y. \\
& (p, x^n+y^n)=(p, x^n-y^n+2y^n) = (p, 2y^n) = 1, \text{Contradiction.} \\
\ \\
\ \\
\ \\
&\text{All of them are contradiction.} \\
&\therefore \not\exists x, y \text{ s.t. } (x, y)=1, (x+y)^3 | (x^n+y^n).
\end{align}
