Let $x, y\in \mathbb{R}$, solve the following system: $\begin{cases}x^9+y^9=1\\x^{10}+y^{10}=1 \end{cases}$

Question

Let $x, y\in \mathbb{R}$, solve the following system: $$\begin{cases}x^9+y^9=1\\x^{10}+y^{10}=1 \end{cases}$$ I have tried to use notations, $S = x+y$ and $P=xy$, but I couldn't continue because of the exponent. I have tried adding the two equations as well as substracting them but no success. Can someone help me with a small hint so I can solve this system successfully? I have found this problem in my 9th grade textbook.

Answer 1

Here is a solution for 9th grade level.

From the second equation, we deduce that $-1\le x,y \le 1$.

If $x$ or $y$ is negative, suppose for example $-1 \le x <0$, then from the first equation we have $$y^9 = 1 - x^9 > 1 \Longrightarrow y >1 \hspace{1cm} \text{ contractory to the fact that } -1\le x,y \le 1$$

Then, we must have $$0 \le x,y \le 1\tag{1}$$ From $(1)$, we deduce that $$x^9 \ge x^{10}$$ $$y^9 \ge y^{10}$$ then from the system of equation $$1 = x^9+ y^9 \ge x^{10} + y^{10} = 1$$

The equality occurs if and only if $(x,y) = (0,1)$ or $(x,y) = (1,0)$. They are then the two solutions of the equation system.

Q.E.D

Answer 2

Let $(x,y)$ be a solution of the system.

Since $x^{10} + y^{10} = 1$, then $x^{10} = 1-y^{10} \leq 1$, so $x \in [-1,1]$. Similarly, one has $y \in [-1,1]$.

Since $x^9 + y^9=1$, then $y = (1-x^9)^{1/9}$. The equation $x^{10} + y^{10} = 1$ then rewrites as $$x^{10} + (1-x^9)^{10/9} = 1$$

Let $f : [-1,1] \rightarrow \mathbb{R}$ defined by $f(x)=x^{10} + (1-x^9)^{10/9}$. By direct computation, one has $$f'(x)=10x^8(x-(1-x^9)^{1/9}) = 10x^8 g(x)$$

where $g(x)=x-(1-x^9)^{1/9}$. But $g'(x)=x^8(1-x^9)^{-8/9}+1 > 0$, so $g$ is increasing. Since $g(-1)=-1-2^{1/9} < 0$, and $g(1)=1>0$, then there exists $a \in (-1,1)$ such that $g(x) \geq 0 \Longleftrightarrow x \geq a$, i.e. such that $f'(x) \geq 0 \Longleftrightarrow x \geq a$.

So $f$ is decreasing over $[-1,a]$ and increasing over $[a,1]$. We deduce that the equation $f(x)=1$ has at most two solutions, and since $x=0$ and $x=1$ are solutions, they are the only ones.

Finally, the system has two real solutions which are $(1,0)$ and $(0,1)$.