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How to solve this equation

  1. Mar 27, 2012 #1
    I have encountered an equation of the form a^n + b^n = c^n. I know this looks like Fermat's equation. How can I solve for n. It doesn't matter if n is not an integer.
  2. jcsd
  3. Mar 28, 2012 #2
    Ok the equation I'm trying to solve is :

    [itex]\left[\sqrt{2+\sqrt{3}}\right]^{n}[/itex]+ [itex]\left[\sqrt{2-\sqrt{3}}\right]^{n}[/itex] = [itex]2^{n}[/itex]

    I have arrived to this point:

    [itex]\left[\sqrt{2+\sqrt{3}}\right]^{2n}[/itex] + [itex]\left[2*\sqrt{2+\sqrt{3}}]\right]^{n}[/itex] - 1 =0

    I think the problem is easy but there is a point I'm missing.

    I know that 2 is the answer from simple observation and with the knowledge that with Fermat's type of equations an integer solution can't be more than 2.
    Last edited: Mar 28, 2012
  4. Mar 28, 2012 #3
    So you want to solve

    [itex]\left[\frac{\sqrt{2+\sqrt{3}}}{2}\right]^{n}+ \left[\frac{\sqrt{2-\sqrt{3}}}{2}\right]^{n}= 1[/itex]

    and you know that n=2 is a solution, and I assume you know that an is decreasing for 0<a<1, so that there can't be more solutions.
  5. Mar 28, 2012 #4


    Staff: Mentor

    okay id try n=0, then n=1 then n=2... and see which ones or one are true. basically you're trying to find n.

    also i think you are on the right track using fermat's last theorem (proved by Andrew Wiles in 1995) as an upper bound for n
  6. Mar 28, 2012 #5
    I need to show that n=2. The answer 2 in this problem was obvious. What would be the general solution to such equations if n is not an integer (i.2 we cannot guess it by simple observation).
  7. Mar 28, 2012 #6


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    Science Advisor

    Do you need an exact analytic answer or will a numerical approximation suffice?
  8. Mar 29, 2012 #7


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    Gold Member

    Does n have to be an integer?
  9. Mar 29, 2012 #8
    It doesn't matter if n is an integer or not. I just want a way to solve the equation because suppose guess the answer was not so easy, how can one solve it without of course a computer.
  10. Mar 29, 2012 #9


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    Gold Member

    It resembles a Binet formula.
  11. Mar 29, 2012 #10
    Yes I think it looks like Binet's formula for Fibonacci sequence.
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