Solve in positive integers:
Source: Bulgarian Math. Competition
For most values of y, the following compound inequality is true:
y^3 < y^3+2y^2+1 < (y+1)^3
If y satisfies this inequality then the expression y^3+2y^2+1 cannot be a perfect cube because it is strictly between two consecutive cubes.
The places when this compound inequality fails to hold are when y^3 >= y^3+2y^2+1 OR y^3+2y^2+1 >= (y+1)^3.
If y^3 >= y^3+2y^2+1 then 0 >= 2y^2+1, but 2y^2+1 is always positive, so no possible cases here.
If y^3+2y^2+1 >= (y+1)^3 then 0 >= y^2+3y. The integers which satisfy this last inequality are 0, -1, -2, and -3
Then for each case:
y=0 -> y^3+2y^2+1=1 -> x=1
y=-1 -> y^3+2y^2+1=2 -> not a perfect cube
y=-2 -> y^3+2y^2+1=1 -> x=1
y=-3 -> y^3+2y^2+1=-8 -> x=-2
So there are three integer solutions: (x,y) = (1,0), (1,-2) and (-2, -3). However none of these are pairs of positive integers.