Equation with an infinite number of solutions

I have the following equation: $x^3+y^3=6xy$. I have two questions: 1. Does it have an infinite number of rational solutions? 2. Which are the solutions over the integers?($x=3$ and $y=3$ is one) Thank you!

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what have you tried so far? over integers another obvious(and trivial) solution is x,y=0 – Sabyasachi Mar 7 '14 at 12:06

If $x$ and $y$ are rational, then so is $y/x=\alpha$. Then $x^3+y^3=6xy$ becomes $$(\alpha^3+1)x^3-6\alpha x^2=0\tag{1}$$ and then $x=0$ or $x=\dfrac{6\alpha}{\alpha^3+1}$. Thus, for any rational $\alpha$, we have the rational solutions $$\left(\frac{6\alpha}{\alpha^3+1},\frac{6\alpha^2}{\alpha^3+1}\right)\tag{2}$$

Since $x+y=\dfrac{6\alpha}{\alpha^2-\alpha+1}$, we have that $$-2\lt x+y\le6\tag{3}$$ Note that $x+y=-2$ only happens when $\alpha=-1$, and that doesn't give a finite $(x,y)$ in $(2)$.

Thus, cubing $(3)$ yields $$-8\le x^3+3x^2y+3xy^2+y^3\le216\tag{4}$$ and applying $x^3+y^3=6xy$, $$-\frac83\le(x+y+2)xy\le72\tag{5}$$ Since $x+y+2\ge1$, inequality $(5)$ leaves only a finite number of $x,y\in\mathbb{Z}$ to check.

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Which are the solutions over the integers?

$(x+y)^3=\underbrace{x^3+y^3}_{6xy}+3xy(x+y)=3xy(x+y+2)\iff6|(x+y)$, since, on one hand, $3$ is a prime, and, on the other hand, x and y being of opposite parity would lead to contradiction. Hence, we have $x=2a+r$, and $y=2b-r$, with $r\in\{0,1\}$, and, at the same time, $x=3A+R$, and $y$$=3B-R, also with R\in\{0,1\}. Can you take it from here ? :-) - Wolfram Alpha says that there are no rational solutions except the one you noted, x=y=3 although. It seems that it chose to skip the trivial x=y=0 though. The link has some irrational solutions too, if you need them. - The equation is symmetric in x and y, suggesting solutions of the form x=y. Plugging that in, we find$$2x^3-6x^2=0$$This gives the solution$x=y=0.$Simplifying by dividing by$2x^2$gives$x-3=0,$leading to another solution$x=y=3.\$

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