(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*}