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?
 A: $$(3+2\sqrt{2})^k= \sum_{j=0}^{k} \binom{k}{j} 3^{^{k-j}} (2\sqrt{2})^{^{j}}$$
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
$$x_k= \sum_{j=0}^{\lfloor \frac{k}{2} \rfloor} \binom{k}{2j} 3^{^{k-2j}} 8^{^{j}}\\
y_k=2\sum_{j=0}^{\lfloor \frac{k-1}{2} \rfloor} \binom{k}{2j+1} 3^{^{k-2j-1}} 8^j $$
A: $(3+2\sqrt{2})^k = x_k + y_k\sqrt{2}$. Thus: $x_{k+1} + y_{k+1}\sqrt{2} = (3+2\sqrt{2})^{k+1} = (3+2\sqrt{2})^k\cdot (3+2\sqrt{2}) = (x_k+y_k\sqrt{2})(3+2\sqrt{2}) = 3x_k + 2x_k\sqrt{2} + 3y_k\sqrt{2} + 4y_k = 3x_k + 4y_k + (2x_k + 3y_k)\sqrt{2}$. Thus we have a recursive linear system relating $x_k$'s and $y_k$'s as follows:
$x_{k+1} = 3x_k + 4y_k$
$y_{k+1} = 2x_k + 3y_k$
with initial value: $(x_1,y_1) = (3,2)$. Using system of ODE or finite difference we can progress to the final answer for an inductive formula about $x_k$ and $y_k$. You can continue...
A: Split up the odd and even terms.
\begin{align}
(3+2\sqrt 2)^n
&=\sum^n_{k=0}\binom{n}{k}3^{n-k}2^{k}2^{k/2}\\
&=\sum^{\lfloor \frac{n}{2}\rfloor}_{k=0}\binom{n}{2k}3^{n-2k}2^{3k}+\sqrt 2 \sum^{\lfloor \frac{n}{2}\rfloor}_{k=0}\binom{n}{2k+1}3^{n-2k-1}2^{3k+1}\\
&=\color\red{\sum_{k \ge0}\binom{n}{2k}3^{n-2k}(2\sqrt 2)^{2k}}+\frac{\sqrt 2}{\color\green{\sqrt 2}}\color\green{ \sum_{k \ge 0}\binom{n}{2k+1}3^{n-2k-1}(2\sqrt 2)^{2k+1}}\\
&=\color\red{\frac{(3+2\sqrt 2)^n+(3-2\sqrt 2)^n}{2}}+\sqrt 2 \color\green{\frac{(3+2\sqrt 2)^n-(3-2\sqrt 2)^n}{2\sqrt 2}}\\
&=\color\red{x_n}+\color\green{y_n}\sqrt 2
\end{align}
Though accurate, this "formula" isn't going to help much in determining the actual value of $(3+2\sqrt 2)^n$. Apply the binomial theorem and compiling the terms straight away would be much faster
