2021/028) Find non -ve integer x and y satisfying $x^3 = y^3 + 2y + 1$

We have $y^3 + 2y + 1 > y^3$

Further $(y+2)^3 = y^3 + 6y^2 + 12y + 8 > y^3 + 2y + 1$ 

So we have $(y+2)^3 > x^3 > y^3$

So $x^3 = (y+1)^3 = y^3 + 3y^2 + 3y +1 = y^3 + 2y + 1$

Or $3y^2 + 3y + 1 = 2y + 1$

Or $3y^2 + y = 0$

Or y = 0 as we are considering non -ve roots

So solution x=1, y = 0