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Math Help - 2n^3+3n^2+n\equiv 0 (mod 6)

  1. #1
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    2n^3+3n^2+n\equiv 0 (mod 6)

    2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}

    6|(2n^3+3n^2+n)\rightarrow 6k=2n^3+3n^2+n

    ???
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  2. #2
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    Hint: 2n^3+3n^2+n = n(n+1)(2n+1)

    This is a rather easy exercise - using either induction or direct proof (unless I am missing something in the question)
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  3. #3
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    Hello, dwsmith!

    I have a back-door approach to this one . . .


    2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}

    As aman-cc pointed out: . 2n^3 +3n^2 + n \:=\:n(n+1)(2n+1)


    I recognized this is as part of the formula for the sum of consecutive squares:

    . . 1^2 + 2^2 + 3^2 + \hdots + n^2 \:=\:\dfrac{n(n+1)(2n+1)}{6}

    Since the left side is an integer, the right side must be an integer.
    . . That is, n(n+1)(2n+1) is a multiple of 6.


    Therefore: . 2n^3 + 3n^2 + n \:\equiv\:0\text{ (mod 6)} .for all integer values of n.

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  4. #4
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    Consider  2n^3 + 3n^2 + n = 2(n^3 - n ) + 3n(n+1) = 2(n-1)(n)(n+1) + 3n(n+1) , use the property of product of few consecutive integers .
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  5. #5
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    2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}

    This a solution I saw once:
    2n^3+3n^2+n=n(2n^2+3n+1)=n(n+1)(2n+1) . One of n and n+1 is divisible by 2 .

    2\equiv -1(mod\ 3);
    n(n+1)(2n+1)\equiv -(n-1)n(n+1)(mod\ 3). Then at least one of n-1 , n , and n+1 is divisible by 3 .

    Since 2 and 3 are relatively prime 2n^3+3n^2+n is divisible by 6 .
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