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Math Help - for PH and others so inclined

  1. #1
    Eater of Worlds
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    for PH and others so inclined

    "Show that \frac{x^{5}}{5}+\frac{x^{3}}{3}+\frac{7x}{15}

    is always an integer for integral values of x".
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  2. #2
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    It is relatively easy (but messy) to show that for each positive integer N
    3N^5  + 5N^3  + 7N is a multiple of 15.
    an other
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  3. #3
    Eater of Worlds
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    That was a fast response, Plato . I noticed, since the semesters have come to a close, there hasn't been much posting.

    Anyway, maybe I am off base, but \frac{x^{5}}{5}+\frac{x^{3}}{3}+\frac{7x}{15}=\fra  c{x^{5}-5}{5}+\frac{x^{3}-3}{3}+x

    Now, use Fermat's theorem:

    x^{5}-x\equiv{0} \;\ mod(5)

    and

    x^{3}-x\equiv{0} \;\ mod (3)

    Therefore, it's an integer.

    Seems rather logical. But, I am not that much of a number theorist. Have been learning it, though.
    Last edited by galactus; December 23rd 2006 at 11:53 AM.
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  4. #4
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    Quote Originally Posted by Plato View Post
    It is relatively easy (but messy) to show that for each positive integer N
    3N^5  + 5N^3  + 7N is a multiple of 15.
    an other
    Hello Plato,

    use mathematical induction to prove this property:

    Prove that

    3n^5+5n^3+7n=15 \cdot k\ , \ k \in \mathbb{Z}

    step\ 1:\ n=1\ :\ 3+5+7=15=15 \cdot 1 \text{ is true}

    step \ 2:\  assumption: 3n^5+5n^3+7n=15 \cdot k \text{ is true}

    step\ 3:\ conclusion: 3(n+1)^5+5(n+1)^3+7(n+1)=

    3n^5+15n^4+30n^3+30n^2+15n+3+5n^3+15n^2+15n+5+7n+7  =

    \underbrace{3n^5+5n^3+7n}_{k\cdot 15}+15(n^4+2n^3+3n^2+2n+1)+3+5+7=

    The 1rst summand is a multiple of 15 according assumption, the 2nd summand is a multiple of 15 because it contains the factor 15 and the last 3 summands add up to 15. Thus 3(n+1)^5+5(n+1)^3+7(n+1)= is multiple of 15 and therefore the asumption is true for all n \in \mathbb{N}

    EB

    Merry Christmas and a happy New Year.
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