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Algebra problem (x+(1/x) problem)

  1. May 18, 2016 #1
    1. The problem statement, all variables and given/known data

    If ##a+\frac{1}{a}=\sqrt{3}##, then ##a^{2016} + (\frac{1}{a})^{2016} ## equals

    A. 0
    B. 1
    C. √2
    D. √3
    E. 2

    2. Relevant equations


    3. The attempt at a solution

    I find no real solution for a.
    What I can do is just find a^2 + (1/a)^2 which equals 1, a^4 + (1/a)^4 = -1, a^8 + (1/a)^8 = -1, and so on...
    Since 2016 is not a square number, I don't know how to determine the value

    Is there any theory about problem like this?
    Please help
     
  2. jcsd
  3. May 18, 2016 #2

    Samy_A

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    Start from ##a²+1/{a²}=1##.
    Multiply by ##a²## to get rid of the denominator, you get ##a^4 +1 =a²##.
    Now notice that 2016 is a multiple of 6. So multiply ##a^4 +1 =a²## again by ##a²## to get ##a^6## in the equation.
     
  4. May 18, 2016 #3

    ehild

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    Find the complex solution. Write it in trigonometric of exponential form, which makes it easy to rise to power 2016. (2016=25*32 *7)
     
  5. May 18, 2016 #4
    a^6 + a^2 = a^4
    a^6 = a^4 - a^2 = (a^2 - 1) - a^2 = -1
    a^12 = 1
    a^24 = 1
    ...
    a^2016 = 1

    So, 1/a^2016 = 1
    a^2016 + (1/a)^2016 = 1 + 1 = 2
    Thank you
     
  6. May 19, 2016 #5
    I'm interested to try this approach..

    The solution for a is ##\frac{\sqrt{3}\pm i}{2}##
    Should I use the plus or the minus as the root?
    If it's plus, ##e^{i\frac{1}{6}\pi}##
    If it's minus, ##e^{i\frac{11}{6}\pi}##

    And, then what should I do??
     
  7. May 19, 2016 #6

    ehild

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    What are their 12th power?
     
  8. May 19, 2016 #7
    The 'plus' root^12 = ##e^{i2\pi}= 1##
    The 'minus' root^12 = ##e^{i22\pi} = 1##

    Okay, since 2016 can be divided by 12, a^2016 = 1
    And 1/a^2016 = 1
    So, a^2016 + 1/a^2016 = 1+1 = 2

    Thanks a lot!
     
  9. May 19, 2016 #8

    ehild

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    The 'minus' root^12 = ##e^{-i2\pi} = 1## :smile:
     
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