Looking for Regular Polygons with a Side to Diagonal ratio Equaling a Metallic Mean Quick backstory: I watched a video last night on the properties of Metallic Mean. The video stated that the ratio of a regular pentagon's side to one of it's diagonals was equal to S1 (1.618033989...). They also stated that the ratio of a regular octagon's side to one of it's diagonals was equal to S2 (2.414213562...).

This all checks out when you plug it into the below equation, but I want to be able to find how many sides of a regular polygon it would take for any Sn to be displayed.
The Variables:

a = Edge length (assume 1; can be changed if needed)
n = Number of vertices
m = Diagonals across (would need to be changed as needed)

The Equation:

dm = a * sin( π * ( m / n ) ) / sin( π / n )

Example for Pentagon:

d2 = 1 * sin( π * ( 2 / 5 ) ) / sin( π / 5 ) = 1.618033989...

What I would like to do is make the equation equal n so that I can change m and dm in order to figure out what regular polygons equal Sn.
I really hope this makes sense, and I appreciate any help in advance.
 A: The answer is that there are no more regular polygons whose diagonal-to-side ratio is a metallic mean. Here is a quick proof (omitting some details since I don't have time right now, I can come back to fill them in later).
The metallic means are quadratic irrationals, i.e. numbers of algebraic degree 2. On the other hand, as stated in the question, for a given number of vertices $n>2$ we have that the $(m-1)$th diagonal-to-side ratio is
$$d_m=\frac{\sin(\pi m/n)}{\sin(\pi/n)}=e^{\pi i(1-m)/n}\frac{e^{2\pi i m/n}-1}{e^{2\pi i/n}-1} =$$
$$=e^{\pi i(1-m)/n} \left( 1 + e^{2\pi i/n} + \cdots + e^{2\pi i(m-1)/n}\right) = e^{\pi i(1-m)/n} + e^{\pi i(3-m)/n} + \cdots + e^{\pi i(m-1)/n},$$
where we take $1<m<n-1$ to avoid trivial cases. First consider the case of $m$ odd, where all the summands are powers of $e^{2\pi i/n}$. Let $G$ be the Galois group of the cyclotomic field $\mathbb{Q}(e^{2\pi i/n})$. Using modular arithmetic, it is straightforward to see that this sum of roots of unity is left invariant by no element of $G$ other than complex conjugation (except in the cases $m=1$ or $m=n-1$, which we excluded). This means that the algebraic degree of $d_m$ is $\varphi(n)/2$, where $\varphi$ is Euler's totient function.
In the case of $m$ even, similar considerations lead us to the algebraic degree of $d_m$ being $\varphi(2n)/2$.
It is easy to check, using Euler's product formula, that the only numbers with $\varphi(n)/2=2$ are $5, 8, 10$ and $12$. For $n=5, 10$ (pentagon and decagon) we get (integral linear combinations of 1 and) the golden mean
$$\phi = \frac{1+\sqrt5}{2} = [1;1,1,1,1,1,1,1,\ldots],$$
where $[a_0;a_1,a_2,\ldots]$ denotes a continued fraction. For $n=8$ (octagon) we get the silver mean
$$\delta_S = 1+\sqrt2 = [2;2,2,2,2,2,2,2,\ldots],$$
and for $n=12$ (dodecagon) we get a nameless gold-silver "alloy"
$$\delta_A = 1+\sqrt3 = [2;1,2,1,2,1,2,1,\ldots],$$
which is not a metallic mean per se. Similarly, the only numbers with $\varphi(2n)/2=2$ are $4, 5$ and $6$. We already dealt with the case $n=5$; in the two remaining cases we just get $\sqrt2$ and $\sqrt3$. This exhausts the list of possibilities.
A: The question uses $n$ to mean two different things, so for clarity I will define the variables used in this answer:


*

*$n$: the number of vertices of the polygon

*$m$: the separation between the vertices whose diagonal we measure (and we assume $1 < m \le \frac{n}{2}$ and $\gcd(m,n) = 1$)

*$k$: the index of the metallic mean (which we assume to be strictly positive)


Then as Roked notes in their answer (and I'm surprised that the video apparently didn't give this), $S_k = \frac{k + \sqrt{k^2 + 4}}{2}$. Now, it's easy to see that $S_k > k$. It's slightly harder to see that $S_k < k+1$, but not too hard: if this were not the case, we would require $\sqrt{k^2+4} \ge k+2$, and since both sides are positive we can square without affecting the inequality to see that $k^2 + 4 \ge k^2 + 4k + 4$, or $k \le 0$, contradicting the assumption that $k$ is strictly positive.
If we want to be more precise, we can expand the square root as a Laurent series to get $$S_k = k + \frac{1}{k} - \left(\frac{1}{k}\right)^3 + \cdots$$, so that for large $k$ it gets closer to $k$.

Now, let's look at extremes on the assumption that $a = 1$. We have the diagonal length $$d_{m,n} = \frac{\sin(m\frac{\pi}n)}{\sin(\frac\pi n)} \le \frac{1}{\sin(\frac\pi n)}$$ So if $S_k = d_{m,n}$, $$k < \frac{1}{\sin(\frac\pi n)}$$
When $n$ is large, $\frac{\pi}{n}$ is small, so we can use the small angle approximation $\sin(\theta) \approx \theta$ to get the heuristic estimate $n > \pi k$. (Note that empirically this actually holds already for the pentagon example, where $n=5$ and $k=1$, because the extreme case is where $m = \frac{n}{2}$ and neither the pentagon nor the octogon example is so extreme).

With the help of a computer algebra package, we can expand the ratio of sines as 
$$\frac{\sin(m\theta)}{\sin(\theta)} = m\left(1 - \frac{m^2 - 1}{6} \theta^2 + \frac{3 m^4 - 10 m^2 + 7}{360} \theta^4 - \frac{3 m^6 - 21 m^4 + 49 m^2 - 31}{15120} \theta^6 + O(m^8\theta^8) \right)$$

So $$\frac{\sin(m \frac{\pi}n)}{\sin(\frac{\pi}n)} \\
= m\left(1 - \frac{m^2 - 1}{n^2} \frac{\pi^2}{6} + \frac{3 m^4 - 10 m^2 + 7}{n^4} \frac{\pi^4}{360} - \frac{3 m^6 - 21 m^4 + 49 m^2 - 31}{n^6} \frac{\pi^6}{15120} + \cdots \right) \\
\approx m\left(1 - 1.645\frac{m^2 - 1}{n^2} + 0.271\frac{3 m^4 - 10 m^2 + 7}{n^4}  - 0.0636\frac{3 m^6 - 21 m^4 + 49 m^2 - 31}{n^6} + \cdots\right)$$
Bearing in mind that $\frac{m}{n} \le \frac{1}{2}$, we see that this will tend to be between about $0.6m$ and $m$. Since $k < S_k < k+1$, that means that we need only consider $k < m < \frac{5}{3}(k+1)$.

Pulling these strands together, we get a heuristic (not fully rigorous) argument that there are unlikely to be more solutions with $a=1$. We have observed that $S_k$ is $k$ plus a bit, and we have observed that $d_{m,n}$ is $m$ minus a bit. Since the "minus a bit" can be more than $1$ it's not impossible for the values to coincide, but there's no good reason that they should. From this perspective, the known solutions look like a case of the law of small numbers (there aren't very many of them, so coincidences are likely).
A: I have searched far and wide, but the best I can do is this:
Any solution--a metallic mean that is the length of a polygon diagonal--must be an integer solution of x, y, and z to this equation:
$$\sin(\pi y/x)/\sin(\pi/x)=(z+\sqrt{z^2+4})/2$$
Assuming all side lengths are 1 WLOG, x is the number of sides on the polygon, y is the position of the diagonal, z is the number corresponding to which metallic mean. (5, 2, 1) works because in a 5 sided shape (x=5), the 2nd diagonal (y=2)(in this instance sides are the 1st "diagonal"), has a length of the golden ratio, the 1st (z=1) metallic ratio. For the same reason, (8, 3, 2) works as well. I have yet to find any more non-trivial integer solutions. I don't know how to prove there aren't any, though.
