Efficiently Solve Equations like a^n + b^n = c^n with These Tips

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Discussion Overview

The discussion revolves around solving equations of the form a^n + b^n = c^n, particularly focusing on a specific case involving square roots and exploring the nature of solutions when n is not necessarily an integer. Participants explore various methods and insights related to this type of equation.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the equation a^n + b^n = c^n and seeks a solution for n, noting its resemblance to Fermat's equation.
  • Another participant reformulates the problem to a specific case involving square roots and suggests that n=2 is a solution based on observation.
  • Some participants discuss the behavior of the function as n varies, with one noting that the function is decreasing for certain ranges of a.
  • There is a suggestion to test various integer values of n (0, 1, 2) to find valid solutions.
  • One participant questions whether n must be an integer, while another clarifies that it does not need to be, emphasizing the need for a method to solve the equation without computational aid.
  • References to Binet's formula for the Fibonacci sequence are made, with some participants drawing parallels to the structure of the equation being discussed.

Areas of Agreement / Disagreement

Participants express differing views on the nature of solutions, particularly regarding whether n must be an integer and the methods to approach solving the equation. There is no consensus on a general solution for non-integer n.

Contextual Notes

Some limitations include the lack of clarity on the assumptions regarding the values of a and b, as well as the dependence on specific forms of the equations presented. The discussion does not resolve the mathematical steps necessary for a complete solution.

Who May Find This Useful

This discussion may be of interest to those exploring advanced mathematical equations, particularly in the context of number theory and algebraic structures.

medwatt
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Hello.
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.
Thanks
 
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Ok the equation I'm trying to solve is :

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

I have arrived to this point:

\left[\sqrt{2+\sqrt{3}}\right]^{2n} + \left[2*\sqrt{2+\sqrt{3}}]\right]^{n} - 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:
So you want to solve

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

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.
 
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
 
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).
 
medwatt said:
Ok the equation I'm trying to solve is :

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

I have arrived to this point:

\left[\sqrt{2+\sqrt{3}}\right]^{2n} + \left[2*\sqrt{2+\sqrt{3}}]\right]^{n} - 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.

Do you need an exact analytic answer or will a numerical approximation suffice?
 
Does n have to be an integer?
 
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.
 
It resembles a Binet formula.
 
  • #10
Yes I think it looks like Binet's formula for Fibonacci sequence.
 

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