# Evaluate a curious determinant with Legendre symbol entries

Let $p$ be an odd prime and let $(\frac{\cdot}p)$ be the Legendre symbol. R. Chapman's conjecture on the exact value of the determinant of $$C_p:=\left[\left(\frac{i-j}p\right)\right]_{0\le i,j\le (p-1)/2}$$ was finally confirmed by M. Vsemirnov [Linear Algebra Appl. 436(2012), 4101-4106; Acta Arith. 159(2013), 331-344]. Here I suggest a variant of this problem.

QUESTION: Let $p>3$ be a prime and let $h(-p)$ be the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$. Define $M_p$ as the matrix obtaining from $C_p$ via replacing all the entries in the first row by $1$. Is the equality $$\det M_p=\begin{cases}(-1)^{(p-1)/4}&\text{if}\ p\equiv 1\pmod4, \\(-1)^{(h(-p)-1)/2}&\text{if}\ p\equiv3\pmod4,\end{cases}$$ always valid?

I have verified the above equality concerning $\det M_p$ for all primes $p$ with $3<p<2000$. In view of this, I conjecture that the question has an affirmative answer.

For any odd prime $p$, let $M_p^+$ be the matrix obtaining from $[(\frac{i+j}p)]_{0\le i,j\le(p-1)/2}$ via replacing all the entries in the first row by $1$. My computation suggests that $$\det M_p^+=2^{(p-1)/2}\det M_p.$$

Similarly, for an odd prime $p$, we define $N_p$ as the matrix obtaining from $[(\frac{i-j}p)]_{1\le i,j\le (p-1)/2}$ via replacing all the entries in the first row by $1$, and also define $N_p^+$ as the matrix obtaining from $[(\frac{i+j}p)]_{1\le i,j\le (p-1)/2}$ via replacing all the entries in the first row by $1$. I guess that $$\det N_p=(-1)^{\lfloor (p+3)/4\rfloor}\det M_p\ \ \text{and}\ \ \det N_p^+=-2^{(p-3)/2}\det M_p.$$

• Chapman's conjecture confirmed by Vsemirnov states that $\det C_p=1$ for any prime $p\equiv3\pmod4$, and $\det C_p=-a_p$ if $p$ is a prime with $p\equiv1\mod 4$ and $\varepsilon_p^{(2-(\frac 2p))h(p)}=a_p+b_p\sqrt p$ with $a_p,b_p\in\mathbb Q$, where $\varepsilon_p$ and $h(p)$ are the fundamental unit and the class number of the real quadratic field $\mathbb Q(\sqrt p)$. Compared with this result, our formula for $\det M_p$ balances the cases $p\equiv1\pmod4$ and $p\equiv3\pmod4$. – Zhi-Wei Sun Sep 13 '18 at 11:29