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Math Help - Proof by Induction?

  1. #1
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    Proof by Induction?

    moved to a new thread for general case
    Last edited by Aquafina; June 10th 2009 at 06:59 AM.
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  2. #2
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    Difficult Problem

    I have not had luck with induction here. However, here are two very unusual observations I have made:

    \lim_{n \rightarrow \infty} frac( \sqrt{\frac{n^5-1}{n-1}}) = \frac{3}{8} for n even and \frac{7}{8} for n odd. In other words, the decimal expansion seems to be tending towards an alternating pattern between .375 and .875. To be a perfect square, this decimal would of course need to be exactly zero, so proving this observation would imply your hypothesis.

    Since it does alternate, you may try using induction where n \rightarrow n+2

    And don't forget the identity \frac{n^5-1}{n-1}=n^4+n^3+n^2+n+1
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    Hi, might seem like a stupid q but go should I prove that using induction?
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    Limit Madness

    So far as I know, induction is not effective for proving a limit exists. Here is what I have gotten so far:

    Let f(n)=n^4+n^3+n^2+n+1

    As n gets large, \sqrt{f(n+2)}-\sqrt{f(n)} \approx 4n+5

    More rigorously, \lim_{n \rightarrow \infty} \frac{\sqrt{f(n+2)}-\sqrt{f(n)}}{4n+5}=1

    The actual value 4n+5 is not so important as the fact that it is an integer. What this essentially means is that if f(n) is not a perfect square, then f(n+2) won't be either, since both of their square roots will have increasingly identical decimal expansions \approx .375 or .875.

    You can actually graph \sqrt{f(n+2)}-\sqrt{f(n)} and  4n+5 and see how astonishingly closely they overlap. I am currently looking around for a method of taking the square root of a polynomial. I'm pretty sure it exists, basically given f(x)=Ax^4+Bx^3+Cx^2+Dx+E , find coefficients a_n that satisfy f(x)=(a_0x^2+a_1x+a_2+a_3x^{-1}+a_4x^{-2}+...)^2 , effectively finding the square root of a polynomial.

    I believe that rewriting \sqrt{f(n+2)}-\sqrt{f(n)} as series, a bunch of stuff will cancel, leaving only 4n+5 and some extra term that goes to zero fairly quickly.

    I have the sneaking suspicion that I am using a cannon to kill a mosquito, and a super member will come along with a painfully obvious elementary proof.
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    i'll try that method thanks

    any 'super member' u have in mind that i could PM to ask for any existing simpler method?
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  6. #6
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    Here are some guys that have put me to shame in other posts: chisigma, Aryth, PaulRS, fardeen_gen, CaptainBlack, Gamma, Soroban. Not sure what their particular areas of interest are, though.
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    Thanks, I PM-ed them. Hopefully one of them has an idea
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  8. #8
    Super Member fardeen_gen's Avatar
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    Quote Originally Posted by Aquafina View Post
    Hi, I need to show that for n > 3:

    (n^5 - 1)/(n-1) ≠ k^2 (does not equal)

    for any integer 'k'

    i.e. it is not a perfect square
    For n > 3

    CASE I: n is even,
    (n^{2}+\frac{n}{2})^{2}<\frac{n^{5}-1}{n-1}< (n^{2}+\frac{n}{2}+1)^{2}

    CASE II: n is odd,
    (n^{2}+\frac{n-1}{2})^{2}<\frac{n^{5}-1}{n-1}< (n^{2}+\frac{n+1}{2})^{2}

    If the above statements are true,
    the value of \frac{n^{5}-1}{n-1} is between two consecutive squares no matter n is even or odd \implies \frac{n^{5}-1}{n-1}\neq k^{2}, for any integer k.


    For verification,

    n = 4\ \Rightarrow\ (18)^2 < 341 < (19)^2.
    n = 5\ \Rightarrow\ (27)^2 < 781 < (28)^2

    I think we are done then!
    Last edited by fardeen_gen; May 14th 2009 at 04:13 AM. Reason: Added a few lines
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  9. #9
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    thanks

    i had a question, how does this prove it for n>3

    because when n=3, k^2 = 121 which is a perfect square

    so how come only when n>3, the proof you showed should work?
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  10. #10
    Super Member fardeen_gen's Avatar
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    Quote Originally Posted by Aquafina View Post
    thanks

    i had a question, how does this prove it for n<=3

    because when n=3, k^2 = 121 which is a perfect square

    so how come only when n>3, the proof you showed should work?
    Quote Originally Posted by Aquafina View Post
    Hi, I need to show that for n > 3...

    The statement is only valid for n > 3 and not n\leq 3. The question mentions it too
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  11. #11
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    yeah but the original question was to find any values of n for which

    n^4+n^3+n^2+n+1 is a perfect square

    i found out that it works for n=3, and guessed that there would be no other value, using excel for very large value of n!

    so i decided to prove


    (n^5 - 1)/(n-1) ≠ k^2 for n>3
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  12. #12
    Super Member fardeen_gen's Avatar
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    Quote Originally Posted by Aquafina View Post
    so i decided to prove

    (n^5 - 1)/(n-1) ≠ k^2 for n>3
    So, what you decided to prove has been proved, right?
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  13. #13
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    haha yes, but can the argument you gave be used towards the original question?
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  14. #14
    Super Member fardeen_gen's Avatar
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    The original question, as far as I am concerned is:
    Hi, I need to show that for n > 3:

    (n^5 - 1)/(n-1) ≠ k^2 (does not equal)

    for any integer 'k'

    i.e. it is not a perfect square


    And of course my argument only holds for the proof you wanted.(make that "initially" wanted )
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  15. #15
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    lol ok, will this proof imply that n=3 is the only solution for n^4+n^3+n^2+n+1 being a perfect square?
    Last edited by Aquafina; May 14th 2009 at 04:15 AM.
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