Expansion of $(1+\sqrt{2})^n$ I was asked to show that $\forall n\in \mathbb N$ there exist a $p\in \mathbb N^\ast$ such that $$(1+\sqrt{2})^n = \sqrt{p} + \sqrt{p-1}$$ 
I used induction but it wasn't fruitful, so I tried to use the binomial expansion of $(1+\sqrt{2})^n$ but it seems I lack some insight to go further. 
Any hint is welcomed.

Note:
I added some lines to 
show this explicit formula:
$p
=1+\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k-1}
$.
If
$(1+\sqrt{2})^n = \sqrt{p} + \sqrt{p-1}
$,
then
$(1+\sqrt{2})^{2n} = 2p-1+2\sqrt{p(p-1)}
$
so that
$\begin{array}\\
4p(p-1)
&=((1+\sqrt{2})^{2n} - (2p-1))^2\\
&=(1+\sqrt{2})^{4n}-2(1+\sqrt{2})^{2n} (2p-1)+(2p-1)^2\\
&=(1+\sqrt{2})^{4n}-2(1+\sqrt{2})^{2n} (2p-1)+4p^2-4p+1\\
\text{so that}\\
-1
&=(1+\sqrt{2})^{4n}-2(1+\sqrt{2})^{2n} (2p-1)\\
\implies\\
2p-1
&=\frac{(1+\sqrt{2})^{4n}+1}{2(1+\sqrt{2})^{2n} }\\
&=\frac12(1+\sqrt{2})^{2n}+\frac{1}{2(1+\sqrt{2})^{2n} }\\
&=\frac12(1+\sqrt{2})^{2n}+\frac12(-1+\sqrt{2})^{2n}
\qquad\text{ since} (1+\sqrt{2})(-1+\sqrt{2})=1\\
&=\frac12\sum_{k=0}^{2n} \binom{2n}{k}(\sqrt{2})^{2n-k}(1+(-1)^k)\\
&=\sum_{k=0}^{n} \binom{2n}{2k}(\sqrt{2})^{2n-2k}\\
&=\sum_{k=0}^{n} \binom{2n}{2k}2^{n-k}\\
&=\binom{2n}{2n}2^{n-n}+\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k}\\
&=1+\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k}\\
&=1+2\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k-1}\\
\text{so}\\
p
&=1+\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k-1}\\
\end{array}
$
