Hệ phương trình tương đương với:
$\left\{\begin{array}{l}y=a+b-x\\x^4+(a+b-x)^4=a^4+b^4\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}y=a+b-x\\2x^4-4(a+b)x^3+6(a+b)^2x^2-4(a+b)^3x+(a+b)^4-a^4-b^4=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}y=a+b-x\\x^4-2(a+b)x^3+3(a+b)^2x^2-2(a+b)^3x+2a^3b+3a^2b^2+3ab^3=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}y=a+b-x\\(x-a)(x-b)(x^2-(a+b)x+2a^2+3ab+2b^2)=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=a\\y=b\end{array}\right.\\\left\{\begin{array}{l}x=b\\y=a\end{array}\right.\end{array}\right.$
Từ đó suy ra: $x^n+y^n=a^n+b^n,\forall n\in\mathbb{N}$.