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Finite series HELP PLEASE!

  1. Dec 13, 2008 #1
    1. The problem statement, all variables and given/known data

    Are there positive integers m and n so that
    n = sqrt(m+sqrt(m+sqrt(m+...sqrt(m))))
    where there are 2008 square root signs.


    2. The attempt at a solution

    n^2=m+sqrt(m+sqrt(m+sqrt(m+...sqrt(m)))).....with 2007 square roots.
    Idk where to go from here. If there were infinite roots, then it would be alot easier to do.
    Any suggestions would be helpful. Thanks.
     
  2. jcsd
  3. Dec 13, 2008 #2
    My first instinct is to get rid of ugly square roots. Square both sides and subtract m 2008 times to get a 2008th degree polynomial in m. You may be able to somehow write this as a binomial.
     
  4. Dec 15, 2008 #3
    I don't see where that would get me tho.

    ((((n^2-m)^2-m)^2-m)^2.....-m)^2-m=0

    Where do i go from there. Its still as complicated as before.
     
  5. Dec 15, 2008 #4

    Dick

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    How about this? If n=sqrt(m+a) with n and m integers, then a is an integer. That tells you that the expression with 2007 square roots is an integer. Etc. Etc. Eventually you get down to sqrt(m+sqrt(m)) is an integer. Is that possible?
     
  6. Dec 15, 2008 #5
    I don't think its possible, b/c ur assuming that m and n are integers.
     
  7. Dec 15, 2008 #6

    Dick

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    Can you show it's not possible? You are talking about k=sqrt(m+sqrt(m)), I assume.
     
  8. Dec 15, 2008 #7
    So it would become:

    ((((a^2-m)^2-m)^2-m)^2....-m)^2=m

    Are u saying that since a and m are integers, this expression would be an integer too, which would equal m??
     
  9. Dec 15, 2008 #8

    Dick

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    No. I'm saying if n=sqrt(m+a) that means 'a' must be an integer. Where 'a' is the expression with 2007 square roots. Then the integer a=sqrt(m+b), where b is the expression with 2006 square roots. b must also be an integer so b=sqrt(m+c) where c is the expression with 2005 square roots. Etc etc. Eventually you get down to z=sqrt(m+sqrt(m)) is an integer. Is THAT possible?
     
  10. Dec 15, 2008 #9
    YES! And then would you say that since sqrt(m+a) is an integer, there exists integers that satisfy this equation??
     
  11. Dec 15, 2008 #10

    Dick

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    No, again. Because I don't think you can solve that final equation, z=sqrt(m+sqrt(m)) where z and m are integers. Try it.
     
  12. Dec 15, 2008 #11
    Thank You Very Much!!!!!! =)
     
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