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

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.
