1. The problem statement, all variables and given/known data Find all solutions x,y∈ℤ to the following Diophantine equation: x^2-x=y^3 2. Relevant equations 3. The attempt at a solution Hello. I am stuck in the last part of finding the solutions. I rearranged x^2-x=y^3 into x(x-1)=y^3. The Fundamental Theorem of Arithmetic tells that since x and x-1 are coprime, and the multiple of the two is a cube, x and x-1 themselves have to be cubes. Note that the difference of the two is 1. Hence, looking through the list of possible cubes, ...-8, -1,0,1,8.... we can see that the cubes with difference of 1 are -1 and 0, and 0 and 1. My question arises here. Do I set (x-1) and x equal to the two pairs? So Pair 1: (x-1)=-1 =>x=0 x=0 and Pair 2: (x-1)=0 => x=1 x=1. Is this correct?