Here's another way to see it:

Consider `x^n+y^n` as a polynomial in the variable `x.` If `x^n+y^n` has a linear factor, then it also has a zero. Conversely, any zero corresponds to a linear factor. But if `n` is even, then `x^n+y^n` is always positive for all `x` (it can only be zero when both `x` and `y` are zero, but any factorization must apply to all choices of `y,` not just the special case `y=0)` , so it can't have a real zero, and thus can't have a linear factor.

If `n` is odd, then we see that if `x=-y,` then `x^n+y^n=(-y)^n+y^n=0,` so the polynomial `x^n+y^n`must have the linear factor `x+y.` The full factorization can be found by polynomial long division.` `

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