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Math Help - Number theory problem

  1. #1
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    Number theory problem

    n^4-n^2=n x (n-1)n(n+1)

    deduce that n^4-n^2 is always divisible by 3 , for any positive integer n.

    Any help starting off this question would be appreciated,

    thanks
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  2. #2
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    Work modulo 3. If n is already divisible by 3 you are done. Otherwise, n is either 1 or 2 modulo 3. In the former case you have that (n-1) must be divisible by 3. In the latter, you have that it is (n+1) the factor of your expression that is a multiple of 3. Done.
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  3. #3
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    not quite with you on this one
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  4. #4
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    n^{4}-n^{2} = n^{2}(n^{2}-1) = n^{2}(n-1)(n+1), right?

    Then, study all possibles remainders of n upon division by 3. If remainder is 0, then the factor n^{2} is a multiple of 3 and you are done. If remainder is 1, then the factor (n-1) is a multiple of 3 and you are also done. Same with remainder 2 and that'll be it.
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    An easy way to say it is that out of three consecutive integers, there is always one divisible by three.
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  6. #6
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    Well, I didn't want to put things that way, esteemed Bruno, 'cause in a sense that's exactly what our friend offahengaway-and-chips has been asked to established.
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  7. #7
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    Quote Originally Posted by offahengaway and chips View Post
    n^4-n^2=n x (n-1)n(n+1)

    deduce that n^4-n^2 is always divisible by 3 , for any positive integer n.

    Any help starting off this question would be appreciated,

    thanks
    I think you can use Division algorithm here.
    ex. If n is an integer, by D.A., n has the following forms:
    "n=2k;n=2k+1" only.

    try to substitute n=2k;n=2k+1 to n, and see if it is divisible by 3. If not try other forms.
    ex. "n=3k; n= 3k+1 ;n= 3k+2" only
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  8. #8
    Super Member PaulRS's Avatar
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    Just think of \binom{n+1}{3}.

    In case you don't know what that is :

    Consider A = \left\{ {\left( {x,y,z} \right) \in \left\{ {1,..,n+1} \right\}^3 /x \ne y;x \ne z;y \ne z} \right\} - the set of all ordered tripes of numbers in <br />
{\left\{ {1,..,n+1} \right\}}<br />
where each entry is different from the other 2. -

    Note that <br />
\left| A \right| = \left( {n + 1} \right) \cdot n \cdot \left( {n - 1} \right)<br />
since x - the first entry- may be given n+1 different values, fixed x, y can take n values, and then fixed the first 2, z can take n-1 values.

    You may partition the set A in subsets of 6 elements by considering the equivalence relation: \left( {x_1 ,x_2 ,x_3 } \right) \sim \left( {a,b,c} \right) if and only if the sequence a,b,c is a permutation of x_1 ,x_2 ,x_3 .

    Then it follows that 6 divides \left| A \right|
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