Contents
1 Introduction
This is a followup to my post on Euler's Identity. After complaining that I think Euler's identity is pretty ugly, I made the claim that the following equation was more beautiful.
$$2\mathrm{cos}(i\mathrm{log}(1+\varphi ))=3.$$ 
Actually, the original was $2\mathrm{sin}(i\mathrm{log}(\varphi ))=i$ but I prefer the above.
The occurrence of two similar but distinct equations involving $\varphi $ led to me, and one or two others, to wonder about more such numbers. I had a bit of fun playing with the resulting identities and would like to record what I found out.
This is not a deep mathematical investigation. The paths it led me down are far from my own mathematical specialisation so I cannot be sure that I am even asking the right questions, let alone knowing which such questions have obvious or wellknown answers and which don't.
The point is to show that even from something silly, it is possible just to have fun with mathematics.
2 Golden Numbers
Let's restate the equations:
$$\begin{array}{rl}2\mathrm{sin}(i\mathrm{log}(\varphi ))& =i,\\ 2\mathrm{cos}(i\mathrm{log}(1+\varphi ))& =3.\end{array}$$ 
My goal was to start from ${e}^{it}=\mathrm{cos}(t)+i\mathrm{sin}(t)$ and get something other than Euler's identity. Rearranging, we get the formulae for $\mathrm{cos}(t)$ and $\mathrm{sin}(t)$ in terms of ${e}^{it}$ and ${e}^{it}$. The breakthrough was the realisation that if $x={e}^{it}$ then ${e}^{it}={x}^{1}$ so I was looking at expressions of the form $x+{x}^{1}$ and $x{x}^{1}$; to wit, the golden ratio. At that point, it is simply a matter of plugging them in and seeing which combinations give nice numbers.
But why stop there? The great thing about $\varphi $ and $1+\varphi $ was that they solve, respectively, $\varphi {\varphi}^{1}=1$ and $(1+\varphi )+(1+\varphi {)}^{1}=3$. The nice property here is that the right hand side consists of integers. But $1$ and $3$ are not the only integers. Let's define:
The Golden Numbers are the solutions of:
$$x\pm {x}^{1}\in \mathbb{Z}$$ 
Specifically, for $n\in \mathbb{Z}$ let ${\varphi}_{n,1}$ be a solution of $x{x}^{1}=n$ and ${\varphi}_{n,1}$ a solution of $x+{x}^{1}=n$.
To link back to the original purpose, we have:
$$\begin{array}{rl}2\mathrm{cos}i\mathrm{log}{\varphi}_{n,1}& =n\\ 2\mathrm{sin}i\mathrm{log}{\varphi}_{n,1}& =ni\end{array}$$ 
Notice that we say "a solution" instead of "the solution". There will usually be two, but knowing one tells us the other (via reciprocation and, for ${\varphi}_{n,1}$, a sign flip) and it is more convenient not to have to distinguish. Thus when we say "$x={\varphi}_{n,\alpha}$" we shall actually mean that $x$ satisfies the equation for ${\varphi}_{n,\alpha}$. The apparent sign flip in the definition is because it gives us the original golden ratio as ${\varphi}_{1,1}$. We can also arrange the defining equation to be such that ${\varphi}_{n,\alpha}$ is a solution of:
$${x}^{2}=nx+\alpha $$ 
which makes the sign convention a little more plausible. From this form, we can derive the formulae:
$${\varphi}_{n,\alpha}=\frac{n\pm \sqrt{{n}^{2}+4\alpha}}{2}$$ 
Of more interest, though, are relations between them.
3 Relating the Golden Numbers
The first observation to make is that ${\varphi}_{n,\alpha}={\varphi}_{n,\alpha}$. Thus we only need to consider $n\in \mathbb{N}$. The second is that ${\varphi}_{n,1}$ is only real for $\leftn\right\ge 2$, and ${\varphi}_{2,1}$ is unique in being single valued (i.e., $1$).
Of more interest is finding relations between them corresponding to the original two equations involving the golden ratio itself. We have ${\varphi}_{1,1}=\varphi $ and so the original equations can be summed up as saying ${\varphi}_{3,1}=1+{\varphi}_{1,1}$.
As our focus has been on integers, it seems reasonable to extend this to consider transformations of the form $a{\varphi}_{n,\alpha}+b$ with $a,b\in \mathbb{Z}$. That is to say, for $a,b\in \mathbb{Z}$, when is $a{\varphi}_{n,\alpha}+b$ another ${\varphi}_{m,\beta}$?
We can test this by simple substitution:
$$\begin{array}{rl}(a{\varphi}_{n,\alpha}+b{)}^{2}& ={a}^{2}{\varphi}_{n,\alpha}^{2}+2ab{\varphi}_{n,\alpha}+{b}^{2}\\ & ={a}^{2}(n{\varphi}_{n,\alpha}+\alpha )+2ab{\varphi}_{n,\alpha}+{b}^{2}\\ & ={a}^{2}\alpha +(2b+na)(a{\varphi}_{n,\alpha}+b)2{b}^{2}nab+{b}^{2}\\ & =(2b+na)(a{\varphi}_{n,\alpha}+b)+\alpha {a}^{2}{b}^{2}nab\end{array}$$ 
Thus if $\alpha {a}^{2}{b}^{2}nab=\pm 1=\beta $ then $a{\varphi}_{n,\alpha}+b={\varphi}_{2b+na,\beta}$. Hence we look for solutions of:
$$\alpha {a}^{2}nab{b}^{2}=\beta $$ 
with $a,b,n\in \mathbb{Z}$ and $\alpha ,\beta \in \{1,1\}$. This is a Diophantine equation. We shall specify a solution by giving $(a,b,n,\alpha )$ and shall always use $\beta $ to refer to the resulting right hand side (thus $\beta =\alpha {a}^{2}nab{b}^{2}$).
There are some particular cases that will need casebycase analysis. To see where they lie, let's start with the general case. This is if $a,b>2$. The point about this restriction is that it allows us to work modulo $a$ and modulo $b$ and still be able to distinguish $\alpha $ and $\beta $. In particular, if we work modulo $a$ we obtain ${b}^{2}\equiv \beta moda$ whilst modulo $b$ shows us that $\alpha {a}^{2}\equiv \beta modb$.
Thus a solution requires $(a,b)$ with ${b}^{2}=\pm 1moda$, and ${a}^{2}=\pm 1modb$. On the other hand, given such a pair, we set $\beta $ to be such that ${b}^{2}=\beta moda$ and $\alpha $ such that $\alpha {a}^{2}=\beta modb$. Then $\alpha {a}^{2}{b}^{2}\beta \equiv 0$ modulo both $a$ and $b$. Since ${b}^{2}=\pm 1moda$, there is some integer $k$ such that $\mp {b}^{2}+ka=1$ whence $\mathrm{gcd}(a,b)=1$. Thus as $\alpha {a}^{2}{b}^{2}\beta \equiv 0$ modulo both $a$ and $b$, it must be that $ab$ divides $\alpha {a}^{2}{b}^{2}\beta $. Hence there is some $n$ such that $\alpha {a}^{2}nab{b}^{2}=\beta $.
This argument still basically holds if either of $a$ or $b$ is $2$ except that when constructing the solution from the pair $(a,b)$ then we get more than one solution because, say, ${b}^{2}\equiv \beta mod2$ is not enough to specify $\beta $ from $b$. As we will see, though, the number of cases with one of $a$ or $b$ being $2$ is sufficiently small that it is easier to consider these as simply special cases.
We shall return to this case after analysing the special cases.
The special cases are when we don't have $a,b>2$. The first observation is that if $(a,b,n,\alpha )$ is a solution then changing the sign of any two of $(a,b,n)$ also yields a solution. So in the search for solutions, we can assume that $a,b\ge 0$. Another useful symmetry of solutions is that if $(a,b,n,\alpha )$ is a solution then so is $(b,a,\alpha n,\alpha )$. Thus it is sufficient to consider the cases $a=0$, $a=1$, and $a=2$, and for each we can assume that $b\ge a$.
If $a=0$ then we have ${b}^{2}=\beta $, whence $\beta =1$ and $b=\pm 1$. Substituting in, we get ${\varphi}_{2,1}=1$ and ${\varphi}_{2,1}=1$. Swapping $a$ and $b$, we get ${\varphi}_{n,\alpha}={\varphi}_{n,\alpha}$ and ${\varphi}_{n,\alpha}={\varphi}_{n,\alpha}$.
Let's now deal with the case of $a=1$. Then $\alpha nb{b}^{2}=\beta $. If also $b=1$ then rearranging gives $n=\alpha 1\beta $. Similarly, if $b=2$ we get $2n=\alpha 4\beta $ (note that $\alpha \beta $ is always even). In these special cases, we find the following identities:
$$\begin{array}{rl}{\varphi}_{1,1}+1& ={\varphi}_{1,1}\\ {\varphi}_{3,1}+1& ={\varphi}_{1,1}\\ {\varphi}_{1,1}+1& ={\varphi}_{3,1}\\ {\varphi}_{1,1}+1& ={\varphi}_{1,1}\\ {\varphi}_{2,1}+2& ={\varphi}_{2,1}\\ {\varphi}_{3,1}+2& ={\varphi}_{1,1}\\ {\varphi}_{1,1}+2& ={\varphi}_{3,1}\\ {\varphi}_{2,1}+2& ={\varphi}_{2,1}\end{array}$$ 
(Note that ${\varphi}_{1,1}$ and ${\varphi}_{1,1}$ are not real.) Our original observation that ${\varphi}_{3,1}={\varphi}_{1,1}+1$ is in this list.
Swapping $a$ and $b$ yields some new identities (and repeats some old ones, not shown):
$$\begin{array}{rl}2{\varphi}_{2,1}+1& ={\varphi}_{2,1}\\ 2{\varphi}_{3,1}+1& ={\varphi}_{4,1}\\ 2{\varphi}_{1,1}+1& ={\varphi}_{4,1}\\ 2{\varphi}_{2,1}+1& ={\varphi}_{6,1}\end{array}$$ 
For $b>2$ we can get some insight from working modulo $b$. In particular (still with $a=1$), working modulo $b$ shows that $\alpha =\beta $ leaving us with ${b}^{2}+nb=0$. As $b\ne 0$, we must have $b=n$ and this yields:
$${\varphi}_{n,\alpha}n={\varphi}_{n,\alpha}$$ 
Combined with the relationship between ${\varphi}_{n,\alpha}$ and ${\varphi}_{n,\alpha}$, this is a restatement of the fact that the two values for ${\varphi}_{n,\alpha}$ add up to $n$. Swapping $a$ and $b$ yields the intriguing identity:
$$n{\varphi}_{n,\alpha}+1={\varphi}_{\alpha {n}^{2}+2,1}$$ 
The last special case is if $a=2$. Then $4\alpha 2nb{b}^{2}=\beta $. We cannot have $b=2$ here, so we assume that $b>2$. Working modulo $b$, we have $4\alpha \equiv \beta modb$. Since $\alpha ,\beta =\pm 1$, this means that $\pm 4\equiv \pm 1modb$ and thus $b=3$ or $b=5$.
If $b=3$ then $\alpha =\beta $ and $3\alpha =2n$, so either $\alpha =1$ and $n=1$ or $\alpha =1$ and $n=2$.
If $b=5$ then $\alpha =\beta $ and $5\alpha =2n$, so either $\alpha =1$ and $n=2$ or $\alpha =1$ and $n=3$.
Thus we have (with swaps):
$$\begin{array}{rl}2{\varphi}_{1,1}+3& ={\varphi}_{4,1}\\ 2{\varphi}_{2,1}+3& ={\varphi}_{2,1}\\ 2{\varphi}_{2,1}+5& ={\varphi}_{6,1}\\ 2{\varphi}_{3,1}+5& ={\varphi}_{4,1}\\ 3{\varphi}_{1,1}+2& ={\varphi}_{7,1}\\ 3{\varphi}_{2,1}+2& ={\varphi}_{2,1}\\ 5{\varphi}_{2,1}+2& ={\varphi}_{14,1}\\ 5{\varphi}_{3,1}+2& ={\varphi}_{11,1}\end{array}$$ 
4 General Relations
Let us return to the general case where $a,b>0$. Here we look for $a,b$ with ${a}^{2}\equiv \pm 1modb$ and ${b}^{2}\equiv \pm 1moda$. Also, using the symmetry it is enough to consider $b\le a$.
Not being a number theorist, the obvious step is to turn to a computer. It is simple to write a function in Sage that does a search up to a specified limit for such pairs and prints the resulting solution.
The $n\ne 0$ and $n\ne 2$ skip some trivial results. In the range $2\le b\le a\le 100$, the results are below.
$$\begin{array}{rlrl}3{\varphi}_{1,1}+5& ={\varphi}_{7,1}& 15{\varphi}_{4,1}+56& ={\varphi}_{52,1}\\ 3{\varphi}_{3,1}+8& ={\varphi}_{7,1}& 17{\varphi}_{4,1}+4& ={\varphi}_{76,1}\\ 3{\varphi}_{3,1}+10& ={\varphi}_{11,1}& 17{\varphi}_{4,1}+72& ={\varphi}_{76,1}\\ 4{\varphi}_{4,1}+15& ={\varphi}_{14,1}& 21{\varphi}_{3,1}+8& ={\varphi}_{47,1}\\ 4{\varphi}_{4,1}+17& ={\varphi}_{18,1}& 21{\varphi}_{1,1}+13& ={\varphi}_{47,1}\\ 5{\varphi}_{2,1}+2& ={\varphi}_{14,1}& 21{\varphi}_{1,1}+34& ={\varphi}_{47,1}\\ 5{\varphi}_{1,1}+3& ={\varphi}_{11,1}& 21{\varphi}_{3,1}+55& ={\varphi}_{47,1}\\ 5{\varphi}_{1,1}+8& ={\varphi}_{11,1}& 24{\varphi}_{5,1}+5& ={\varphi}_{110,1}\\ 5{\varphi}_{3,1}+13& ={\varphi}_{11,1}& 26{\varphi}_{5,1}+5& ={\varphi}_{140,1}\\ 5{\varphi}_{5,1}+24& ={\varphi}_{23,1}& 29{\varphi}_{2,1}+12& ={\varphi}_{82,1}\\ 5{\varphi}_{5,1}+26& ={\varphi}_{27,1}& 33{\varphi}_{3,1}+10& ={\varphi}_{119,1}\\ 6{\varphi}_{6,1}+35& ={\varphi}_{34,1}& 34{\varphi}_{3,1}+13& ={\varphi}_{76,1}\\ 6{\varphi}_{6,1}+37& ={\varphi}_{38,1}& 34{\varphi}_{1,1}+21& ={\varphi}_{76,1}\\ 7{\varphi}_{7,1}+48& ={\varphi}_{47,1}& 34{\varphi}_{1,1}+55& ={\varphi}_{76,1}\\ 7{\varphi}_{7,1}+50& ={\varphi}_{51,1}& 34{\varphi}_{3,1}+89& ={\varphi}_{76,1}\\ 8{\varphi}_{3,1}+3& ={\varphi}_{18,1}& 35{\varphi}_{6,1}+6& ={\varphi}_{198,1}\\ 8{\varphi}_{1,1}+5& ={\varphi}_{18,1}& 37{\varphi}_{6,1}+6& ={\varphi}_{234,1}\\ 8{\varphi}_{1,1}+13& ={\varphi}_{18,1}& 48{\varphi}_{7,1}+7& ={\varphi}_{322,1}\\ 8{\varphi}_{3,1}+21& ={\varphi}_{18,1}& 50{\varphi}_{7,1}+7& ={\varphi}_{364,1}\\ 8{\varphi}_{8,1}+63& ={\varphi}_{62,1}& 55{\varphi}_{3,1}+21& ={\varphi}_{123,1}\\ 8{\varphi}_{8,1}+65& ={\varphi}_{66,1}& 55{\varphi}_{1,1}+34& ={\varphi}_{123,1}\\ 9{\varphi}_{9,1}+80& ={\varphi}_{79,1}& 55{\varphi}_{1,1}+89& ={\varphi}_{123,1}\\ 9{\varphi}_{9,1}+82& ={\varphi}_{83,1}& 56{\varphi}_{4,1}+15& ={\varphi}_{194,1}\\ 10{\varphi}_{3,1}+3& ={\varphi}_{36,1}& 63{\varphi}_{8,1}+8& ={\varphi}_{488,1}\\ 10{\varphi}_{3,1}+33& ={\varphi}_{36,1}& 65{\varphi}_{8,1}+8& ={\varphi}_{536,1}\\ 10{\varphi}_{10,1}+99& ={\varphi}_{98,1}& 70{\varphi}_{2,1}+29& ={\varphi}_{198,1}\\ 12{\varphi}_{2,1}+5& ={\varphi}_{34,1}& 72{\varphi}_{4,1}+17& ={\varphi}_{322,1}\\ 13{\varphi}_{3,1}+5& ={\varphi}_{29,1}& 80{\varphi}_{9,1}+9& ={\varphi}_{702,1}\\ 13{\varphi}_{1,1}+8& ={\varphi}_{29,1}& 82{\varphi}_{9,1}+9& ={\varphi}_{756,1}\\ 13{\varphi}_{1,1}+21& ={\varphi}_{29,1}& 89{\varphi}_{3,1}+34& ={\varphi}_{199,1}\\ 13{\varphi}_{3,1}+34& ={\varphi}_{29,1}& 89{\varphi}_{1,1}+55& ={\varphi}_{199,1}\\ 15{\varphi}_{4,1}+4& ={\varphi}_{52,1}& 99{\varphi}_{10,1}+10& ={\varphi}_{970,1}\end{array}$$ 
5 The Golden Ratio
What is particularly interesting are those involving ${\varphi}_{1,1}$ since this is the original golden ratio. Running up to $1000$, we have:
$$\begin{array}{rl}5{\varphi}_{1,1}+3& ={\varphi}_{11,1}\\ 8{\varphi}_{1,1}+5& ={\varphi}_{18,1}\\ 13{\varphi}_{1,1}+8& ={\varphi}_{29,1}\\ 21{\varphi}_{1,1}+13& ={\varphi}_{47,1}\\ 34{\varphi}_{1,1}+21& ={\varphi}_{76,1}\\ 55{\varphi}_{1,1}+34& ={\varphi}_{123,1}\\ 89{\varphi}_{1,1}+55& ={\varphi}_{199,1}\\ 144{\varphi}_{1,1}+89& ={\varphi}_{322,1}\\ 233{\varphi}_{1,1}+144& ={\varphi}_{521,1}\\ 377{\varphi}_{1,1}+233& ={\varphi}_{843,1}\\ 610{\varphi}_{1,1}+377& ={\varphi}_{1364,1}\\ 987{\varphi}_{1,1}+610& ={\varphi}_{2207,1}\end{array}$$ 
Unsurprisingly, the numbers appearing are the Fibonacci and Lucas numbers.
Taking two of these at random and substituting them in to the formulae for $\mathrm{cos}$ and $\mathrm{sin}$, we see that:
$$\begin{array}{rl}2\mathrm{cos}i\mathrm{log}(987\varphi +610)& =2207,\\ 2\mathrm{sin}i\mathrm{log}(610\varphi +377)& =1364i.\end{array}$$ 
More generally, the obvious conjecture from this is that
$${F}_{n}\varphi +{F}_{n1}={\varphi}_{{L}_{n},(1{)}^{n}}$$ 
(starting with ${F}_{0}=0$ and ${F}_{1}=1$).
Let us look for more patterns. If we take those involving ${\varphi}_{2,1}$ we get the following list (again, looking for $a,b\le 1000$):
$$\begin{array}{rl}5{\varphi}_{2,1}+2& ={\varphi}_{14,1}\\ 12{\varphi}_{2,1}+5& ={\varphi}_{34,1}\\ 29{\varphi}_{2,1}+12& ={\varphi}_{82,1}\\ 70{\varphi}_{2,1}+29& ={\varphi}_{198,1}\\ 169{\varphi}_{2,1}+70& ={\varphi}_{478,1}\\ 408{\varphi}_{2,1}+169& ={\varphi}_{1154,1}\\ 985{\varphi}_{2,1}+408& ={\varphi}_{2786,1}\end{array}$$ 
Again, the coefficients on the left follow a kind of Fibonacci pattern, only this time the recurrence seems to be ${c}_{n}=2{c}_{n1}+{c}_{n2}$. The numbers in the sequence on the right are the Companion Pell numbers (OEIS A002203).
This investigation suggests that if we have found a solution $(a,b,n,1)$, then we get a new solution with $(an+b,a,n,1)$. A little more investigation suggests that this generalises to arbitrary $\alpha $ so that from $(a,b,n,\alpha )$ we should get another solution with $(\alpha (an+b),a,n,\alpha ,)$. We can prove this:
$$\begin{array}{rl}\alpha (\alpha (an+b){)}^{2}n\alpha (an+b)a{a}^{2}& =\alpha ({a}^{2}{n}^{2}+2anb+{b}^{2})\alpha {a}^{2}{n}^{2}\alpha nab{a}^{2}\\ & =\alpha anb+\alpha {b}^{2}{a}^{2}\\ & =\alpha (\alpha {a}^{2}nab{b}^{2})\\ & =\alpha \beta \\ & \in \{1,1\}.\end{array}$$ 
This provides a way to generate a lot of relations because we have obvious starting points: $(1,0,n,\alpha )$. The first few terms are:
$$\begin{array}{rlrl}(1,0,n,\alpha )& & {\varphi}_{n,\alpha}& ={\varphi}_{n,\alpha}\\ (\alpha n,1,n,\alpha )& & \alpha n{\varphi}_{n,\alpha}+1& ={\varphi}_{2+{n}^{2},1}\\ ({n}^{2}+\alpha ,\alpha n,n\alpha )& & ({n}^{2}+\alpha ){\varphi}_{n,\alpha}+\alpha n& ={\varphi}_{3\alpha n+{n}^{3},\alpha}\end{array}$$ 
The second of these was the "intriguing identity".
The sequence of coefficients is a generalised Fibonacci sequence. The initial data is the pair $(n,\alpha )$ and the resulting sequence is formed by the recurrence:
$${a}_{k}^{(n,\alpha )}=\alpha (n{a}_{k1}^{(n,\alpha )}+{a}_{k2}^{(n,\alpha )}),\phantom{\rule{1em}{0ex}}$$ 
The $k$th relation has right hand side ${\varphi}_{m,\beta}$ where $\beta =(\alpha {)}^{k}$ and $m=2{a}_{k1}^{(n,\alpha )}+n{a}_{k}^{(n,\alpha )}$. This simplifies as:
$$\begin{array}{rl}m& =2{a}_{k1}^{(n,\alpha )}+n{a}_{k}^{(n,\alpha )}\\ & ={a}_{k1}^{(n,\alpha )}+({a}_{k1}^{(n,\alpha )}+n{a}_{k}^{(n,\alpha )})\\ & ={a}_{k1}^{(n,\alpha )}+\alpha {a}_{k+1}^{(n,\alpha )}\end{array}$$ 
Thus our resulting identity is:
$${a}_{k}^{(n,\alpha )}{\varphi}_{n,\alpha}+{a}_{k1}^{(n,\alpha )}={\varphi}_{{a}_{k1}^{(n,\alpha )}+\alpha {a}_{k+1}^{(n,\alpha )},(\alpha {)}^{k}}.$$ 
Note that we can run the recurrence backwards. The one before $(a,b,n,\alpha )$ is $(b,\alpha abn,n,\alpha )$. In particular, the one before $(1,0,n,\alpha )$ is $(0,\alpha ,n,\alpha )$ and the one before that is $(\alpha ,\alpha n,n,\alpha )$ which becomes
$$\alpha {\varphi}_{n,\alpha}\alpha n={\varphi}_{\alpha n,\alpha}.$$ 
Here's the list for $(n,\alpha )=(3,1)$.
$$\begin{array}{rl}3{\varphi}_{3,1}+1& ={\varphi}_{7,1}\\ 8{\varphi}_{3,1}+3& ={\varphi}_{18,1}\\ 21{\varphi}_{3,1}+8& ={\varphi}_{47,1}\\ 55{\varphi}_{3,1}+21& ={\varphi}_{123,1}\\ 144{\varphi}_{3,1}+55& ={\varphi}_{322,1}\\ 377{\varphi}_{3,1}+144& ={\varphi}_{843,1}\\ 987{\varphi}_{3,1}+377& ={\varphi}_{2207,1}\end{array}$$ 
For $(n,\alpha )=(4,1)$, it starts as follows:
$$\begin{array}{rl}4{\varphi}_{4,1}+1& ={\varphi}_{14,1}\\ 15{\varphi}_{4,1}+4& ={\varphi}_{52,1}\\ 56{\varphi}_{4,1}+15& ={\varphi}_{194,1}\\ 209{\varphi}_{4,1}+56& ={\varphi}_{724,1}\\ 780{\varphi}_{4,1}+209& ={\varphi}_{2702,1}\end{array}$$ 
Our special cases give us other starting points. For example, from ${\varphi}_{1,1}+1={\varphi}_{1,1}$ we actually get a cycle of length $6$:
$$\begin{array}{rl}{\varphi}_{1,1}+1& ={\varphi}_{1,1}\\ +1& ={\varphi}_{2,1}\\ {\varphi}_{1,1}& ={\varphi}_{1,1}\\ {\varphi}_{1,1}1& ={\varphi}_{1,1}\\ 1& ={\varphi}_{2,1}\\ {\varphi}_{1,1}& ={\varphi}_{1,1}\end{array}$$ 
Feeding in our other starting points, we find the following. (We ignore any sequences with ${\varphi}_{\pm 2,1}$ since they are $\pm 1$).

${\varphi}_{3,1}+1={\varphi}_{1,1}$ produces the sequence:
$$\begin{array}{rl}{\varphi}_{3,1}+1& ={\varphi}_{1,1}\\ 2{\varphi}_{3,1}+1& ={\varphi}_{4,1}\\ 5{\varphi}_{3,1}+2& ={\varphi}_{11,1}\\ 13{\varphi}_{3,1}+5& ={\varphi}_{29,1}\\ 34{\varphi}_{3,1}+13& ={\varphi}_{76,1}\\ 89{\varphi}_{3,1}+34& ={\varphi}_{199,1}\\ 233{\varphi}_{3,1}+89& ={\varphi}_{521,1}\\ 610{\varphi}_{3,1}+233& ={\varphi}_{1364,1}\end{array}$$ What is interesting about this sequence is that between this and the sequence beginning ${\varphi}_{3,1}={\varphi}_{3,1}$ then we get all the Fibonacci numbers. This sequence also has the property that when run backwards, the terms are the same as forwards except with $(a,b)$ swapped.

${\varphi}_{1,1}+1={\varphi}_{3,1}$ is the second identity in the sequence generated by ${\varphi}_{1,1}$.

${\varphi}_{1,1}+1={\varphi}_{1,1}$ becomes ${\varphi}_{1,1}={\varphi}_{1,1}$ after two iterations.

${\varphi}_{2,1}+2={\varphi}_{2,1}$ becomes ${\varphi}_{2,1}={\varphi}_{2,1}$ after two iterations.

${\varphi}_{3,1}+2={\varphi}_{1,1}$ becomes ${\varphi}_{3,1}+1={\varphi}_{1,1}$ on the next iteration.

${\varphi}_{1,1}+2={\varphi}_{3,1}$ becomes ${\varphi}_{1,1}={\varphi}_{1,1}$ after three iterations.

${\varphi}_{2,1}+2={\varphi}_{2,1}$ becomes ${\varphi}_{2,1}={\varphi}_{2,1}$ after two iterations.

$2{\varphi}_{3,1}+1={\varphi}_{4,1}$ is the term after ${\varphi}_{3,1}+1={\varphi}_{1,1}$.

$2{\varphi}_{1,1}+1={\varphi}_{4,1}$ is the result of two iterations from ${\varphi}_{1,1}={\varphi}_{1,1}$.

$2{\varphi}_{2,1}+1={\varphi}_{6,1}$ is the term after ${\varphi}_{2,1}={\varphi}_{2,1}$.

$2{\varphi}_{1,1}+3={\varphi}_{4,1}$ appears in the reverse sequence for ${\varphi}_{1,1}={\varphi}_{1,1}$.

$2{\varphi}_{2,1}+5={\varphi}_{6,1}$ becomes ${\varphi}_{2,1}={\varphi}_{2,1}$ after three iterations.

$2{\varphi}_{3,1}+5={\varphi}_{4,1}$ is in the backward family from ${\varphi}_{3,1}+1={\varphi}_{1,1}$.

$3{\varphi}_{1,1}+2={\varphi}_{7,1}$ is in the family for ${\varphi}_{1,1}={\varphi}_{1,1}$.

$5{\varphi}_{2,1}+2={\varphi}_{14,1}$ is in the family for ${\varphi}_{2,1}={\varphi}_{2,1}$.

$5{\varphi}_{3,1}+2={\varphi}_{11,1}$ is in the family for ${\varphi}_{3,1}+1={\varphi}_{1,1}$.
Thus all of our "special cases" are not special save for one:
$${\varphi}_{3,1}+1={\varphi}_{1,1}$$ 
However, even this one is not so special since ${\varphi}_{n,\alpha}={\varphi}_{n,\alpha}$ so we can rewrite it as:
$${\varphi}_{3,1}1={\varphi}_{1,1}$$ 
whereupon it is revealed as a rearrangement of:
$${\varphi}_{1,1}+1={\varphi}_{3,1}.$$ 
The point being that if ${\varphi}_{n,\alpha}+b={\varphi}_{m,\beta}$ then also ${\varphi}_{m,\beta}+b={\varphi}_{n,\alpha}$ and this gives us another starting point. However, of the various identities with leading term $1$ that we found, only ${\varphi}_{3,1}+1={\varphi}_{1,1}$ actually gives us a new sequence. So it is pretty special, and the original golden ratio, $\varphi $, retains its spot as the most glittery of the golden numbers.
6 Golden Primes
At this stage, an interesting question is: how many golden numbers do we need to generate the rest? One approach to answering this is to use a sieve method.
There are two sieve methods that could be used. One is to start with the tautologies ${\varphi}_{n,\alpha}={\varphi}_{n,\alpha}$ and run the generalised Fibonacci generations forwards and backwards and mark all the ${\varphi}_{m,\beta}$ that are so produced. (To make this efficient, we would first want to prove that if ${\varphi}_{n,\alpha}$ produces ${\varphi}_{m,\beta}$ then all of the ones that ${\varphi}_{m,\beta}$ generates are also in the sequence for ${\varphi}_{n,\alpha}$.)
The problem with this sieve is that although it provides a simple rule for generating relations between the ${\varphi}_{n,\alpha}$, it doesn't guarantee that we have found all of them.
The other sieve method would be to use our original method for finding relations: look for pairs $(a,b)$ with ${a}^{2}\equiv \pm 1modb$ and ${b}^{2}\equiv \pm 1moda$. This is guaranteed to produce all the relations, but not necessarily in a useful order for crossing the ${\varphi}_{n,\alpha}$ off our list.
Thus to produce a definite list of the "golden primes", we would need to prove either that all relations could be generated by the Fibonacci method, or that for a given ${\varphi}_{n,\alpha}$ then there was a bound on $(a,b)$ beyond which we couldn't find a relation that produced ${\varphi}_{n,\alpha}$. Both of these seem reasonable things to investigate further, but this article is already long enough. So we will, instead, run the $(a,b)$â€“sieve and cross off all generated ${\varphi}_{n,\alpha}$ that we find.
Note that with the sieve then it is sufficient to continue with our assumption that $a,b\ge 0$. A quick implementation in Sage reveals that the overwhelming majority of pairs $(n,\alpha )$ cannot be generated by others.
7 Golden Conclusion
This was nothing more than a merry jaunt with numbers. It started with a simple observation and led on from there. As promised, nothing deep was discovered but I did stumble upon a nice link from the Golden Numbers to a certain type of Diophantine Equations, a topic that I've heard mentioned from time to time but never found a reason to look more closely at. It might be interesting to use them as a way to learn a bit more about Diophantine Equations.
So while this might not qualify as genuine mathematical gold, it certainly has its sparkle. And to the right mind, iron pyrites is far more interesting than dull old gold.