if you substitute $a=x-y$ and $b=xy$ then $b=\frac{a^3-8}{2-3a}$ and from $3a-2 | a^3-8$ and $3a\equiv 2 \mod 3a-2$ we get $2^4 \cdot 13 \equiv 0 \mod 3a-2$ and it all comes to cases banging which I left to the reader

Multipying the equation by $27$ and using that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$, it follows that \[ (3x)^3+(-3y)^3+(-2)^3-54xy=8\cdot 27-8 \] or \[ (3x-3y-2)(9x^2+9y^2+4+6x-6y+9xy)=208=2^4\cdot 13 \] and we've arrived to the cases Megus said we should consider.

i have a different approach: we have $x>y$ but if $x\geq y+2$ then $2xy+8\geq 6y^{2}+12y+8\rightarrow x>3y$. from our initial equation we have $y^{3}+8=x(x^{2}-2y)>3y.(9y^{2}-2y)=27y^{3}-18y$ which is really absurd cuz $y\geq 1$. so $x=y+1$ . ...

If $x = y$, we have no solution in Z, then let us consider $x \neq y$. Setting $d = x-y$, $p = xy$, we have $d(d^{2}+3p) = 2(4+p)$ From this relation, we have these 2 systems to solve \[(1)\left\{\begin{array}{rcl}d&=&2\\ d^{2}+3p&=&4+p \end{array}\right. \] \[(2)\left\{\begin{array}{rcl}d^{2}+3p&=&2\\ d&=&4+p \end{array}\right. \] (1) gives $d = 2$, $p = 0$, so $(x,y)=(2,0)$ (2) gives no solution in Z. Finaly, only $(x,y) = (2,0)$ is solution Is it correct ?

zweig wrote: If $x = y$, we have no solution in Z, then let us consider $x \neq y$. Setting $d = x-y$, $p = xy$, we have $d(d^{2}+3p) = 2(4+p)$ From this relation, we have these 2 systems to solve \[(1)\left\{\begin{array}{rcl}d&=&2\\ d^{2}+3p&=&4+p \end{array}\right. \] \[(2)\left\{\begin{array}{rcl}d^{2}+3p&=&2\\ d&=&4+p \end{array}\right. \] (1) gives $d = 2$, $p = 0$, so $(x,y)=(2,0)$ (2) gives no solution in Z. Finaly, only $(x,y) = (2,0)$ is solution Is it correct ? Unfortunately, we're not given statements like $d$, $d^{2}+3p$, or $4+p$ prime. The factors might break up another way.

Case1:x=0 or y=0, solution (0,-2) and (2,0) Case 2: x>0, y>0 Then x>y, we have $ x^3 - y^3 = (x - y)(x^2 + xy + y^2) = 2xy + 8 \ge x^2 + xy + y^2 \ge 3xy$, so $ xy \le 8$. Since we cannot have $ x\ge 3$, $ y \ge 3$, we only need to test out y=1,2, no solution. Case 3: x<0, y<0 Then let x=-p, y=-q, we get case 2 Case 4: x>0, y<0 Then let y=-q, we have $ x^3 + q^3 = 8 - 2xq$, so $ x \le 2$, $ q \le 2$, no solution Case 5: x<0, y>0 Let x=-p, again we have $ p^3 + y^3 = 8 - 2py$, so $ p \le 2$, $ y\le 2$ , no solution.

Xixas wrote: Solve in integers $ x$ and $ y$ the equation $ x^3 - y^3 = 2xy + 8$. we have that $ (x-y)(x^2+xy+y^2)=2(xy+4)$ if $ x-y\geq{2}$, we have that $ 4\geq{x^2+y^2}$, which, when testing, gives us $ (2,0),(0,-2)$. Notice that $ x^2+xy+y^2$ is always positive.So if $ (x-y)$ is negative than we would have $ xy\leq{-4}$, absurd ,just break into cases: $ x-y$ positive and $ x-y$ negative and then just use inequalities: 1:$ x-y\geq{1}$ $ (x-y)(x^2+xy+y^2)=2(xy+4)$ LHS positive ,so we need only consider $ x-y=1$ ( $ x-y\geq{2}$ has already been done.) to do this case , just note that we have that $ (x-y)^2+x^2+y^2=8$ which is absurd. 2:$ x-y\leq{-1}$ $ (x-y)(x^2+xy+y^2)=2(xy+4)$ LHS is negative, so $ xy\leq{-4}$ if $ x-y\leq{-2}$, we´ll have $ -2(x+y)^2\geq8$,absurd. so we need only consider $ x-y=-1$, which leads to $ -(x+y)^2=xy+8$, and then $ 3y^2-5y=7$ , since $ x-y=-1$.Easy to check that it is absurd. So the only possible solutions are $ (2,0)$ and $ (-2,0)$

Denote $ x=k+y$. Then solve as quadratic toward $ y$ and we get $ D=-\left(3k-2\right)\left(k^{3}+2k^{2}-32\right)$ which is very esy to check when is positive.

scorpius119 wrote: zweig wrote: If $x = y$, we have no solution in Z, then let us consider $x \neq y$. Setting $d = x-y$, $p = xy$, we have $d(d^{2}+3p) = 2(4+p)$ From this relation, we have these 2 systems to solve \[(1)\left\{\begin{array}{rcl}d&=&2\\ d^{2}+3p&=&4+p \end{array}\right. \] \[(2)\left\{\begin{array}{rcl}d^{2}+3p&=&2\\ d&=&4+p \end{array}\right. \](1) gives $d = 2$, $p = 0$, so $(x,y)=(2,0)$ (2) gives no solution in Z. Finaly, only $(x,y) = (2,0)$ is solution Is it correct ? Unfortunately, we're not given statements like $d$, $d^{2}+3p$, or $4+p$ prime. The factors might break up another way.

good solution

But this is IMO Longlist 1982 #7.

Let $x = \frac{u+v}{2}, y = \frac{u-v}{2}$ where $u,v$ are integers. Then $\left(\frac{u+v}{2}\right)^3-\left(\frac{u-v}{2}\right)^3 = 2\left(\frac{u+v}{2}\right)\left(\frac{u-v}{2}\right)+8$ which is equivalent to \begin{align*} u^2(3v-2) &= -v^3+2v^2+32 \\ u^2 &= \frac{-v^3+2v^2+32}{3v-2} \end{align*}Clearly, $\frac{-v^3+2v^2+32}{3v-2} \geq 0$ only when $v = 1,2$. Checking, $v = 2$ produces $u = \pm 2$. Thus the solutions are $(2,0)$ and $(0,-2)$.

{0, -2}, {2, 0}

If $x-y=D$ then we have: $x^3-(x-D)^3-2x(x-D)-8=0\implies(3D-2)x^2-(3D^2-2D)x+(D^3-8)=0\implies x=\frac{D}{2}\pm\frac{\sqrt{-3D^4-4D^3+4D^2+96D-64}}{2(3D-2)}$ \[-3D^4-4D^3+4D^2+96D-64\geq 0\implies (3D-2)(D^3+2D^2-32)\leq 0\implies 0.\bar{6}=\frac{2}{3}\leq D\leq\frac{-2+2 \sqrt[3]{53-6 \sqrt{78}}+2 \sqrt[3]{53+6\sqrt{78}}}{3}\approx 2.62919\] Although we don't actually need to solve the cubic in closed form. Instead, we can easily show the cubic only has one real root and then show it is n the $(2,3)$ interval using IVT. Therefore, since we need $D\in\mathbb{Z}$, we are forced into $D\in\{1,2\}$. If $D=1$ then $x=\frac{1\pm\sqrt{29}}{2}$, which is no good. If $D=2$, then $x\in\{0,2\}$ and so we get the pairs $(0,-2)$ and $(2,0)$.