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.
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1$\begingroup$ There are obvious solutions involving $0$ and $1$. The interesting question is whether there are more and whether they are real $\endgroup$– HenryCommented May 18, 2023 at 10:51
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2$\begingroup$ $(x,y)=(1,0),(0,1)$ are the only real solutions. There are $72$ further complex solutions, parametrized by the roots of the (resultant) polynomial $2t^{72}+18y^{71}+\cdots +81t + 10$, which has no real roots. $\endgroup$– Dietrich BurdeCommented May 18, 2023 at 10:56
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1$\begingroup$ How can I prove that $(x,y) = (1,0),(0,1)$ are the only real solutions? $\endgroup$– David399Commented May 18, 2023 at 10:59
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1$\begingroup$ How about methods involving calculus? Also edit your question with your background, and the source of this question (textbook/PDF/Olympiad question with correct name and link). Having said that, your approach makes sense when the exponents are smaller (and will take time if they are bigger, as the answer below shows). $\endgroup$– Sarvesh Ravichandran IyerCommented May 18, 2023 at 11:02
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1$\begingroup$ Interesting looking graphs! $\endgroup$– PaulCommented May 18, 2023 at 11:04
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2 Answers
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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
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$\begingroup$ Nice. From (1) could you argue that since the two curves are convex (easy to show, may need calculus?) they can meet at two points at most. Since you have 2 points already you are done. $\endgroup$– PaulCommented May 18, 2023 at 11:25
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1$\begingroup$ Thank you for the solution, it's easier for me to understand this than doing a resultant. $\endgroup$– David399Commented May 18, 2023 at 11:27
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1$\begingroup$ @Paul, yes, you can do that. However, a quicker way without using calculus is to observe that $x^9(1-x) \ge 0$ (the equality occurs if and only if $x = 0$ or $x = 1$). $\endgroup$– NN2Commented May 18, 2023 at 11:28
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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 $(0,1)$.
answered May 18, 2023 at 11:04