<s>[INST] 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
 [/INST] 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 $$
a_n  = \frac{{3^n  + 1}}{2} + \sum\limits_{k = 1}^{\left\lfloor {n/2} \right\rfloor } \binom{n}{2k}2^{3k - 1} 3^{n - 2k}  . 
$$
</s>