# 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.

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What type of numbers are $x,y$? – Sandeep Silwal May 4 '14 at 1:38
$x,y$ are real numbers – cwk709394 May 4 '14 at 1:41
@recursiverecursion Um I think that's a little off topic. :) This is definitely not related to Fermat's Theorem, it has $x,y\in\mathbb{R}$. – NotNotLogical May 4 '14 at 1:56
@cwk709394 Can you show that the region is convex? Then the tangent line could not intersect it anywhere else. – NotNotLogical May 4 '14 at 2:00

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.

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Thanks so much. This is what I want to find a simple way to solve the question. – cwk709394 May 4 '14 at 2:30
You're welcome. Glad to have been of help. – Macavity May 4 '14 at 2:33

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.

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I'm sorry I didn't tell you, I also tried this method. Sorry! /_\ – cwk709394 May 4 '14 at 2:32