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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.

user19405892
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    This means $x+y\alpha$ is a unit in the ring of integers of $\mathbb{Q}(\sqrt{5})$, which are of the form $\pm\left(\frac{1+\sqrt{5}}{2}\right)^k$ by Dirichlet's unit theorem. – Tob Ernack Jun 05 '17 at 17:46
  • I think the proof is something like: if $x+y\alpha \neq 1$, then either $(x+y\alpha) \cdot \alpha^{-1} = (x+y\alpha) \cdot (\alpha-1)$ or $(x+y\alpha) \cdot \alpha$ has a smaller measure than $x+y\alpha$, for a suitable choice of "measure". – Daniel Schepler Jun 05 '17 at 17:49
  • Do you know about the norm in number fields? – Michael Burr Jun 05 '17 at 18:08
  • I think something must be missing. Assume $x=1,y=3$. Then $z=\frac{1-(4\alpha + 3)w}{1+3\alpha}$ and $(1+3\alpha)(z+w\alpha)=1$ and $1+3\alpha>0$. However $1+3\alpha \neq \alpha^k$ for all integer $k$. $\alpha^3 \approx 4.2360679775$ $\alpha^4\approx 6.85410196625$, whereas $1+3\alpha\approx 5.85410196625$. BTW, FWIW, $\alpha^n=\alpha^{n-1}+\alpha^{n-2}$. – Χpẘ Jun 06 '17 at 00:49
  • @Χpẘ but is the $z$ that you obtain an integer? – Tob Ernack Jun 06 '17 at 02:09
  • @Tob Good point - I missed that. After looking at this more, I see that $z=\frac{1}{1+3\alpha}-\alpha w$. For $z,w$ to both be integer, then for some integer $k$:, $\alpha w-k-\frac{1}{1+3\alpha}=0$. Since $\frac{1}{1+3\alpha} < 1$ then $k=\lfloor\alpha w\rfloor$. Towards that end I graphed all the integers, $w$, $-5000...5000$ using the function $\alpha w-\lfloor\alpha w\rfloor$, here. Blue line is $y=\frac1{1+3\alpha}$. Some points come very close. $w=3026$ gives $z=-4896.00002956$. So it seems plausible that another $w$ gives integer $z$. – Χpẘ Jun 07 '17 at 16:22

1 Answers1

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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.

mathlove
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