Prove that there exists an integer $k$ such that $x+y\alpha={\alpha}^k$ 
Let $x,y,z,w$ be integers such that $(x+y\alpha)(z+w\alpha)=1$ and  $x+y\alpha>0$. Prove that there exists an integer $k$ such that $x+y\alpha={\alpha}^k$ where $\alpha=\frac{1+\sqrt{5}}{2}$.

Note that $\alpha^2 = \alpha+1$. We have $$(x+y\alpha)(z+w\alpha) = xz+wx\alpha+yz\alpha+yw\alpha^2-1 = 0.$$ This is the same as $xz+yw-1+\alpha(wx+yz+yw) = 0$. Therefore, $wx+yz+yw = 0$ and $xz+yw = 1$. Thus from the first equation we get $w = -\dfrac{yz}{x+y}$. Then substituting this into the equation $xz+yw = 1$ we get $$xz-\dfrac{y^2z}{x+y} = \dfrac{z(x(x+y)-y^2)}{x+y} = 1.$$ Therefore, $z(x(x+y)-y^2) = x+y$.
I didn't see how to continue.
 A: As you've already written, we have
$$wx+yz+yw = 0\quad\text{and}\quad xz+yw = 1$$
Representing $z,w$ by $x$ and $y$, we get
$$z=\frac{x+y}{x^2+xy-y^2},\quad w=\frac{-y}{x^2+xy-y^2}$$
implying that we have to have
$$\frac{x}{x^2+xy-y^2},\frac{y}{x^2+xy-y^2}\in\mathbb Z$$
which implies that
$$\small\left(\frac{x}{x^2+xy-y^2}\right)^2+\frac{x}{x^2+xy-y^2}\cdot \frac{y}{x^2+xy-y^2}-\left(\frac{y}{x^2+xy-y^2}\right)^2=\frac{1}{x^2+xy-y^2}\in\mathbb Z$$
from which we have to have
$$|x^2+xy-y^2|=1\tag1$$
Now, it is known that $(x,y)=(F_n,F_{n+1})$ with $n\ge 1$, i.e. two successive terms of the Fibonacci sequence, are the only positive solutions for $(1)$. (see, for example, here for a proof)


*

*Case 1 : If both $x,y$ are positive, then we get $(x,y)=(F_n,F_{n+1})$ where $F_n=\frac{\alpha^n-\left(-\frac{1}{\alpha}\right)^n}{\sqrt 5}$ with $n\ge 1$, so$$x+y\alpha=\frac{\alpha^n-\left(-\frac{1}{\alpha}\right)^n}{\sqrt 5}+\frac{\alpha^{n+1}-\left(-\frac{1}{\alpha}\right)^{n+1}}{\sqrt 5}\cdot \alpha=\alpha^n\cdot\frac{\alpha^2+1}{\sqrt 5}=\alpha^{n+1}$$

*Case 2 :  If $xy=0$, then with $x+y\alpha\gt 0$, we have $(x,y)=(0,1),(1,0)$, and$$x+y\alpha=0+1\cdot\alpha=\alpha^1,\qquad x+y\alpha=1+0\cdot\alpha=\alpha^0$$respectively.

*Case 3 : If $x\gt 0$ and $y\lt 0$, then we get$$(1)\iff |(-y)^2+x(-y)-x^2|=1$$So, $(x,y)=(F_{n+1},-F_n)$ are the only solutions for $(1)$ in this case, and we get$$x+y\alpha=\frac{\alpha^{n+1}-\left(-\frac{1}{\alpha}\right)^{n+1}}{\sqrt 5}-\frac{\alpha^{n}-\left(-\frac{1}{\alpha}\right)^{n}}{\sqrt 5}\cdot \alpha=(-1)^n\alpha^{-n}$$From $x+y\alpha\gt 0$, $n$ has to be even, and if $n$ is even, then $x+y\alpha$ is of the form $\alpha^k$ where $k$ is an integer.

*Case 4 : If $x\lt 0$ and $y\gt 0$, then we get$$(1)\iff |y^2+(-x)y-(-x)^2|=1$$So, $(x,y)=(-F_{n+1},F_n)$ are the only solutions for $(1)$ in this case, and we get $$x+y\alpha=-\frac{\alpha^{n+1}-\left(-\frac{1}{\alpha}\right)^{n+1}}{\sqrt 5}+\frac{\alpha^{n}-\left(-\frac{1}{\alpha}\right)^{n}}{\sqrt 5}\cdot \alpha=(-1)^{n+1}\alpha^{-n}$$From $x+y\alpha\gt 0$, $n$ has to be odd, and if $n$ is odd, then $x+y\alpha$ is of the form $\alpha^k$ where $k$ is an integer.
