Solve all $n$ variables $x_0,...,x_{n-1}$ given
$$ x_{j}-x_{j-1}=x_{j-1}-x_{j-2},\quad j\ge 2\\ x_{1}-x_{0}=x_{0}\\ x_{n-1}=1 $$
this looks like recurrence but the initial condition is at $n-1$ rather than $0$, also there's only one such condition, so I don't think traditional characteristic root method of guessing the solution is of the form $r^j$ would work.
This maybe irrelavant but actually $\pi(x_j-x_{j-1})$ is the number of points on a unit circle with distance to origin in the range of $(x_{j-1}^{1/2},x_{j}^{1/2}]$ where we've $n$-partitioned $(0,1]$ into $(0,x_{0}^{1/2}],...,(x_{n-1}^{1/2},1]$. Above is really trying to associate each partition to same number of points.
As dxiv points out, the differences between consecutive terms is constant. The second equation says this constant is $x_0$. Therefore $x_j = (j+1) x_0$, and, in particular, $1 = x_{n-1} = n x_0$. Therefore $x_0 = 1/n$ and $x_j = (j+1)/n$.