$x+y$ divides $x^n+y^n$ exactly when $n$ is odd.Suggest a slogan
Let \(n\ge1\). If \(n\) is odd, then in \(\ZZ[x,y]\) \[ x^n+y^n=(x+y)\sum_{k=0}^{n-1}(-1)^kx^{n-1-k}y^k . \] If \(n\) is even, then \(x+y\) does not divide \(x^n+y^n\) in \(\ZZ[x,y]\).
For odd \(n\), \((-y)^n=-y^n\), so substituting \(a=-y\) in \(x^n-a^n=(x-a)\sum_{k=0}^{n-1}a^kx^{n-1-k}\) (FF-ED3CD) gives the factorization.
Since \(x+y\) is monic in \(x\) over \(\ZZ[y]\), division by \(x+y\) in \((\ZZ[y])[x]\) shows that \(x+y\) divides \(f\in\ZZ[x,y]\) if and only if \(f(-y,y)=0\). For even \(n\), \((-y)^n+y^n=2y^n\ne0\).
When does $x + y$ divide $x^n + y^n$, and what is the factorization?