<s>[INST] Solving $y^3=x^3+8x^2-6x+8$ Solve for the equation $y^3=x^3+8x^2-6x+8$ for positive integers x and y.
My attempt- $$y^3=x^3+8x^2-6x+8$$
            $$\implies y^3-x^3=8x^2-6x+8$$
            $$\implies (y-x)(y^2+x^2+xy)=8x^2-6x+8$$
Now if we are able to factorise $8x^2-6x+8$ then we can compare LHS with RHS.Am I on the right track?Please help.
 [/INST] Using Gerry Myerson's clever inequality observation, we have two possibilities.
Case 1: $y=x+1$. Then by substitution and simplification, we must solve
$$5x^2-9x+7=0,$$
which has no integer solutions, as seen by using the quadratic equation.
Case 2: $y=x+2$. This yields the [factored] quadratic
$$2x(x-9)=0.$$
Evidently, the only two integer solutions are $x=0$ and $x=9$, yielding the solutions $(x,y)=(0,2)$ and $(x,y)=(9,11)$.
There are no other solutions.
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