Motivated by Question 316142 of mine, I consider the new sum $$S(n):=\sum_{\pi\in S_{n}}e^{2\pi i\sum_{k=1}^{n}k\pi(k)/n}$$ for any positive integer $n$, where $S_n$ is the symmetric group of all the permutations of $\{1,\ldots,n\}$. Note that $S(n)$ is the permanent of the matrix $[e^{2\pi ijk/n}]_{1\le j,k\le n}$. As $S(n)\equiv\det[e^{2\pi ijk/n}]_{1\le j,k\le n}\pmod2$, it is easy to see that $S(n)\equiv n\pmod2$ in the ring of all algebraic integers.

Suppose that $S(n)\not=0$. Then the coefficient of $x_1^{n-1}\ldots x_n^{n-1}$ in the polynomial $$P(x_1,\ldots,x_n):=\prod_{1\le j<k\le n}(x_k-x_j)\left(e^{2\pi ik/n}x_k-e^{2\pi ij/n}x_j\right)$$ is $\text{per}[(e^{2\pi ij/n})^{k-1}]\not=0$. Applying Alon's Combinatorial Nullstellensatz to the subset $A=\{z\in\mathbb C:\ z^n=1\}$ of the complex field $\mathbb C$, we see that there is a permutation $\sigma\in S_n$ such that $j+\sigma(j)\not\equiv k+\sigma(k)\pmod n$ for all $1\le j<k\le n$. Thus $$\sum_{k=1}^n(k+\sigma(k))\equiv\sum_{j=1}^n j\pmod n$$ and hence $\sum_{k=1}^nk=n(n+1)/2\equiv0\pmod n$. This shows that $n$ must be odd.

Via a computer I find that \begin{gather}S(1)=1,\ S(3)=-3,\ S(5)=-5,\ S(7)=-105,\ S(9)=81,、 \\S(11)=6765=3\cdot5\cdot11\cdot41,\ S(13)=175747=11\cdot13\cdot1229, \\ S(15)=30375=3^5\cdot 5^3,\ S(17)=25219857=3\cdot13\cdot17\cdot38039.\end{gather} Thus it is natural for me to formulate the following conjecture.

Conjecture. (i) For each $n=1,3,5,\ldots$, the sum $S(n)$ is an integer divisible by $n$.

(ii) For any odd prime $p$, we have $S(p)\equiv-p\pmod{p^2}$.

Any ideas towards the solution?

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    $\begingroup$ I don't mean to be rude, but why do you always state your posts as "QUESTION: …? I conjecture that this question has a positive answer" rather than just "CONJECTURE: …."? $\endgroup$ – LSpice Dec 4 '18 at 3:12
  • $\begingroup$ @LSice Thank you for your inquiry. Indeed, I prefer to state it as a conjecture rather than a question. But Mathoverflow prefers questions and answers. $\endgroup$ – Zhi-Wei Sun Dec 4 '18 at 3:16
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    $\begingroup$ $S(n)$ is an integer because it is invariant under the Galois group $\mathrm{Gal}(\mathbf{Q}(\zeta_n)/\mathbf{Q})$. If $p$ is an odd prime then $S(p)$ is a sum of $p!$ terms each congruent to 1 mod $1-\zeta_p = 1-e^{2\pi i/p}$. Since the norm of $1-\zeta_p$ is equal to $p$, it follows that $S(p)$ is divisible by $p$. $\endgroup$ – François Brunault Dec 4 '18 at 5:52

The conjectures are true, and can be recovered from the action of $({\bf Z} / n {\bf Z})^2$ on $S_n$ by composition from the right and left with translations $k \mapsto k + a \bmod n$.

Translation by $1$ from either side multiplies $\exp \frac{2\pi i}{n} \sum_k k \phi(k)$ by $\exp\bigl(\pm 2\pi i {n \choose 2} / n\bigr) = \exp(\pm \pi i (n-1))$, which is $-1$ if $n$ is even and $+1$ if $n$ is odd. The even case immediately shows that $S(n) = 0$ if $n$ is even.

In any case $S(n) \in \bf Z$ because $S(n)$ is an algebraic integer in the $n$-th cyclotomic field $K_n$ and the Galois group of $K_n$ permutes the summands (conjugate $\pi$ by the map $k \mapsto mk \bmod n$ for $m \in ({\bf Z} / n {\bf Z})^*$).

Since the action of ${\bf Z} / n {\bf Z}$ from the right is free, the summands come in batches of $n$ equal $n$-th roots of unity, so $S(n)/n$ is also an algebraic integer in $K_n$, and thus a rational integer. (We could have used the action from the left to the same effect.)

If $p$ is an odd prime then the action of $({\bf Z} / p {\bf Z})^2$ is not free but the only permutations with nontrivial stabilizer are the affine-linear maps $k \mapsto mk + c \bmod p$ with $m,c \in {\bf Z} / p {\bf Z}$ and $m \neq 0$. For each of these $p^2-p$ permutations, $\sum_k k \pi(k) \equiv 0 \bmod p$. Therefore $S(p)$ is $p^2 - p$ plus a multiple of $p^2$, which gives the desired congruence. This argument fails for $p=3$ because it uses $\sum_{k=1}^p k^2 \equiv 0 \bmod p$, but we already know that $S(3) = -3$. This completes the proof.

P.S. this action of $({\bf Z} / p {\bf Z})^2$ on $S_p$ also gives one combinatorial proof of Wilson's theorem: since the action is free on all but $p^2-p$ permutations, we deduce $p! = \#S_p \equiv p^2-p \bmod p^2$, which is equivalent to $(p-1)! \equiv -1 \bmod p$.

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The conjecture is true.

(i) $S(n)=0$ when $n$ is even. Proof: We identify $S_n$ as permutations of the elements of $\mathbb Z/n\mathbb Z$. Denote by $\sigma$ the permutation which satisfies $\sigma(i)=i-1$ for all $i$. Then we have $$\frac{\sum_{k=1}^n k\pi(\sigma(k))}{n}=\frac{\sum_{k=1}^n k\pi(k)}{n}+\frac{1+2+\cdots+n}{n}=\frac{\sum_{k=1}^n k\pi(k)}{n}+\frac{n+1}{2}.$$ This means that when $n$ is even we have $$e^{2\pi i\sum_{k=1}^{n}k\pi(k)/n}+e^{2\pi i\sum_{k=1}^{n}k\pi(\sigma(k))/n}=e^{2\pi i\sum_{k=1}^{n}k\pi(k)/n}(1+e^{\pi i})=0$$ as desired.

(ii) $S(n)$ is an integer divisible by $n$ when $n$ is odd. Proof: With the same definition for $\sigma$ as above we have: $$\frac{\sum_{k=1}^n k\pi(\sigma(k))}{n}=\frac{\sum_{k=1}^n k\pi(k)}{n}+\frac{n+1}{2}\equiv\frac{\sum_{k=1}^n k\pi(k)}{n} \pmod{\mathbb Z}$$ therefore $$\sum_{j=0}^{n-1} e^{2\pi i\sum_{k=1}^{n}k\pi(\sigma^j(k))/n}=ne^{2\pi i\sum_{k=1}^{n}k\pi(k)/n}.$$ From here we can make a choice of coset representatives of $\langle \sigma\rangle$ in $S_n$. Say, $T_n$, the permutations which fix $n$. Then we can write $$\frac{S(n)}{n}=\sum_{\pi \in T_n}e^{2\pi i\sum_{k=1}^{n-1}k\pi(k)/n}$$ which shows $S(n)/n$ is an algebraic integer. Moreover since multiplication by some number $a$ relatively prime to $n$, is a bijection on elements of $T_n$ we have: $$\varphi(n)\frac{S(n)}{n}=\sum_{\gcd(a,n)=1}\sum_{\pi \in T_n}e^{2\pi i\sum_{k=1}^{n-1}ka\pi(k)/n}\in \mathbb Z$$ which means $S(n)/n$ is rational and therefore an integer.

(iii) $S(p)=-p\pmod{p^2}$. Proof: Similarly to above we have $$S(p)/p=\sum_{\pi\in T_p}e^{2\pi i\sum_{k=1}^{p-1}k\pi(k)/p}$$ and from here we have $$(p-1)\frac{S(p)}{p}=\sum_{a=1}^{p-1}\sum_{\pi\in T_p}e^{2\pi i\sum_{k=1}^{p-1}ka\pi(k)/p}=(p-2)!(p-1)=-1\pmod{p}$$ which gives $\frac{S(p)}{p}=-1\pmod{p}$ as desired.

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