TITLE: Permuting elements of a set around a circle
QUESTION [12 upvotes]: Given $15$ objects placed around a circle, is it possible to permute their order such that the distance between any two elements is different in the second permutation from that in the original state?

REPLY [2 votes]: This sci.math post says that the literature contains several independent proofs of the result that the toroidal $n$-queens problem has a solution iff $\gcd(n,6)=1$ and gives six references. I have access to one of them, Torleiv Kløve, The modular $n$-queen problem, in Discrete Mathematics 19 (1977), 289--291. The proof there is short enough to be worth typing out here.
We want a function $f:\Bbb Z/n\Bbb Z\to\Bbb Z/n\Bbb Z$ such that $f$, $f+\mathrm{id}_{\Bbb Z/n\Bbb Z}$, and $f-\mathrm{id}_{\Bbb Z/n\Bbb Z}$ are all injections. The theorem is that such an $f$ exists iff $\gcd(n,6)=1$.

Proof. Suppose that $f$ is such a function, and let
$$S=\sum_{k\in\Bbb Z/n\Bbb Z}k^2\quad\text{and}\quad T=\sum_{k\in\Bbb Z/n\Bbb Z}kf(k)\;.$$
Then $\displaystyle\sum_{k\in\Bbb Z/n\Bbb Z}f(k)^2=S$, since $f$ is injective. Since $f+\mathrm{id}_{\Bbb Z/n\Bbb Z}$ and $f-\mathrm{id}_{\Bbb Z/n\Bbb Z}$ are also injective,
$$S=\sum_{k\in\Bbb Z/n\Bbb Z}\big(f(k)\pm k\big)^2=\sum_{k\in\Bbb Z/n\Bbb Z}f(k)^2\pm 2\sum_{k\in\Bbb Z/n\Bbb Z}kf(k)+\sum_{k\in\Bbb Z/n\Bbb Z}k^2=2S\pm 2T\;,$$
$2S=4S$, and $2S=0$ in $\Bbb Z/n\Bbb Z$. In $\Bbb Z$, therefore, we have
$$2\sum_{k=0}^{n-1}k^2=\frac{n(n-1)(2n-1)}3\equiv 0\pmod n$$
and hence $\gcd(n,3)=1$.
He now appeals to N.J. Fine’s solution of problem E 1699 (Amer. Math. Monthly 72 (1965), 552-3). Specialized to the present setting, the argument (which may have given Kløve the idea for his proof) is that
$$\sum_{k\in\Bbb Z/n\Bbb Z}\big(f(k)+k\big)=2\sum_{k\in\Bbb Z/n\Bbb Z}k$$
in $\Bbb Z/n\Bbb Z$, but if $n$ is even, then $$\sum_{k=0}^{n-1}k=\frac{n(n-1)}2\equiv-\frac{n}2\equiv\frac{n}2\pmod n\;,$$
so in $\Bbb Z/n\Bbb Z$ we have $$\sum_{k\in\Bbb Z/n\Bbb Z}\big(f(k)+k\big)=2\cdot\frac{n}2=0\ne\sum_{k\in\Bbb Z/n\Bbb Z}k\;,$$ and $f+\mathrm{id}_{\Bbb Z/n\Bbb Z}$ cannot be injective. Thus, $n$ must be odd, and $\gcd(n,6)=1$.
Conversely, if $\gcd(n,6)=1$, then $f(k)=2k$ is a solution. $\dashv$