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Show by induction that a given polynomial is an integer

  1. Oct 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Show with mathematical induction that [tex]\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30} \in {Z}[/tex] for all [tex]n\ge 1[/tex].

    2. Relevant equations

    Probably.

    3. The attempt at a solution

    Inductive statement: [tex]Q(n)[/tex]: [tex]\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30} \in {Z}[/tex]

    [tex]Q(1)[/tex]: [tex]\frac{1}{5} + \frac{1}{2} + \frac{1}{3} - \frac{1}{30} = 1 \in {Z}[/tex]

    Since [tex]Q(1)[/tex] is true, assume that [tex]Q(n)[/tex] is true. Show that [tex]Q(n) \Rightarrow Q(n+1)[/tex].

    [tex]Q(n+1)[/tex]: [tex]\frac{(n+1)^5}{5} + \frac{(n+1)^4}{2} + \frac{(n+1)^3}{3} - \frac{(n+1)}{30} = \frac{(n+1)(6(n+1)^4 + 15(n+1)^3 + 10(n+1)^2 - 1)}{30} = ... =

    \frac{6n^5 + 45n^4 + 130n^3 + 119n}{30} + 6n^2 + 1[/tex]

    I'm not getting anywhere. I tried to assume that [tex]Q(n+1)[/tex] is true and to subsequently show that [tex]Q(n+1) \Rightarrow Q(n+2)[/tex]. That attempt yielded no results. I also tried to show this with modular arithmetic, but it made the induction seem redundant. Furthermore, I wasn't successful in using modular arithmetic to show the validity of [tex]Q(n+1)[/tex].

    I'm just not sure how to attack this. Help will be greatly appreciated.

    Also, hi. :)
     
    Last edited: Oct 23, 2009
  2. jcsd
  3. Oct 23, 2009 #2

    Hurkyl

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    How have you compared f(n) against f(n+1)? (where I'm using f to denote the function in n you're studying)
     
  4. Oct 23, 2009 #3
    In my example above: I guess I have not. I have considered to rewrite [tex]{Z}[/tex] as 2k and 2k+1 ([tex]k \in {Z}[/tex]) (or as 2n and 2n+1?) and study the two separate cases of f(n)=2n and f(n)=2n+1 (e.g. two equations rather than f(n) being defined by the corresponding R.H.S.). However, I couldn't think of a mathematically valid reason to use n as a variable, and introducing k as a new variable seems like a bad idea.

    Edit: Updated a few instances of P(n) and P(n+1) to Q(n) and Q(n+1), respectively, in the original post.
     
    Last edited: Oct 23, 2009
  5. Oct 23, 2009 #4
    When all else fails, just expand and see what you get.
     
  6. Oct 23, 2009 #5

    Dick

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    Look at f(n+1)-f(n) as Hurkyl suggested. Can you it's an integer?
     
  7. Oct 23, 2009 #6
    Thanks; that worked out. Everything is so easy once you know what to do, surprisingly enough. frown.gif
     
    Last edited by a moderator: Apr 24, 2017
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