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

  1. Apr 16, 2015 #1
    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?
     
  2. jcsd
  3. Apr 16, 2015 #2

    BvU

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    Yes. You get two answers: x,y = 1, 0 and x,y = 0,0. Pretty boring.
     
  4. Apr 19, 2015 #3
    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.
     
  5. Apr 19, 2015 #4

    BvU

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    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 "
     
  6. Apr 20, 2015 #5
    Thank you! I understand perfectly now.
     
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