Lemma 1.6.1. of the book "Catalan Numbers" (R. P. Stanley) states what follows.
Let $\alpha = (a_0,...,a_{2n})$ be a sequence of $2n+1$ terms containing exactly $n+1$ $1$'s and $n$ $(-1)$'s. Then the fact that all the $2n+1$ cyclic shifts $(a_j,a_{j+1},...,a_{j-1})$ are distinct, easily follows from $n+1$ and $n$ being relatively prime.
The problem is I don't see the implication.. am I missing something clear? Can anyone enlighten me?
Thanks a lot
Suppose two cyclic shifts $(a_j,...,a_{j-1})$ and $(a_k,...,a_{k-1})$ are the same, with $0 \le j < k \lt 2n+1$. Then
$a_j=a_k=a_{2k-j}=a_{3k-2j}=\dots\\a_{j+1}=a_{k+1}=a_{2k-j+1}=a_{3k-2j+1}=\dots$
etc. (where all indices are modulo $2n+1$).
Let $g=\gcd(k-j,2n+1)$. Then we can partition the $a_i$ into $g$ sets such that the $\frac{2n+1}{g}$ members of each set have the same value. But then $n$ and $n+1$ must both be multiples of $\frac{2n+1}{g}$, which is only possible if $\frac{2n+1}{g}=1$ i.e. $k-j$ is a multiple of $2n+1$ which contradicts $0 \le j < k \lt 2n+1$.
