Completing Cube

Prove or disprove that There exists a positive integer M, such that 1^3+2^3+…+n^3=M^3, where n>1.

Thank you for helping.

Simon YAU.

1 個解答

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  • 5 年 前
    最佳解答

    The answer is NO that

    "There exists a positive integer M,

     such that 1³ + 2³ + … + n³ = M³, where n > 1."

    First of all, we know a common result about the sum of cubes:

    1³ + 2³ + … + n³ = [ n(n+1)/2 ]²

    Therefore, the statement is equivalent to

    "There exists a positive integer M,

     such that [ n(n+1)/2 ]² = M³, where n > 1."

    If a number is both a perfect square and a perfect cube, it must be a perfect sixth power.

    That is, [ n(n+1)/2 ]² = M³ = A⁶ where A is a positive integer.

    In other words, we need that

    n(n+1)/2 = A³ is a perfect cube, and M = A² is a perfect square.

    (Actually, the only triangular number n(n+1)/2 which is also a perfect cube is only 1.)

    Consider

    n(n+1)/2 = A³

    n² + n = 2A³

    4n² + 4n = 8A³

    4n² + 4n + 1 = 8A³ + 1

    (2n + 1)² = (2A)³ + 1

    (2n + 1)² - (2A)³ = 1 ...(✿)

    Among all integers p and q such that p² - q³ = 1,

    the ONLY solution is p = 3 and q = 2.

    This is called the Catalan's conjecture (now a theorem).

    In other words, the only solution to (✿) is that

    2n + 1 = 3 and 2A = 2M² = 2

    That is, n = 1 and M = 1.

    Therefore, the ONLY possible case is n = 1.

    That means, there does not exists a positive integer M

     such that 1³ + 2³ + … + n³ = M³, where n > 1.

    References:

    http://en.wikipedia.org/wiki/Squared_triangular_nu...

    http://mathworld.wolfram.com/CubicTriangularNumber...

    http://mathworld.wolfram.com/CatalansConjecture.ht...

    Basic references:

    http://www.themathpage.com/arith/appendix.htm

    http://mathcentral.uregina.ca/QQ/database/QQ.09.06...

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