A relation between $3-\sqrt 5$ and $\sqrt 5-1$

Question

Let $a=\sqrt{1+\cfrac{\sqrt{5}+1}{2}}$ and let $b=\sqrt{1+\cfrac{\color{red}{\sqrt{5}-1}}{2}}$. Then $$\bigg(a+\frac 1a + b + \frac 1b\bigg)\bigg(a-\frac 1a+b-\cfrac 1b\bigg)=\color{blue}{3-\sqrt 5}.$$ Note that, curiously, $$\color{blue}{3-\sqrt 5} = \cfrac{1}{1+\cfrac{1}{4-\cfrac{1}{1+\ddots}}}\quad\text{and}\quad \color{red}{\sqrt 5 - 1}=\cfrac{1}{1-\cfrac{1}{4+\cfrac{1}{1-\ddots}}}$$

Is there an explanation for these strange relationships? In fact, does this mean anything at all?

The structure of the continued fractions probably have to do with the following fact, which isn't too difficult to prove.

$$\frac 2a=\cfrac{1}{a+\cfrac{1}{b-\cfrac{1}{a+\ddots}}}+\cfrac{1}{a-\cfrac{1}{b+\cfrac{1}{a-\ddots}}}\tag*{$\forall a,b$}$$ If so, the continued fraction relationship between $3-\sqrt 5$ and $\sqrt 5 -1$ is merely a special case of the theorem above when $(a,b)=(1,4)$, in which the appearance of $\sqrt 5$ is related to the following similarly structured continued fractions. $$\frac{1+\sqrt 5}{2}=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ddots}}}\quad\text{and}\quad \sqrt {5}=2+\cfrac{1}{4+\cfrac{1}{4+\cfrac{1}{4+\ddots}}}$$ However, it all seems a bit too vague for my understanding. I'm sure I could find a way to relate every number to each other in maths, but are these relationships particularly special? To me, they definitely are of special interest, but that is subjective.

Answer 1

To me, the relation between them is $(\sqrt5-1)^2=2(3-\sqrt5)$.

Answer 2

Actually there's a typo at your first Eq, $$\left(a+\frac{1}{a}+b+\frac{1}{b}\right)\left(a-\frac{1}{a}+b-\frac{1}{b}\right)≠3-\sqrt{5}$$

The correct one will be, $$\left(a+\frac{1}{a}+b+\frac{1}{b}\right)\left(a\color{red}{+}\frac{1}{a}\color{red}{-}b-\frac{1}{b}\right) \tag{1}$$

Take $\varphi=(\sqrt{5}+1)/2$. We know $\varphi$ is a root of $x^2-x-1=0$. From this we can say that; $$\varphi+1=\varphi^2\tag{2}$$ $$\varphi-1=1/\varphi=(\sqrt{5}-1)/2 \tag{3}$$

Your $a,b$ can be simplified as, $$(a,b)=(\varphi,\sqrt{\varphi})$$ So Eq $1$, will be, (by solely using the relations $2,3$) \begin{align} & = \left(a+\frac{1}{a}\right)^{2}-\left(b+\frac{1}{b}\right)^{2} \\ & = \left(2φ-1\right)^{2}-\left(\sqrt{φ}+\frac{1}{\sqrt{φ}}\right)^{2} \\ & = 5-φ^{3} \\ & = 5-\left(2φ+1\right) \\ & = 2\left(2-φ\right) \\ \end{align} Since, $φ=\frac{1}{φ}+1$, \begin{align} & = 2\left(1-\frac{1}{φ}\right) \\ & = \frac{2}{φ^{2}} \\ & = 2\left(\sqrt{5}-1\right)^{2} \\ \end{align}

There's also no need for criminally expanding your question, when the answer you seek is just $2\left(\sqrt{5}-1\right)^{2}$. Also it might be good to see other little or two things on golden ratio which can help like,

$$φ^{n}=φ^{n-1}+φ^{n-2}$$ $$φ^{n}=F_{n-1}+F_{n}φ$$ $$\sqrt{1+\sqrt{φ}}=\frac{1}{\sqrt{2}}\left(\sqrt[4]{φ^{3}}+\sqrt[4]{φ^{-3}}\right)$$ And etc. Where $F_{n}$ is the $n^{th}$ Fibonacci number, and $F_3=2$.