I have this problem which says $$ x=3\sqrt{y\sqrt{3\sqrt{y\cdots}}}\\ y=3\sqrt{x\sqrt{3\sqrt{x\cdots}}} $$ What is $x+y$ I tried $$ x^2=9y\sqrt{x}\\ x^3=81y^2\\ y^3=81x^2\\ x^3+y^3=81(x^2+y^2)\\ (x+y)(x^2-xy+y^2)=81(x^2+y^2) $$ I don't know what to do or if I am do that right
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Use proper Latex codification otherwise this post will be closed. It is not at all clear what you are asking. – Nicky Hekster Aug 03 '17 at 09:12
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sorry, I've edited it – AbdulQader Qassab Aug 03 '17 at 09:14
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Much better, thanks! – Nicky Hekster Aug 03 '17 at 09:18
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I changed the second line to $y = {}$, since that makes more sense with what comes after, and it was in the original version of the question, – Arthur Aug 03 '17 at 10:04
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From second equality we obtain $$x^2=9x\sqrt{x},$$ which gives $x=0$ or $x=81$. The rest is smooth.
Michael Rozenberg
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Suppose $x \neq 0 \neq y$. From $x^3=81y^2$ and $y^3=81x^2$, it follows that $\frac{x^3}{y^3}=\frac{y^2}{x^2}$, hence $x^5=y^5$, so $x=y$. I think you can take it from there.
Nicky Hekster
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OK! If you think this is the right answer, please tick it as such! – Nicky Hekster Aug 03 '17 at 10:02