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I imagine that this should be done in the following way: There is a polynomial $P$ such that:

$$P(x^{\frac{2}{3}}+y^{\frac{2}{3}})=P$$

My first guess (obvious observation?) is that it can't be a polynomial in one variable, otherwise one of the variables will have a rational power. So the number of variables must be $2$. I've tried this:

$$(x^n+y^m)(x^{\frac{2}{3}}+y^{\frac{2}{3}})=(x^n+y^m)$$

$$(x^n+y^m)x^{\frac{2}{3}}+(x^n+y^m)y^{\frac{2}{3}}=x^n+y^m$$

$$x^nx^{\frac{2}{3}}+y^mx^{\frac{2}{3}}+x^ny^{\frac{2}{3}}+y^my^{\frac{2}{3}}-x^n-y^m=0$$

$$x^nx^{\frac{2}{3}}+\overbrace{y^m(x^{\frac{2}{3}}-1)+x^n(y^{\frac{2}{3}}-1)}^{\text{Annoying cross-terms }}+y^my^{\frac{2}{3}}=0$$

So this seems to show that we need to have at least one of the sums $n+2/3,m+2/3$ to be equal to $6$. But what should be the value of the other? And how do we eliminate the annoying cross-terms?

I understand also that $P$ could be $x^{a_1}+y^{b_1}+x^{a_2}+y^{b_2}+\dots +x^{a_t}+y^{b_t}$ for a certain $t$ but I don't know how to look into the entire collection of P's to find for it and also I don't know how to cut the general annoying cross-terms that this would give me. If possible, I'd like to have only some hints.

Red Banana
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1 Answers1

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Hint:

$$ \begin{align} 1=\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)^3 & = \left(\sqrt[3]{x^2}\right)^3 + 3\;\sqrt[3]{x^2}\;\sqrt[3]{y^2}\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)+\left(\sqrt[3]{y^2}\right)^3 \\ & =x^2 + y^2 + 3\;\sqrt[3]{x^2\,y^2} \end{align} $$

Regroup, and raise to the $3^{rd}$ power one more time.

dxiv
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    (As noted in my now-deleted answer: when you manipulate implicit equations you always have to be careful about introducing spurious solutions; it's important to understand why that doesn't happen here. Hint: it actually might! What's your base field?) – Steven Stadnicki Dec 01 '16 at 02:13
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    @StevenStadnicki That's a good point. I sort of assumed the question was about real polynomials, since that's what "algebraic curves" most commonly refer to. The real cube root happens to be a bijection on $\mathbb{R}$ but that's not necessarily the case on arbitrary fields, and the question didn't actually spell out the base field. – dxiv Dec 01 '16 at 02:33
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    Yes, it is about real polynomials. – Red Banana Dec 01 '16 at 02:39