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Let $p$ be an odd prime and let $S_p$ denote the determinant $$\det\left[\left(\frac{i^2+j^2}p\right)\right]_{1\le i,j\le (p-1)/2}$$ with $(\frac{\cdot}p)$ the Legendre symbol. By Theorem 1.2 of my paper arXiv:1308.2900 available from http://arxiv.org/abs/1308.2900, $-S_p$ is a quadratic residue modulo $p$. Here I ask a further question.

QUESTION. Is it true that for each prime $p\equiv3\pmod4$ the number $-S_p$ is always a positive square divisible by $2^{(p-3)/2}$?

Define $a_p=\sqrt{-S_p}/2^{(p-3)/4}$ for any prime $p\equiv3\pmod4$. Then \begin{gather*}a_3=a_7=a_{11}=1,\ a_{19}=2,\ a_{23}=1,\ a_{31}=29,\ a_{43}=254, \\a_{47}=367,\ a_{59}=9743,\ a_{67}=305092,\ a_{71}=29,\ a_{79}=1916927. \end{gather*} I have computed the values of $a_p$ for all primes $p\equiv3\pmod4$ with $p<2000$. Based on the numerical data, I conjecture that the above question has an affirmative answer but I'm unable to prove this.

Any ideas towards the solution? Your comments are welcome!

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  • $\begingroup$ This is not all that closely related, but I wonder whether some techniques for a determinant problem of Robin Chapman's might not apply. empslocal.ex.ac.uk/people/staff/rjchapma/etc/evildet.pdf $\endgroup$ – Gerry Myerson Sep 9 '18 at 11:22
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    $\begingroup$ I assume you also noted the corresponding conjecture for $p\equiv1\pmod4$: write uniquely $p=a^2f^4+b^2$ with $f$ odd and $a\equiv1\pmod4$ squarefree. Then your determinant is conjecturally equal to $a$ times a square. $\endgroup$ – Henri Cohen Sep 9 '18 at 19:25
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    $\begingroup$ Here I mention comments from Maxim Vseminov at St. Petersburg (yesterday): "The 2-part of the conjecture is rather easy. Since $p\equiv3\pmod4$, $i^2+j^2$ vanishes modulo $p$ only when $p$ divides both $i$ and $j$. Therefore, all matrix entries are $\pm$1's. If we add the first row to the remaining rows we obtain $(p-3)/2$ rows with even entries. Hence, the determinant is divisible by $2^{(p-3)/2}$. However, finding the 2-adic valuation might be a difficult problem. E.g., the 2-adic valuation is strictly greater than $(p-3)/2$ for $p=19, 43, 67,\ldots$". $\endgroup$ – Zhi-Wei Sun Sep 10 '18 at 6:30
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    $\begingroup$ More comments from Maxim Vsemirnov at St. Petersburg (yesterday): "It seems that the case $p\equiv1\pmod 4$ is also interesting. Now the determinant is "almost a square". Moreover, my computation for small primes suggest the following conjecture: when $p\equiv1\pmod4$, the determinant is $\pm a \cdot d^2$, where $ a^2+4b^2=p$. However, I have no idea about the nature of the square factor." $\endgroup$ – Zhi-Wei Sun Sep 10 '18 at 6:36
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    $\begingroup$ The arguments of Max Alekseyev and Dmitry Krachun prove your conjecture. $\endgroup$ – GH from MO Sep 10 '18 at 19:29
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It can be seen that $S_p$ is not divisible by $p$. The argument about the decomposition of the matrix as $\frac{2}{i\sqrt{p}}A^2$ suggested above implies that $-S_p$ is a square in $\mathbb{Q}[\zeta_p][\sqrt{\lambda_p}]$, where $\lambda_p=-2i\sqrt{p}$. Writing $-S_p=(a\sqrt{\lambda_p}+b)^2$ with $a,b\in\mathbb{Q}[\zeta_p]$, we have $ab=0$ (since $S_p^2\in\mathbb{Z}\subset\mathbb{Q}[\zeta_p]$ and $\sqrt{\lambda_p}\not\in\mathbb{Q}[\zeta_p]$). The case $b=0$ is impossible as then $S_p$ is divisible by $p$, which follows from the fact that the norm of $\lambda_p$ is divisible by an odd power of $p$. Thus $-S_p=b^2$ with $b\in\mathbb{Q}[\zeta_p]$, which implies $-S_p$ is a square since $S_p$ is not divisible by $p$.

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  • $\begingroup$ Very nice argument, welcome to MO! $\endgroup$ – GH from MO Sep 10 '18 at 19:34
  • $\begingroup$ @MaxAlekseyev: Your $\alpha$ equals $(1-I\sqrt{p}\pm\sqrt{\lambda_p})/(p-1)$. It lies in $\mathbb{Q}(\zeta_p,\sqrt{\lambda_p})$. $\endgroup$ – GH from MO Sep 10 '18 at 22:20
  • $\begingroup$ Very nice indeed! I had a feeling that my approach simplifies the problem in a right direction, although my hope was to compute $\det(A)$ explicitly. The non-constructive argument comes as a surprise. $\endgroup$ – Max Alekseyev Sep 10 '18 at 23:23
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This is not an answer, but a reduction to supposedly simpler problem. (edited with more details)

Using quadratic Gauss sums, we can express Legendre symbol $\left(\frac{i^2+j^2}p\right)$ as \begin{split} \left(\frac{i^2+j^2}p\right) &= \frac{1}{I\sqrt{p}}\sum_{n=0}^{p-1} \zeta^{(i^2+j^2)n^2} \\ &= \frac{1}{I\sqrt{p}}\left(1+2\sum_{n=1}^{(p-1)/2} \zeta^{i^2n^2}\zeta^{j^2n^2}\right),\qquad(\star) \end{split} where $\zeta := \exp(\frac{2\pi I}p)$ and $I$ is the imaginary unit. From $(\star)$ it follows that $$\left[\left(\frac{i^2+j^2}p\right)\right]_{i,j=1}^{(p-1)/2} = \frac{1}{I\sqrt{p}}(J + 2B^2),$$ where $J$ is the matrix of all ones and $B:=[\zeta^{i^2j^2}]_{i,j=1}^{(p-1)/2}$. This enables application of pavl0's approach in evaluating the determinant in order to show that it is a square.


Alternatively, $(\star)$ allows us to extract the square root from the determinant without computing its value. To do so, we need to embed "$1+$" into the quadratic sum in $(\star)$ by introducing a parameter $\alpha$: $$1+2\sum_{n=1}^{(p-1)/2} \zeta^{i^2n^2}\zeta^{j^2n^2} = 2\sum_{n=1}^{(p-1)/2} (\zeta^{i^2n^2}+\alpha)(\zeta^{j^2n^2}+\alpha).$$ Since $2\sum_{n=1}^{(p-1)/2} \zeta^{i^2n^2} = I\sqrt{p} - 1$ for any $i\not\equiv 0\pmod{p}$, we need $\alpha$ to satisfy the quadratic equation: $$1 = 2(I\sqrt{p} - 1)\alpha + (p-1)\alpha^2.$$ So, we can set $\alpha := ((p^{1/4}-1)(1+Ip^{1/4}))^{-1}$, which is a root of this equation. Then $(\star)$ turns into $$\left(\frac{i^2+j^2}p\right) = \frac{2}{I\sqrt{p}}\sum_{n=1}^{(p-1)/2} (\zeta^{i^2n^2}+\alpha)(\zeta^{j^2n^2}+\alpha),$$ implying that $$\left[\left(\frac{i^2+j^2}p\right)\right]_{i,j=1}^{(p-1)/2} = \frac{2}{I\sqrt{p}} A^2,$$ where matrix $A := \left[ \zeta^{i^2j^2}+\alpha \right]_{i,j=1}^{(p-1)/2}$. It follows that $$-S_p = T_p^2,\quad \text{where}\quad T_p := \det\left(\frac{1+I}{p^{1/4}}A\right).$$ The original question reduces to showing that $T_p$ is an integer.

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  • $\begingroup$ Looks nice. Can you give the details for the first display? It is routine, of course, but having the details would be useful. $\endgroup$ – GH from MO Sep 10 '18 at 4:33
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    $\begingroup$ @GHfromMO: I've added the details. $\endgroup$ – Max Alekseyev Sep 10 '18 at 12:55
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This is a partial answer, as to why the determinant is always divisible by $2^{(p-3)/2}.$ We can write matrix $M_p:=\left[\left(\dfrac{i^2+j^2}{p}\right)\right]_{1\le i,j\le (p-1)/2}$ as $J+2A$, where $J$ is a matrix of ones, and $A$ is some symmetric matrix with integer entries. If $A$ is invertible, then we can apply the matrix determinant lemma, in particular, we find that $\det(M_p)=\det(2A)+1^t\mathrm{adj}(2A)1=2^{(p-1)/2}\det(A)+2^{(p-3)/2}1^t\mathrm{adj}(A)1.$

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    $\begingroup$ Along these lines, we can explicitly express $M_p = \frac{1}{I\sqrt{p}}(J+2B^2)$, where $B:=[\zeta^{i^2 j^2}]_{i,j=1}^{(p-1)/2}$ and $\zeta:=\exp(\frac{2\pi I}p)$. $\endgroup$ – Max Alekseyev Sep 10 '18 at 1:24

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