<s>[INST] How to proceed solving $(7+4\sqrt3)^m+(7-4\sqrt3)^m=14$? I have$$(7+4\sqrt3)^m+(7-4\sqrt3)^m=14$$
By noticing that $7+4\sqrt3=\frac1{7-4\sqrt3}$ one way to solve the equation is using substitution $(7+4\sqrt3)^m=t$ and solve for $t$ in $t+\frac1t=14$.

But I'm trying to use a little different approach:
We have $7+4\sqrt3+7-4\sqrt3=14$ So by using the substitution $u=7+4\sqrt3$ we have :
$$u^m+\frac1{u^m}=u+\frac1u$$
But from here how can I prove mathematically that the only answers are $m=1$ and $m=-1$ ?
 [/INST] $m = 1$ is a solution, which is easily seen. Suppose that $m \neq 1$. Clearly, $m = 0$ is not a solution Because $u + \dfrac{1}{u} \neq 2$.
First, suppose that $m > 0$. Then, we would have
$$u^m - u = \dfrac{1}{u} - \dfrac{1}{u^m}.$$
Here, we can use $a^k - b^k = \left( a - b \right) \left( a^{k - 1} + a^{k - 2}b + \cdots + b^{k - 1} \right)$ after taking $u$ (respectively $\dfrac{1}{u}$) common.
Hence, we get
$$u \left( u - 1 \right) \left( 1 + u + u^2 + \cdots u^{m - 2} \right) = \dfrac{1}{u} \left( 1 - \dfrac{1}{u} \right) \left( 1 + \dfrac{1}{u} + \dfrac{1}{u^2} + \cdots + \dfrac{1}{u^{m - 2}} \right).$$
The right-hand side can be simplified as
$$u \left( u - 1 \right) \left( 1 + u + u^2 + \cdots + u^{m - 2} \right) = \dfrac{1}{u^m} \left( u - 1 \right) \left( 1 + u + u^2 + \cdots + u^{m - 2} \right).$$
Now, we know that $u \neq 0$, $u \neq 1$ and $u > 0$. Hence, most of the terms cancel off and we get
$$u = \dfrac{1}{u^m}.$$
This tells us that $m = -1$, which is a contradiction to our assumption.
Now, for $m < 0$, we let $m = -n$ for some $n > 0$. Then, we have
$$u^{-n} - u = \dfrac{1}{u} - \dfrac{1}{u^{-n}}.$$
Then,
$$u^{-n} \left( 1 - u^{n + 1} \right) = \dfrac{1}{u} \left( 1 - u^{n + 1} \right).$$
This gives us $\dfrac{1}{u^n} = \dfrac{1}{u}$, and since $u > 0$, we conclude that $n = 1$. That is, $m = -1$.
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