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

  • Thread starter Thread starter medwatt
  • Start date Start date
AI Thread Summary
The discussion revolves around solving the equation a^n + b^n = c^n, specifically for the case where a = √(2+√3), b = √(2-√3), and c = 2. The user has identified that n=2 is a solution and is exploring whether there are other solutions, particularly for non-integer values of n. They reference Fermat's Last Theorem to suggest that integer solutions are limited, and they seek a method to solve the equation analytically or numerically without relying on computational tools. The conversation highlights the challenge of finding general solutions for non-integer n and draws parallels to Binet's formula related to the Fibonacci sequence.
medwatt
Messages
122
Reaction score
0
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
 
Mathematics news on Phys.org
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.
 

Similar threads

3
Replies
105
Views
6K
Replies
1
Views
1K
Replies
7
Views
2K
Replies
19
Views
2K
Replies
4
Views
2K
Replies
2
Views
2K
Back
Top