<s>[INST] Use the binomial theorem to give a formula for positive integers $x_{k}$ and $y_{k}$ such that $(3 + 2\sqrt{2})^{^{x}} = x_{k} + y_{k}\sqrt{2}$. Use the binomial theorem to give a formula for positive integers $x_{k}$ and $y_{k}$ such that

$$(3 + 2\sqrt{2})^{^{x}} = x_{k} + y_{k}\sqrt{2}.$$

Is this simply just applying the binomial theorem?
I get something like 
$$(3 + 2\sqrt{2})^{^{k}} = (1 + (2 + 2\sqrt{2}))^{^{k}} $$
$$= \sum_{j=0}^{k} \binom{k}{j} 2^{^{j}} (1 + \sqrt{2})^{^{j}}$$
but I am struggling to proceed from here. I don't know how to arrive to the last part of the equality, $x_{k} + y_{k}\sqrt{2}$. Any help would be greatly appreciated.
Edit: I guess I'm rather confused about what the question is asking me to do exactly. It's saying that I should find one formula for $x_{k}$ and one formula for $y_{k}$ individually, right?
 [/INST] $$(3+2\sqrt{2})^k= \sum_{j=0}^{k} \binom{k}{j} 3^{^{k-j}} (2\sqrt{2})^{^{j}}$$
Now split the sum in terms with $j$ odd and $j$ even. The part where all $j$ are even is an integer, while the art where $j$ is odd has the form integer times $\sqrt{2}$.
Added: 
$$ \sum_{j=0}^{k} \binom{k}{j} 3^{^{k-j}} (2\sqrt{2})^{^{j}}=\\
=\sum_{j=0}^{\lfloor \frac{k}{2} \rfloor} \binom{k}{2j} 3^{^{k-2j}} (2\sqrt{2})^{^{2j}}+\sum_{j=0}^{\lfloor \frac{k-1}{2} \rfloor} \binom{k}{2j+1} 3^{^{k-2j-1}} (2\sqrt{2})^{^{2j+1}}\\
=\sum_{j=0}^{\lfloor \frac{k}{2} \rfloor} \binom{k}{2j} 3^{^{k-2j}} 8^{^{j}}+\sum_{j=0}^{\lfloor \frac{k-1}{2} \rfloor} \binom{k}{2j+1} 3^{^{k-2j-1}} 8^j (2\sqrt{2})\\
=\sum_{j=0}^{\lfloor \frac{k}{2} \rfloor} \binom{k}{2j} 3^{^{k-2j}} 8^{^{j}}+2\sqrt{2}\sum_{j=0}^{\lfloor \frac{k-1}{2} \rfloor} \binom{k}{2j+1} 3^{^{k-2j-1}} 8^j \\$$
Thus
$$x_k= \sum_{j=0}^{\lfloor \frac{k}{2} \rfloor} \binom{k}{2j} 3^{^{k-2j}} 8^{^{j}}\\
y_k=2\sum_{j=0}^{\lfloor \frac{k-1}{2} \rfloor} \binom{k}{2j+1} 3^{^{k-2j-1}} 8^j $$
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