Prove or disprove the system about $n$th power has only one solution $x=y=1$ $$\begin{cases}x^n+y^n=2\\x+y=2\end{cases}\;,\;n\in\mathbb{N}\;,\;x,y\in\mathbb{R}\;,\;n>2$$
I have tried to show that $\displaystyle y'=-\frac{x^{n-1}}{y^{n-1}}=-1$
$$......$$
therefore $x=y=1$ is the only one solution.
Is there any method(s) to prove the question ?
It looks like Fermat's Last Theorem.
 A: Here is one way: Make the change of variables $x= 1+a, y=1+b$.  Then the second equation gives $a = -b$.  Thus it remains to show that the following equation has only one solution $a=0$:
$$(1+a)^n + (1-a)^n -2 = 0$$
The LHS is a polynomial with all coefficients non-negative.  Hence it cannot have a positive root.  As the LHS is even, similarly there can be no negative root.  Hence $a=0$ is the only solution.
A: I'm not sure if this is rigorous, but we can maximize the function $$f(x,y)=x+y$$ over the constraint $x^n+y^n=2$. Lagrange Multipliers gives $$(1,1)=\lambda (x^{n-1},y^{n-1})$$ Therefore we have $x=\pm y$ depending on the parity of $n$. Plugging into the equation we have $$2x^n=2\implies x,y=\pm1$$ which gives the values $(-2,0,2)$. Hence $2$ is the maximum of $x+y$ over that family of curves. 
There cannot be another local maximum, so the only possibility is that $x+y$ increases without bound over the constraint. But it is not hard to see that this is not the case.
Comments about the validity of this argument would be appreciated. Specifically, I am not sure if it is valid to apply Lagrange over this non-compact region $x^n+y^n=2$ when $n$ is odd.
