6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I don't think this qualifies as a Diophantine equation. $\endgroup$ – Shraddheya Shendre Jan 16 '17 at 15:49
  • $\begingroup$ Oh! I have now known that a Diophantine has only integer solutions but it involves reals. $\endgroup$ – Lokesh Sangewar Jan 16 '17 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 '17 at 16:01
  • $\begingroup$ It looks cleaner as $x^{10}+y^{10}=10xy-8$ but I am not sure that is progress. $\endgroup$ – Ross Millikan Jan 16 '17 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 '17 at 16:03
9
$\begingroup$

Hint: Write it as:

$x^{10}+y^{10}=10xy-8$

Then use AM-GM:

$10xy-5=x^{10}+y^{10}+1+1+1\ge5\sqrt[5]{x^{10}y^{10}}=5x^2y^2$

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 $x=-1,y=-1$

$\endgroup$
  • $\begingroup$ Wow! I never thought of using the AM-GM. What a brain! Thank for the answer. $\endgroup$ – Lokesh Sangewar Jan 16 '17 at 16:19
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.