Relationship between Chebyshev polynomials and square roots ($\sqrt{3}+\sqrt{2}=\frac{1}{\sqrt{T_1(5)-\sqrt{T_1(5)^2-1}}}$ etc.) (If my English is strange, I would appreciate it if you could correct it.)
There seems to be a property about the sum of square roots. (This is almost self-explanatory.)
let $ a,\ b,\ t \in \mathbb{N}^0,\ a\geq b$
$$ \sqrt{a}+\sqrt{b} = 
    \begin{cases}
         A  &(t\ mod\ 2 = 0)&\\
        \frac{a-b}{A} &(t\ mod\ 2= 1)&
    \end{cases}
$$
$$ A = \sqrt[2^t]{B+(-1)^{t}\sqrt{B^2-(a-b)^{2^t}}}$$
$$ B = \frac{(\sqrt{a}+\sqrt{b})^{2^t}+(\sqrt{a}-\sqrt{b})^{2^t}}{2}$$

Interestingly, $B$ seems to be equal to $T_{2^{t-1}}(a+b)$ when $a-b = 1$.
$T_n(x)$ is Chebyshev polynomials that is expressed by following equation.
$$T_n(x) = n\sum_{k=0}^{n}{(-2)^k\frac{(n+k-1)!}{(n-k)!(2k)!}(1-x)^k}$$
Originally, the range of $x$ is $-1$ to $1$, but this property can be seen when extended to natural numbers.
Please comment if you know anything.

According to wiki, $T_n(x)$ can be expressed by the following equation too.
$$ T_n(x) = \frac{\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n}{2} $$
Hence,
$$ T_{2^{t-1}}(a+b) = \frac{(a+b+\sqrt{(a+b)^2-1})^{2^{t-1}}+(a+b-\sqrt{(a+b)^2-1})^{2^{t-1}}}{2} $$
On the other hands,
\begin{eqnarray*}
  B &=& \frac{(\sqrt{a}+\sqrt{b})^{2^t}+(\sqrt{a}-\sqrt{b})^{2^t}}{2}\\
&=& \frac{(a+b+\sqrt{4ab})^{2^{t-1}}+(a+b-\sqrt{4ab})^{2^{t-1}}}{2}
\end{eqnarray*}
That is, if the square roots are equal, then $T_{2^{t-1}}(a+b)$ and $B$ are equal.
Therefore,
\begin{eqnarray*}
 (a+b)^2-1 &=& 4ab\\
a^2+2ab+b^2-1 -4ab &=& 0\\
(a-b)^2 &=& 1\\
a-b &=&\pm1
\end{eqnarray*}
This indicates that $T_{2^{t-1}}(a+b)$ and $B$ are equal when $a-b =\pm1$.

For example, let $a=3,b=2$,
\begin{eqnarray*}
  \sqrt{3}+\sqrt{2} &=& \frac{1}{\sqrt{T_1(5)-\sqrt{T_1(5)^2-1}}} = \frac{1}{\sqrt{5-\sqrt{24}}}\\
&=& \sqrt[4]{T_2(5)+\sqrt{T_2(5)^2-1}}= \sqrt[4]{49 +\sqrt{2400}}\\
&=& \frac{1}{\sqrt[8]{T_4(5)-\sqrt{T_4(5)^2-1}}} = \frac{1}{\sqrt[8]{4801-\sqrt{23049600}}}\\
&\vdots&
\end{eqnarray*}
