Diophantine equation: $2(x^3+xy+y^3)=3(x+y)$

Question

Here is a nice equation: $2(x^3+xy+y^3)=3(x+y)$ over $ \mathbb{Z}$ x $\mathbb{Z}$. Any nice way to approach this?

Answer 1

Since $x+y$ is even, let $x+y=2a,xy=b$ where $a,b\in\mathbb Z$.

Notice that we have $$(2a)^2-4b\ge 0\iff b\le a^2\tag1$$ since $x,y$ are roots of $X^2-2aX+b=0$.

Now, since we have $$\begin{align}2(x^3+xy+y^3)=3(x+y)&\iff 2\{(x+y)((x+y)^2-3xy)+xy\}=3(x+y)\\&\iff 2\{2a(4a^2-3b)+b\}=6a\\&\iff b(6a-1)=8a^3-3a\end{align}$$ with $6a-1\not=0$, we have $$b=\frac{8a^3-3a}{6a-1}\tag2$$ From $(1)$, we have $$\frac{8a^3-3a}{6a-1}\le a^2\tag3$$ Here, let us separate it into cases :

(Case 1) If $6a-1\ge 0\iff a\ge 1/6$, then $$(3)\iff 8a^3-3a\le a^2(6a-1)\iff a(a-1)(2a+3)\le 0.$$ Hence, we have $1/6\le a\le 1$.

(Case 2) If $6a-1\lt 0\iff a\lt 1/6$, then $$(3)\iff 8a^3-3a\ge a^2(6a-1)\iff a(a-1)(2a+3)\ge 0.$$ Hence, we have $-3/2\le a\le 0$.

Since we can see every integer we get from $(3)$ is $a=0,\pm1$, the answer is $$(a,b)=(0,0),(1,1)\Rightarrow (x,y)=(0,0),(1,1).$$ (Here, note that $b=5/7$ is not an integer for $a=-1$.)