The origin of this question is that I found a way to 'eliminate' vertex weights from weighted $K_n$ graphs, i.e. if one assumes that the weight $w_{ij}$ of edge $e_{ij}$ can be expressed as $\pi_i+\omega_{ij}+\pi_j$ with vertex potentials $\pi_i$ and $pi_j$, then it is possible to recover the $\omega_{ij}$.

I then was curious about revovering the vertex potentials and to that end set up system of linear equations obtained from the edges on a Hamilton cycle.

After an appropriate relabeling of the vertices the equations are of the form
$\pi_i+\pi_{i+1}\ =\ w_{i,i+1}-\omega_{i,i+1};\quad 0\le\ i\ \le n,\quad n+1:=0$

Initially much to my surprise, it turned out that for the determinant of the associated matrix, one has $$\begin{vmatrix} 1 & 1 & 0 & 0&0&\cdots &0 & 0\\ 0 & 1& 1& 0&0&\cdots& 0&0\\ 0 & 0&1&1&0&\cdots&0&0\\ \vdots &\vdots & \ddots & \ddots & \ddots&\ddots&\vdots&\vdots\\ 0 &0& \cdots &0&1&1&0&0\\ 0 & 0 & \cdots &0 & 0&1&1&0\\ 1 & 0 &\cdots&0&0&0&0&1 \end{vmatrix}\ =\ 0\iff n\equiv0\mod2$$ which means that the unknown vertex potentials can only be determined, if their number is odd.

After a bit of thinking about that observation it became clear that the determinant of analogous matrices that have $m$ consecutive $1$s instead of two shifted cyclically by 1 to the right from one row to the next, is apparently non-zero exactly if $m$ and $n$ are relatively prime.

Question:is there a non-trivial explanation for the phenomen that the solvability of related problems, whose only discernible essential difference is the number $n$ of equations, depends on number theoretic properties of the relation between $n$ and a fixed set of parameters, that are independent of $n$?

The arguments I have in mind would only be based on the "original" statement of the problem and not on their "translation" to a system of linear equations.

Ideally the arguments would be understandable without a background in linear algebra.