Let $p$ be an odd prime. For $a\in\mathbb Z$ let $\{a\}_p$ denote the least nonnegative residue of $a$ modulo $p$. The list $\{1^2\}_p,\ldots,\{((p-1)/2)^2\}_p$ is a permutation of all the quadratic residues modulo $p$ among $1,\ldots,p-1$, and I used Galois theory to determine the sign of this permutation for $p\equiv3\pmod4$ in my preprint http://arxiv.org/abs/1809.07766.

Now let us consider a similar problem for triangular numbers. Recall that the triangular numbers are those integers $T_n=n(n+1)/2$ $(n=0,1,2,\ldots)$. It is easy to see that for any odd prime $p$ those $\{T_k\}_p\ (k=1,\ldots,(p-1)/2)$ are pairwise distinct.

QUESTION: Is my following conjecture true?

Conjecture. Let $p>3$ be a prime. If $p\equiv3\pmod4$, then $$(-1)^{|\{(j,k):\ 1\leqslant j<k\leqslant(p-1)/2\ \&\ \{T_j\}_p>\{T_k\}_p\}|} =(-1)^{(h(-p)+1)/2+|\{1\leqslant k\leqslant\lfloor\frac{p+1}8\rfloor:\ (\frac kp)=1\}|},$$ where $(j,k)$ is an ordered pair, $(\frac kp)$ is the Legendre symbol, and $h(-p)$ is the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$. Also, \begin{align}&(-1)^{|\{(j,k):\ 1\leqslant j<k\leqslant(p-1)/2\ \&\ \{T_j\}_p+\{T_k\}_p>p\}|} \\=&\begin{cases}(-1)^{(p-1)/8}&\text{if}\ p\equiv1\pmod8, \\(-1)^{|\{1\leqslant k<\frac p4:\ (\frac kp)=-1\}|}&\text{if}\ p\equiv5\pmod 8, \\(-1)^{(h(-p)+1)/2+|\{1\leqslant k\leqslant\lfloor\frac{p+1}8\rfloor:\ (\frac kp)=-1\}|}&\text{if}\ p\equiv3\pmod4. \end{cases}\end{align}

I have checked this conjecture via a computer. It should be valid in my opinion. Any comments are welcome!

Let $p$ be an odd prime. For $a\in\mathbb Z$ let $\{a\}_p$ denote the least nonnegative residue of $a$ modulo $p$. The list $\{1^2\}_p,\ldots,\{((p-1)/2)^2\}_p$ is a permutation of all the quadratic residues modulo $p$ among $1,\ldots,p-1$, and I used Galois theory to determine the sign of this permutation for $p\equiv3\pmod4$ in my preprint arXiv:1809.07766 available from http://arxiv.org/abs/1809.07766.

Now let us consider a similar problem for triangular numbers. Recall that the triangular numbers are those integers $T_n=n(n+1)/2$ $(n=0,1,2,\ldots)$. It is easy to see that for any odd prime $p$ those $\{T_k\}_p\ (k=1,\ldots,(p-1)/2)$ are pairwise distinct.

QUESTION: Is my following conjecture true?

Conjecture. Let $p>3$ be a prime. If $p\equiv3\pmod4$, then $$(-1)^{|\{(j,k):\ 1\leqslant j<k\leqslant(p-1)/2\ \&\ \{T_j\}_p>\{T_k\}_p\}|} =(-1)^{(h(-p)+1)/2+|\{1\leqslant k\leqslant\lfloor\frac{p+1}8\rfloor:\ (\frac kp)=1\}|},$$ where $(j,k)$ is an ordered pair, $(\frac kp)$ is the Legendre symbol, and $h(-p)$ is the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$. Also, \begin{align}&(-1)^{|\{(j,k):\ 1\leqslant j<k\leqslant(p-1)/2\ \&\ \{T_j\}_p+\{T_k\}_p>p\}|} \\=&\begin{cases}(-1)^{(p-1)/8}&\text{if}\ p\equiv1\pmod8, \\(-1)^{|\{1\leqslant k<\frac p4:\ (\frac kp)=-1\}|}&\text{if}\ p\equiv5\pmod 8, \\(-1)^{(h(-p)+1)/2+|\{1\leqslant k\leqslant\lfloor\frac{p+1}8\rfloor:\ (\frac kp)=-1\}|}&\text{if}\ p\equiv3\pmod4. \end{cases}\end{align}

I have checked this conjecture via a computer. It should be valid in my opinion. Any comments are welcome!