Simple diophantine: solutions to x^2-x=y^3

[spacetimedude]

Homework Statement

Find all solutions x,y∈ℤ to the following Diophantine equation:
x^2-x=y^3

Homework Equations

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?

 

[BvU]

Yes. You get two answers: x,y = 1, 0 and x,y = 0,0. Pretty boring.

 

[spacetimedude]

Thank you. I am still a bit confused.
How do we know if x-1 corresponds to the smaller value of the cube and x the higher value? Can we just say x-1 is always smaller than x?

There is another similar problem: 4x^2=y^3+1

Rearranging the equation:
y^3=(2x+1)(2x-1)
(2x+1) and (2x-1) are coprime and their difference is 2.
Looking at the list of possible cubes, the two cubes that are different by 2 are 1 and -1.

So then,
(2x+1)=1 =>x=0
(2x-1)=-1 =>x=0

But can we also do
(2x+1)=-1 => x=-1
(2x-1)=1 => x=1.

So the possible solutions are x=-1 ,x=1, and x=0 (and y's accordingly)? The textbook I'm using says the only solution is x=0.

 

[BvU]

Can we just say x-1 is always smaller than x?

Yes we can: $-1 < 0 \quad \Rightarrow \quad -1 + x < x$.

Re the other exercise (a thread extension, but never mind: same difference):
Temporarily replace (2x+1) by a and (2x-1) by b.

a and b have to be cubes AND differ by two. There are two cubes that differ by two: -1 and +1.

First case: a = 1 & b = -1
Try to solve (2x+1) = 1 & (2x-1) = -1
Solution: x = 0. Answer to the exercise: x,y = 0, -1

Second case: a = -1 & b = 1
Try to solve (2x+1) = -1 & (2x-1) = 1
No solution !Point is that your "But can we also do" does not lead to x = 1 OR x = -1 but to "No value for x that satisfies (2x+1) = -1 AND (2x-1) = 1 "

 

[spacetimedude]

Thank you! I understand perfectly now.