What is the pattern in the powers of $\sqrt{2}-\sqrt{1}$? What is the pattern in this?
$$\begin{align}
\left(\sqrt{2}-\sqrt{1}\right)^1 &= \sqrt{2}-\sqrt{1}\\
\left(\sqrt{2}-\sqrt{1}\right)^2 &= \sqrt{9}-\sqrt{8}\\
\left(\sqrt{2}-\sqrt{1}\right)^3 &= \sqrt{50}-\sqrt{49}\\
\left(\sqrt{2}-\sqrt{1}\right)^4 &= \sqrt{289}-\sqrt{288}\\
\end{align}$$
I thought of applying the binomial theorem
 A: Given that
$$\left(\sqrt{2}-\sqrt{1}\right)^k = \sqrt{A_k}-\sqrt{A_k-1}$$
then
$$\left(\sqrt{2}-\sqrt{1}\right)^{k+1} = \left(\sqrt{2}-\sqrt{1}\right)(\sqrt{A_k}-\sqrt{A_k-1})=\\=\sqrt 2\sqrt{A_k}-\sqrt 2\sqrt{A_k-1}-\sqrt{A_k}+\sqrt{A_k-1}=$$
$$=\sqrt{A_{k+1}}-\sqrt{A_{k+1}-1}$$
with

*

*$\sqrt{A_{k+1}}=\sqrt 2\sqrt{A_k}+\sqrt{A_k-1} \implies A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)}$


*$\sqrt{A_{k+1}-1}=\sqrt 2\sqrt{A_k-1}+\sqrt{A_k}\implies A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)}$
that is

$$A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)}$$

and $A_1=2$.

Using an approach similar to this one, we have
$$A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)} \\\iff 2A_{k+1}-1=3(2A_{k}-1)+2\sqrt2\sqrt{(2A_k-1)^2-1}$$
and by $2A_{k+1}-1=\frac12\left(t_k+\frac1{t_k}\right) \implies t_1=3+2\sqrt 2$ we obtain
$$t_{k+1}+\frac1{t_{k+1}}=3\left(t_k+\frac1{t_k}\right)+2\sqrt 2\sqrt{\left(t_k+\frac1{t_k}\right)^2-4}$$
$$t_{k+1}+\frac1{t_{k+1}}=3\left(t_k+\frac1{t_k}\right)+2\sqrt 2\left(t_k-\frac1{t_k}\right)$$
$$t_{k+1}+\frac1{t_{k+1}}=(3+2\sqrt 2)t_k+\frac{3-2\sqrt 2}{t_k}$$
$$t_{k+1}+\frac1{t_{k+1}}=(3+2\sqrt 2)t_k+\frac{1}{(3+2\sqrt 2)t_k}$$
$$t_{k}=(3+2\sqrt 2)^k$$
and then

$$A_k=\frac{(3+2\sqrt 2)^k+(3-2\sqrt 2)^k+2}{4}$$

A: If $(\sqrt 2  - 1)^n  = \sqrt {a_n }  - \sqrt {a_n  - 1}$, then $
(\sqrt 2  + 1)^n  = \sqrt {a_n }  + \sqrt {a_n  - 1}$. Hence,
$$
a_n  = \left( {\frac{{(\sqrt 2  + 1)^n  + (\sqrt 2  - 1)^n }}{2}} \right)^2  = \frac{{2 + (3 + 2\sqrt 2 )^n  + (3 - 2\sqrt 2 )^n }}{4}.
$$
This is A115599 in the OEIS. See also A055997. You can show using the binomial theorem that $a_n$ is always an integer. In fact $$
a_n  = \frac{{3^n  + 1}}{2} + \sum\limits_{k = 1}^{\left\lfloor {n/2} \right\rfloor } \binom{n}{2k}2^{3k - 1} 3^{n - 2k}  . 
$$
A: We have $$(\sqrt{1}-\sqrt{2})^k = \sqrt{a_k^2} - \sqrt{2b_k^2}$$
where $a_k,b_k$ obey the following recurrence relationships:
$$a_{k+1}=a_{k}+2b_{k}\\
b_{k+1}=a_{k}+b_{k}$$
with $a_{1}=1$ and  $b_1 = 1$.  This can be proved inductively. To understand the recurrence relationship we can recast it as a matrix equation:
$${\begin{pmatrix}
a_{k+1}\\
b_{k+1}
\end{pmatrix}} = \begin{pmatrix}
1 & 2 \\
1 & 1 
\end{pmatrix} \begin{pmatrix}
a_k\\
b_k
\end{pmatrix} $$
The eigenvectors are $\begin{pmatrix}
\sqrt{2} \\
1 
\end{pmatrix}$, $\begin{pmatrix}
\sqrt{2} \\
-1 
\end{pmatrix}$ with coresponding eigenvalues: $1+\sqrt{2}$ and $1-\sqrt{2}$. 
This allows us to solve:
$${\begin{pmatrix}
a_{n}\\
b_{n}
\end{pmatrix}} = {\begin{pmatrix}
1 & 2 \\
1 & 1 
\end{pmatrix}}^{n-1} \begin{pmatrix}
1\\
1
\end{pmatrix} $$ with: $$a_{n} = \frac{(1+\sqrt{2})^n + (1-\sqrt{2})^n}{2}\\
b_{n} = \frac{(1+\sqrt{2})^n - (1-\sqrt{2})^n}{2\sqrt{2}}$$
A recurrence relationship can also be written for each term separately as:
$$a_{n} = 2a_{n-1} + a_{n-2}\\
b_{n} = 2 b_{n-1} + b_{n-2}$$
with starting terms $a_1 = 1$, $a_2 = 3$, $b_1 = 1$, $b_2 = 2$.
We can write it as a recurrence relationship or a matrix equation if we wish to use only integers, or we can write the terms individually, but that causes the reappearance of $\sqrt{2}$ in the formula.
