SELECTION EXAMINATION

The given equation can be written as $$\begin{array}{rl}8x^3+y^3-6xy& =4\iff (2x{)}^3+y^3+1^3-3\cdot 2x\cdot y\cdot 1=5\\ & \iff (2x+y+1)(4x^2+y^2+1-2xy-2x-y)=5\end{array}(1)$$ $$\iff \frac{1}{2}(2x+y+1)[(2x-y{)}^2+(2x-1{)}^2+(y-1{)}^2]=5(2)$$ From (2), since $(2x-y{)}^2+(2x-1{)}^2+(y-1{)}^2>0$, we have $2x+y+1>0$, and hence from (1) we get $$2x+y+1=1\text{ or }2x+y+1=5\iff 2x+y=0\text{ or }2x+y=4.$$ From (1), for $2x+y=4$, we have $$4x^2+y^2+1-2xy-2x-y=1\iff (2x+y{)}^2-6xy-(2x+y)=0\iff xy=2$$ From the system $2x+y=4$, $xy=2$ we get the solution $(x,y)=(1,2)$. Also, from (1) for $2x+y=0$ we have $$4x^2+y^2+1-2xy-2x-y=5\iff (2x+y{)}^2-6xy-(2x+y)=4\iff xy=-\frac{2}{3}.$$ However, from the system $2x+y=0$, $xy=-\frac{2}{3}$ we get no solutions. We also can work in the following way: The equation can be written as $$8x^3+y^3-6xy=4\iff (2x+y{)}^3-3\cdot 2x\cdot y\cdot (2x+y)-6xy=4,$$ and so by putting $2x+y=s$, $2xy=p$, we get $$s^3-3ps-3p=4\Rightarrow p=\frac{s^3-4}{3(s+1)}.$$ Since $3p\in \mathbb{Z}$, we have that $$\frac{s^3-4}{s+1}=\frac{s^3+1-5}{s+1}=s^2-s+1-\frac{5}{s+1}\in \mathbb{Z}.$$ --- Thus $s+1$ must be a divisor of 5, that is $$s+1\in \{-1,1,-5,5\}\text{ or }s\in \{-2,0,-6,4\},$$ And hence we find the pairs $$(s,p)=(-2,4),(s,p)=\left(0,-\frac{4}{3}\right),(s,p)=\left(-6,\frac{44}{3}\right),(s,p)=(4,4),$$ From which only the last gives integer values for $x,y$, i. e. $x=1,y=2$.