Find all positive reals $x,y \in \mathbb{R}^+$ satisfying: $$\frac{x^9}{y} + \frac{y^9}{x} = 10-\frac{8}{xy}$$

Since this involves higher exponents I am unable to tackle this problem. Please help me.

  • $\begingroup$ I don't think this qualifies as a Diophantine equation. $\endgroup$ Jan 16, 2017 at 15:49
  • $\begingroup$ Oh! I have now known that a Diophantine has only integer solutions but it involves reals. $\endgroup$ Jan 16, 2017 at 15:50
  • $\begingroup$ Well, if you multiply it by $xy$ you will get a polynomial equation in $2$ variables of degree $10$. Generally there is no algorithm to solve such equations. But in that particular case? Who knows, doesn't look easy. $\endgroup$
    – freakish
    Jan 16, 2017 at 16:01
  • $\begingroup$ It looks cleaner as $x^{10}+y^{10}=10xy-8$ but I am not sure that is progress. $\endgroup$ Jan 16, 2017 at 16:02
  • $\begingroup$ How about you tell us some story about the equation? Where did you get it from? Perhaps there's some hint in it. $\endgroup$
    – freakish
    Jan 16, 2017 at 16:03

2 Answers 2


Hint: Write it as:


Then use AM-GM:


Which gives $5(xy-1)^2\le0$, so $xy=1$

Then initial equation can be rewritten as $x^{10}+\frac{1}{x^{10}}=2$

Which by AM-GM again, (or by writing it as a $(x^{10}-1)^2=0$) has the solution, $x^{10}=1$, so $x=\pm1$

So $x=1,y=1$, or $x=-1,y=-1$

  • $\begingroup$ Wow! I never thought of using the AM-GM. What a brain! Thank for the answer. $\endgroup$ Jan 16, 2017 at 16:19

Just to give a non-AM-GM answer, let $u=xy$, which must be positive in order for $x^{10}+y^{10}=10xy-8$ to have a solution, and note that $x^{10}+y^{10}=(x^5-y^5)^2+2(xy)^5$. Thus

$$\begin{align}x^{10}+y^{10}-10xy+8 &=(x^5-y^5)^2+2(u^5-5u+4)\\ &=(x^5-y^5)^2+2(u-1)(u^4+u^3+u^2+u-4)\\ &=(x^5-y^5)^2+2(u-1)^2(u^3+2u^2+3u+4)\\ &\ge0 \end{align}$$

with equality if and only if $x^5=y^5$ and $xy=u=1$ (since $u\gt0$ implies $u^3+2u^2+3u+4\gt0$). From this we see that $x=y=\pm1$ are the only possibilities. In particular, $(x,y)=(1,1)$ is the only solution with $x,y\in\mathbb{R}^+$.

Remark: I showed the factorization of $u^5-5u+4$ in two steps for ease of checking.


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