Let $p$ be an odd prime. For $d\in\mathbb Z$ we define $$T(d,p):=\det\left[\left(\frac{i^2+dj^2}p\right)\right]_{0\le i,j\le(p-1)/2},$$ where $(\frac{\cdot}p)$ is the Legendre symbol. By (1.17) of my paper arXiv:1308.2900, if$(\frac dp)=-1$ then $(\frac{T(d,p)}p)=1$.

Suppose that $p\equiv3\pmod4$. Then, by (1.14) of arXiv:1308.2900, $T(d,p)=T(-1,p)$ for any $d\in\mathbb Z$ with $(\frac dp)=-1$. As $T(-1,p)$ is a skew-symmetric determinant of even order, it is an integer square.

In the case $p\equiv1\pmod4$, if $d$ and $d'$ are both quadratic nonresidues modulo $p$, then we clearly have $T(d,p)=\pm T(d',p)$.

I have the following conjecture which seems quite challenging.

Conjecture. Let $p\equiv1\pmod4$ be a prime and write $p=x^2+4y^2$ with $x$ and $y$ positive integers. Then, for any integer $d\in\mathbb Z$ with $(\frac dp)=-1$, there is a positive integer $t(p)$ (not depending on $d$) such that $$|T(d,p)|=2^{(p-1)/2}t(p)^2y.$$

Via Mathematica, I find that \begin{gather}t(5)=1,\ t(13)=3,\ t(17)=4,\ t(29)=91,\ t(37)=81,\ t(41)=180, \\t(53)=1703,\ t(61)=87120,\ t(73)=16104096,\ t(89)=3947892146, \\ t(97)=19299520512,\ t(101)=885623936875,\ t(109)=36548185365.\end{gather}

Your comments are welcome!

PS: I have verified the conjecture for all primes $p<5000$ with $p\equiv1\pmod4$.

  • 4
    $\begingroup$ Out of curiosity - do all those questions of yours come out of some other research, or are they just a result of some random numerical experimentation? $\endgroup$
    – Wojowu
    Commented Jan 6, 2019 at 12:01
  • $\begingroup$ @Wojowu If one has good mathematical feeling or intuition like Ramanujan, he/she can have lots of new discoveries in math. My questions posted to Mathoverflow are just small parts of my mathematical conjectures, of course they are not results of random numerical experiments, they come from combination of my philosophy, intuition, inspiration, experience and computation. $\endgroup$ Commented Jan 6, 2019 at 15:07
  • $\begingroup$ I have verified the conjecture for all primes $p<5000$ with $p\equiv1\pmod4$. $\endgroup$ Commented Jan 6, 2019 at 15:39

1 Answer 1


I have obtained some partial results that seem promising, but not a full solution:

Let $\chi$ be a nontrivial Dirichlet character mod $p$ with $\chi(-1)=1$. Then

$\sum_{j=0}^{(p-1)/2} \left( \frac{ i^2+ dj^2}{p} \right) \chi(j) = \frac{1}{2} \sum_{j=1}^{p-1 } \left( \frac{ i^2+ dj^2}{p} \right) \chi(j) $

If $i=0$, then this sum is zero. Otherwise, we can perform a substitution replacing $j$ by $ij$, getting

$ \frac{1}{2} \sum_{j=1}^{p-1 } \left( \frac{ i^2+ di^2 j^2}{p} \right) \chi(i j) = \chi(i) \frac{1}{2}\sum_{j=1}^{p-1 } \left( \frac{ 1+ d j^2}{p} \right) \chi( j) $. So $\chi$ is an eigenvector of this matrix, with eigenvalue $ \frac{1}{2}\sum_{j=1}^{p-1 } \left( \frac{ 1+ d j^2}{p} \right) \chi( j) $.

The complement to these eigenvectors has a basis consisting of the function that is $1$ on $0$ and the function that is $1$ on all nonzero $j$. The matrix preserves the space generated by this basis, and acts on it by $$\begin{pmatrix} 0 & - (p-1)/2 \\ 1 & \frac{1}{2} \sum_{j=1}^{p-1} \left( \frac{1+ dj^2}{p} \right)\end{pmatrix} $$ and therefore with determinant $(p-1)/2$. So the determinant of your matrix is

$$\frac{p-1}{2} \prod_{\chi \textrm { nontrivial, } \chi(-1)=1} \left( \frac{1}{2}\sum_{j=1}^{p-1 } \left( \frac{ 1+ d j^2}{p} \right) \chi( j) \right)$$

We can write $\sum_{j^2 = t} \chi(j)$ as $\chi_1(t) + \chi_2(t)$ where $\chi_1$ and $\chi_2$ are the squareroots of $t$ in the ring of characters, so $$\sum_{j=1}^{p-1 } \left( \frac{ 1+ d j^2}{p} \right) \chi( j) = \sum_{t=1}^{p-1} \left( \frac{ 1+ d t}{p} \right) \chi_1( t)+ \sum_{t=1}^{p-1} \left( \frac{ 1+ d t}{p} \right) \chi_2( t) $$

If we focus attention on the eigenvector associated to $\chi$ the Legendre symbol, then $\chi_1$ and $\chi_2$ are the two characters of order $4$, so the left term is of the form $a+bi$ and the right term is its complex conjugate $a-bi$. By the evaluation of the absolute value of Jacobi sums, $a^2+b^2=p$, and because the number of $t$ with $\chi_1(t)= \pm 1$ (i.e. $t$ a quadratic residue) and $1+dt \neq 0$ (implied by $t$ a quadratic residue) is $p-1/2$, which is even, $a$ is even, so in fact $a=\pm y,b=\pm x$, and this eigenvalue is $\pm y$.

So it seems when the eigenvalue associated to the Legendre symbol is removed, you would like to have a square determinant. But I don't yet quite see how to obtain that.

I tried to get a square by finding matching pairs of eigenvalues, or by removing this eigenvector and conjugating to a skew-symmetric matrix, but neither approach quite worked.


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