# How to solve this equation

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

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.

jedishrfu
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

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).

chiro
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?

coolul007
Gold Member
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.

coolul007
Gold Member
It resembles a Binet formula.

Yes I think it looks like Binet's formula for Fibonacci sequence.