square root / factor problem $(A/B)^{13} - (B/A)^{13}$ Let
$A=\sqrt{13+\sqrt{1}}+\sqrt{13+\sqrt{2}}+\sqrt{13+\sqrt{3}}+\cdots+\sqrt{13+\sqrt{168}}$ and
$B=\sqrt{13-\sqrt{1}}+\sqrt{13-\sqrt{2}}+\sqrt{13-\sqrt{3}}+\cdots+\sqrt{13-\sqrt{168}}$.
Evaluate $(\frac{A}{B})^{13}-(\frac{B}{A})^{13}$.
By Calculator, I have $\frac{A}{B}=\sqrt{2}+1$ and $\frac{B}{A}=\sqrt{2}-1$.
But, I don't know how. Has someone any idea about this.
 A: Let
$$A=\sum_{n=1}^{168}\sqrt{13+\sqrt{n}},B=\sum_{n=1}^{168}\sqrt{13-\sqrt{n}}$$
since
$$\sqrt{2}A=\sum_{n=1}^{168}\sqrt{26+2\sqrt{n}}=\sum_{n=1}^{168}\left(\sqrt{13+\sqrt{169-n}}+\sqrt{13-\sqrt{169-n}}\right)=A+B$$ so we have $x=\dfrac{A}{B}=\sqrt{2}$,then we have
$$x=\sqrt{2}+1,\dfrac{1}{x}=\sqrt{2}-1\Longrightarrow x+\dfrac{1}{x}=2\sqrt{2}$$
let
$$a_{n}=x^n-x^{-n}$$ use this well know indentity
$$a_{n+2}=(x+\dfrac{1}{x})a_{n+1}-a_{n}\Longrightarrow a_{n+2}=2\sqrt{2}a_{n+1}-a_{n}$$
$$a_{1}=2,a_{2}=4\sqrt{2}$$
so 
$$a_{3}=2\sqrt{2}a_{2}-a_{1}=16-2=14$$
$$a_{4}=2\sqrt{2}a_{3}-a_{2}=28\sqrt{2}-4\sqrt{2}=24\sqrt{2}$$
$$a_{5}=2\sqrt{2}a_{4}-a_{3}=96-14=82$$
$$\cdots$$
A: This took me some time to solve. Here you go:
First, we find this:
$$\begin{aligned}
(\sqrt{13+\sqrt{a}}-\sqrt{13-\sqrt{a}})^2
&=13+\sqrt{a}+13-\sqrt{a}-2\sqrt{13+\sqrt{a}}\sqrt{13-\sqrt{a}}\\
&=2(13-\sqrt{169-a})
\end{aligned}$$
So,
$$\sqrt{13+\sqrt{a}}-\sqrt{13-\sqrt{a}}=\sqrt{2}\sqrt{13-\sqrt{169-a}}$$
By what we have, we write,
$A-B=\sqrt{2}B$
What I used here is the fact that I've summed over all $a$ from $1$ to $168$, and that summing with $\sqrt{169-a}$ is the same as summing with $\sqrt{a}$ in this question.
Now, we have $A=(1+\sqrt{2})B$
Thus, $\frac{A}{B}=1+\sqrt{2}$ and $\frac{B}{A}=\sqrt{2}-1$
We just calculate $(\frac{A}{B})^{13}$ and $(\frac{B}{A})^{13}$ which I believe is okay to be done using calculator. Else, comment so I can edit my answer.
