Elementary congruences and L-functions In a recent article, Emmanuel Lecouturier proves a generalization of the following surprising result: for a Mersenne prime $N = 2^p - 1 \ge 31$, the element 
$$ S = \prod_{k=1}^{\frac{N-1}2} k^k $$ 
is a $p$-th power modulo $N$, and observed that he did not know an elementary proof. Neither do I.
Numerical experiments suggest that $s_p$ is actually a $6p$-th power
modulo $N$. I can't even see why it is a quadratic residue, i.e., why
the following result (not proved in the article cited) should hold:
$$ T = \prod_{k=1}^{\frac{N+1}4} (2k-1) $$ 
is a square mod N.
For arbitrary primes $N \equiv 3 \bmod 4$, the following seems to hold: 
$$ \Big(\frac{T}{N} \Big) = \begin{cases}
    - (-1)^{(h-1)/2} & \text{ if } N \equiv  3 \bmod 16, \\
    - 1      & \text{ if } N \equiv  7 \bmod 16, \\
      (-1)^{(h-1)/2} & \text{ if } N \equiv 11 \bmod 16, \\
    + 1      & \text{ if } N \equiv 15 \bmod 16, 
\end{cases} $$
where $h$ is the class number of the complex quadratic number field with discriminant $-N$. This suggests a possible proof using L-functions (i.e. using methods in (Congruences for L-functions, Urbanowicz, K.S. Williams) and explains the difficulty of finding an elementary proof. 
My questions:


*

*Is this congruence for $(T/N)$ known?

*How would I start attacking this result using known results on L-functions?

 A: 1. First we show that
$$\left(\frac{T}{N}\right) = \begin{cases}
    - (-1)^{(h-1)/2} & \text{ if } N \equiv  3 \bmod 16, \\
      (-1)^{(h-1)/2} & \text{ if } N \equiv 11 \bmod 16. \\
\end{cases}$$
Consider the sets
$$A_0:=\{1,2,\dots,\tfrac{N-1}{2}\},\quad A_1:=\{1,3,\dots,\tfrac{N-1}{2}\},\quad A_2:=\{2,4,\dots,\tfrac{N-3}{2}\},$$
$$ A_3:=\{1,2,\dots,\tfrac{N-3}{4}\},\quad A_4:=\{\tfrac{N+1}{4},\tfrac{N+5}{4},\dots,\tfrac{N-1}{2}\}.$$
Consider also the corresponding character sums
$$S_i:=\sum_{a\in A_i}\left(\frac{a}{N}\right),\qquad i=0,1,2,3,4.$$
Modulo $N$, we have $2A_3=A_2$ and $-2A_4=A_1$. Therefore, $S_3=-S_2$ and $S_4=S_1$, and hence $S_1-S_2=S_3+S_4=S_0$. But also $S_1+S_2=S_0$, hence $S_1=S_0$ and $S_2=0$. On the other hand, it is known that $S_0=3h$, see e.g. Theorem 70 in Fröhlich-Taylor: Algebraic number theory. This means that $S_1=3h$, i.e. the number of quadratic nonresidues in $A_1$ equals
$$ n=\frac{N+1}{8}-\frac{3h}{2}=\frac{N+5}{8}-2h+\frac{h-1}{2}.$$
We conclude
$$ \left(\frac{T}{N}\right)=(-1)^n=(-1)^{(N+5)/8}(-1)^{(h-1)/2}. $$
This is the claimed formula (with Jeremy Rouse's correction).
2. Now we show that
$$\left(\frac{T}{N}\right) = \begin{cases}
    - 1      & \text{ if } N \equiv  7 \bmod 16, \\
    + 1      & \text{ if } N \equiv 15 \bmod 16.
\end{cases}$$
We proceed as before, but this time we get $S_3=S_2$ and $S_4=-S_1$, so that $S_2-S_1=S_3+S_4=S_0$. Together with $S_2+S_1=S_0$, this yields $S_1=0$ and $S_2=S_0$. Hence the formula for $n$ simplifies to $n=(N+1)/8$, and we conclude
$$\left(\frac{T}{N}\right)=(-1)^n=(-1)^{(N+1)/8}.$$
A: I answer only for the $3$-part of your question (I have to think more for the $2$-part).
Let $N$ be a prime  $\equiv 1  \text{ (mod } 3\text{)}$. Let $\text{log} : (\mathbf{Z}/N\mathbf{Z})^{\times} \rightarrow \mathbf{F}_3$ be a discrete logarithm. We have the following identity:
$$ \sum_{k=1}^{\frac{N-1}{2}} k \cdot \log(k) \equiv \frac{-\log(2)}{8} \cdot \frac{N-1}{3} \text{ (mod } 3\text{)}$$
The idea is to rewrite the left hand side as a sum over $(\mathbf{Z}/N\mathbf{Z})^{\times}$ in terms of Bernoulli polynomials (which, I think, is an old idea of my advisor Loïc Merel).
The proof is the following (essentially Lemme $11$ of my preprint). Let $f : \mathbf{R} \rightarrow \mathbf{\mathbf{R}}$ be the $1$-periodic function defined by $f(x) = x$ if $x \in  [0,\frac{1}{2}]$, $f(\frac{1}{2}) = \frac{1}{4}$ and $f(x)=0$ if $x \in ]\frac{1}{2},1[$. Let $E(x)$ be the integer part of $x$. Let $\overline{B_2}(x) = (x-E(x))^2-(x-E(x))$ and $\overline{B}_1(x) = (x-E(x))-\frac{1}{2}$ (except that $\overline{B}_1(0):=0$) be (essentially) the second and first periodic Bernoulli functions respectively. Then we have :
$$f(x) = \frac{1}{4}\cdot \overline{B}_2(2x) - \overline{B}_2(x) + \frac{1}{2}\cdot \overline{B}_1(x-\frac{1}{2}) $$
(to find such identities systematically, it is convenient to compute the Fourier coefficients of $f$ and remark that the Bernoulli polynomials have nice Fourier coefficients).
Then we find :
$$\sum_{k=1}^{\frac{N-1}{2}} k\cdot \log(k) \equiv  \sum_{k \in (\mathbf{Z}/N\mathbf{Z})^{\times}} f(\frac{k}{N})\cdot \log(k) \text{ (mod } 3\text{)}$$
which you can simplify greatly with the Bernoulli functions (the $\log(2)$ comes from $2x$ inside $\overline{B_2}(2x)$).
In particular if $N \equiv 1  \text{ (mod } 9\text{)}$ then your $S$ is a $3$-rd power. If $N = 2^p-1 \geq 31$ is a Mersenne prime , then $\log(2) \equiv 0 \text{ (mod } 3\text{)}$ since $\log(2^p) \equiv p\log(2) \equiv0 \text{ (mod } 3\text{)}$ and $p>3$.
Although I didn't state it, I think I can also prove the following regarding the $3$-th power part: if $N = a^2-ab+b^2$ is a prime $\equiv 1  \text{ (mod } 9\text{)}$ then $b^{-2}\cdot(ab^2+ba^2)$ is a $3$-rd power modulo $N$. It follows from the same methods as in the preprint.
