It might be quite a simple one but I am confused here.
Let $y \in \mathbb{R}-\{ 0 \}$ and $n \in \mathbb{2N}$, even. Then prove that $x^n+y^n=(x+y)^n$ only when $x=0$.
I thought it is a good idea to implement Rolle's theorem here for $f(x)=x^n+y^n-(x+y)^n$ in the domain $[0,x_0]$. But then $0=f(0) \neq f(x_0)=x_0^n+y^n-(x_0+y)^n$.
How can I apply the theorem then? Also, if it can be applied how does it really prove the above statement?
If there were another root of $f$, then by Rolle's Theorem we have $f'(x)=0$ for some $x$. But \begin{align*} f'(x)=nx^{n-1}-n(x+y)^{n-1}=n(x^{n-1}-(x+y)^{n-1}), \end{align*} so $x^{n-1}-(x+y)^{n-1}=0$. We know that $n-1$ is odd, so $x=x+y$, then $y=0$, a contradiction.