Multiple Choice Question (MCQ)

How many integer solution sets (x,y) for the equation x^3+6x^2+5x=y^3-y+2?

  1. 0

  2. ×

    1

  3. ×

    3

  4. ×

    infinite

Collected in the board: Number Theory

Steven Zheng posted 10 hours ago

Answer

  1. x^3+6x^2+5x=y^3-y+2
    (1)

    x^3+6x^2+5x

    =x(x+1)(x+5)

    =x(x+1)(x+2+3)

    Therefore there must be one out of x, (x+1),(x+5) that is divisible by 3

    In other words, x(x+1)(x+5) is a multiple of 3.

    On the other hand,

    y^3-y=y(y-1)(y+1)

    which is the product of 3 consecutive numbers.

    Therefore there must be one that is divisible by 3

    Dividing the equation (1) by 3 results in

    \dfrac{x(x+1)(x+2+3)}{3} = \dfrac{y(y-1)(y+1)}{3} -\dfrac{2}{3}

    which shows the remainder -\dfrac{2}{3} is not an integer. Hence, there's no integer solution to hold the equation equal.


Steven Zheng posted 10 hours ago

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