Equations are ($n \ge 3$):
$\forall i=0 \dots n-1, a_i a_{i+1} = 2s_i / \sin(\frac {2\pi} n)$, with $a_n = a_0$.
Taking the $\log$ transforms these into linear equations: if $l_i = \log a_i$, the system is equivalent to
$\forall i=0 \dots n-1, l_i + l_{i+1} = \log(2s_i / \sin(\frac {2\pi} n))$.
The matrix has its diagonal made of $1$, plus a second diagonal above of $1$, and a $1$ at the lower left corner, and zeroes everywhere else.
Exemple for $n=4$: $\begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 1\\ 1 & 0 & 0 & 1 \end{pmatrix}$
When developing the determinant by the first line:
- If we take the $1$ on $(0,0)$, then on column $1$ we cannot take the $1$ on $(0,1)$ so we take the one on $(1,1)$, and then we are forced to take the $1$ on $(2,2)$, etc.: this is the principal diagonal.
- If we take the $1$ on $(0,1)$, then on line $1$ we cannot take the $1$ on $(1,1)$ so we take the $1$ on $(1,2)$, and then, etc.: we take the second diagonal of $1$s, including the $1$ on the lower left corner as last element.
The first diagonal has the identity permutation, so it has even parity, which makes a $+1$.
The second diagonal above has a cycle permutation, so its parity is even if $n$ is odd, and odd if $n$ is even - this is because an $n$ cycle decomposes into $n-1$ transpositions. Which makes $(-1)^{n-1}$
So the determinant is $1 + (-1)^{n-1}$, i.e. $0$ when $n$ is even and $2$ when $n$ is odd: the system has a unique solution for $n$ odd, and either no solution or an infinity, for $n$ even.
For $n$ even, another way to see it is to use an alternate sum of equations $i=1 \dots n-1$. This gives the left handside of equation $i=0$, so the system is underdetermined: either no solution or an infinity.
The case $n=4$ allows still another proof: one can apply a scaling by $\alpha > 0$ on the $x$ axis and a scaling by $1/ \alpha$ on the $y$ axis, this changes the $(a_i)$ but not the $(s_i)$.
