Prove or disprove the system about $n$th power has only one solution $x=y=1$

Question

$$\begin{cases}x^n+y^n=2\\x+y=2\end{cases}\;,\;n\in\mathbb{N}\;,\;x,y\in\mathbb{R}\;,\;n>2$$

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.

Answer 1

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.

Answer 2

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.