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Binomial theorem-related proof

  1. Sep 13, 2014 #1
    I've been thinking about this for some time. Now I'm coming on here in the hope of getting some help;

    Prove that an+3 + (a + 1)2n+3 is divisible by a2 + a + 1.

    I can't quite remember the restrictions on n, though I'd imagine it'd be "for all real n ≥ -1" or something similar.

    Thanks in advance! :)

    EDIT: The way to do this would probably be through the method of mathematical induction, since that was the subject of class the day this problem was given.
    Last edited: Sep 13, 2014
  2. jcsd
  3. Sep 13, 2014 #2

    Char. Limit

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    Mathematical induction is definitely the way to go, yeah. Proving that it's true for n=-1 is rather simple, as I'm sure you've already seen. Then you just need to take the expression:

    [tex]a^{(n+1)+3} + (a + 1)^{2(n+1)+3}[/tex]

    And rewrite it into an expression of the form:

    [tex]f(a) \left(a^{n+3} + (a+1)^{2n+3}\right)[/tex]

    Where f(a) can be any function. Since you know the latter is divisible by [itex]a^2+a+1[/itex] by assumption, then the full expression will be divisible by [itex]a^2+a+1[/itex] as well, and that will conclude the proof!
  4. Sep 13, 2014 #3
    Can I really assume that an+3 + (a + 1)2n+3 is divisible by a2 + a + 1, when that is what I am aiming to prove?

    Sorry, just trying to get into the whole induction thinking...

    Thanks for your time, btw :) really appreciate it!
  5. Sep 13, 2014 #4

    Char. Limit

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    Yes, you can. In fact, that's the whole point of induction! You prove a certain base case (here, n=-1), then you make the assumption that it's true for an unspecified number n, and using that assumption, prove that it's also true for n+1. Both of those facts together show that it's true for all n≥-1, as long as n's an integer!
  6. Sep 13, 2014 #5


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    No, that's NOT what you are trying to prove. What you are trying to prove is that "an+3 + (a + 1)2n+3 is divisible by a2 + a + 1 for all n". What you are assuming with induction is "an+3 + (a + 1)2n+3 is divisible by a2 + a + 1 for a single value of n.
    (I prefer to use a different letter, say "k", rather than "n" when stating induction hypothesis to avoid that mistake: "if a statement is true for n= 1 (or 0 or -1) and whenever the statement is true for a number, k, it I also true for k+1 then the statement is true for all n.")

    That is, once you have proved it true for -1, you then show it is true for =1+1= 0. Now that you know it is true for 0, you show it is true for 0+1= 1. Now that you know it is true for 1, you show it is true for 1+ 1= 2.

    Showing that "if it is true for k then it is true for k+1" collapses all of those into one proof.
  7. Sep 13, 2014 #6
    Ah, yes, I shall think about that in the future.

    As for this problem, though. I still can not solve it :(
  8. Sep 13, 2014 #7
    I'm going to need some hint, I've tried some different ways; Writing (a + 1)^(2n+3) as a sum, futile long division.
    Thanks in advance.
  9. Sep 24, 2014 #8
    In = an+3 + (a+1)2n+3 , when n=-1 and n=0 it can be shown that it is divisible by a2+a+1
    In+1 = an+4 + (a+1)2n+5
    In+1 - In = (an+4 - an+3) + [(a+1)2n+5 - (a+1)2n+3]
    which can be ultimately reduce to In+1 = a.In + (a2+a+1).(a+1)2n+3
    So, by induction method u can prove. Not true for n less than -1.
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