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    The sum of cubes of three consecutive positive integers is divisible by $9$

    P-FKEN5

    Area
    Prelim
    Topics
    Induction, Number Theory
    Seen at
    UGA
    Years
    2003
    Status
    draft

    Prove by induction that the sum of the cubes of 3 consecutive positive integers is divisible by 9.

    Solution

    For \(n\ge1\), let \[ S_n=n^3+(n+1)^3+(n+2)^3. \] We prove by induction that \(9\mid S_n\).

    For \(n=1\), \[ S_1=1+8+27=36, \] which is divisible by \(9\).

    Assume \(9\mid S_n\). Then \[ S_{n+1}-S_n=(n+3)^3-n^3 =9(n^2+3n+3), \] which is divisible by \(9\). Hence \(S_{n+1}\) is divisible by \(9\). By induction, the sum of the cubes of any three consecutive positive integers is divisible by \(9\).

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