Let <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="41980_2025_982_Article_IEq1.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">$p>3$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>p</mi> <mo>></mo> <mn>3</mn> </mrow> </math></EquationSource> </InlineEquation> be a prime, and let <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="41980_2025_982_Article_IEq2.gif" Format="GIF" Height="15" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">$d\in {\mathbb {Z}}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>d</mi> <mo>∈</mo> <mi mathvariant="double-struck">Z</mi> </mrow> </math></EquationSource> </InlineEquation> with <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="41980_2025_982_Article_IEq3.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="37" /> </InlineMediaObject> <EquationSource Format="TEX">$p\not \mid d$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>p</mi> <mo>∤</mo> <mi>d</mi> </mrow> </math></EquationSource> </InlineEquation>. For <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="41980_2025_982_Article_IEq4.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="48" /> </InlineMediaObject> <EquationSource Format="TEX">$m\in {\mathbb {Z}}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>m</mi> <mo>∈</mo> <mi mathvariant="double-struck">Z</mi> </mrow> </math></EquationSource> </InlineEquation> with <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="41980_2025_982_Article_IEq5.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="166" /> </InlineMediaObject> <EquationSource Format="TEX">$(p-1)/2\leqslant m\leqslant p-1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">(</mo> <mi>p</mi> <mo>-</mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">/</mo> <mn>2</mn> <mo>⩽</mo> <mi>m</mi> <mo>⩽</mo> <mi>p</mi> <mo>-</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>, Sun considered the determinant <Equation ID="Equ11"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="41980_2025_982_Article_Equ11.gif" Format="GIF" Height="27" Rendition="HTML" Resolution="72" Type="Linedraw" Width="285" /> </MediaObject> <EquationSource Format="TEX">$\begin{aligned} S_m(d,p)=\det \left[ (i^2+dj^2)^{m}\right] _{1\leqslant i,j \leqslant (p-1)/2}, \end{aligned}$</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <msub> <mi>S</mi> <mi>m</mi> </msub> <mrow> <mo stretchy="false">(</mo> <mi>d</mi> <mo>,</mo> <mi>p</mi> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <mo movablelimits="true">det</mo> <msub> <mfenced close="]" open="["> <msup> <mrow> <mo stretchy="false">(</mo> <msup> <mi>i</mi> <mn>2</mn> </msup> <mo>+</mo> <mi>d</mi> <msup> <mi>j</mi> <mn>2</mn> </msup> <mo stretchy="false">)</mo> </mrow> <mi>m</mi> </msup> </mfenced> <mrow> <mn>1</mn> <mo>⩽</mo> <mi>i</mi> <mo>,</mo> <mi>j</mi> <mo>⩽</mo> <mo stretchy="false">(</mo> <mi>p</mi> <mo>-</mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">/</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>and determined <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="41980_2025_982_Article_IEq6.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="60" /> </InlineMediaObject> <EquationSource Format="TEX">$S_m(d,p)$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>S</mi> <mi>m</mi> </msub> <mrow> <mo stretchy="false">(</mo> <mi>d</mi> <mo>,</mo> <mi>p</mi> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> modulo p when <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="41980_2025_982_Article_IEq7.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="140" /> </InlineMediaObject> <EquationSource Format="TEX">$m\in \{p-2,p-3\}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>m</mi> <mo>∈</mo> <mo stretchy="false">{</mo> <mi>p</mi> <mo>-</mo> <mn>2</mn> <mo>,</mo> <mi>p</mi> <mo>-</mo> <mn>3</mn> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="41980_2025_982_Article_IEq8.gif" Format="GIF" Height="25" Rendition="HTML" Resolution="72" Type="Linedraw" Width="75" /> </InlineMediaObject> <EquationSource Format="TEX">$(\frac{-d}{p})=-1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">(</mo> <mfrac> <mrow> <mo>-</mo> <mi>d</mi> </mrow> <mi>p</mi> </mfrac> <mo stretchy="false">)</mo> <mo>=</mo> <mo>-</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>. In this paper, we obtain <InlineEquation ID="IEq9"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="41980_2025_982_Article_IEq9.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="71" /> </InlineMediaObject> <EquationSource Format="TEX">$S_{p-2}(d,p)$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>S</mi> <mrow> <mi>p</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mrow> <mo stretchy="false">(</mo> <mi>d</mi> <mo>,</mo> <mi>p</mi> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> modulo p in the remaining case <InlineEquation ID="IEq10"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="41980_2025_982_Article_IEq10.gif" Format="GIF" Height="25" Rendition="HTML" Resolution="72" Type="Linedraw" Width="62" /> </InlineMediaObject> <EquationSource Format="TEX">$(\frac{-d}{p})=1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">(</mo> <mfrac> <mrow> <mo>-</mo> <mi>d</mi> </mrow> <mi>p</mi> </mfrac> <mo stretchy="false">)</mo> <mo>=</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>, and determine the Legendre symbols <InlineEquation ID="IEq11"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="41980_2025_982_Article_IEq11.gif" Format="GIF" Height="29" Rendition="HTML" Resolution="72" Type="Linedraw" Width="66" /> </InlineMediaObject> <EquationSource Format="TEX">$(\frac{S_{p-3}(d,p)}{p})$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">(</mo> <mfrac> <mrow> <msub> <mi>S</mi> <mrow> <mi>p</mi> <mo>-</mo> <mn>3</mn> </mrow> </msub> <mrow> <mo stretchy="false">(</mo> <mi>d</mi> <mo>,</mo> <mi>p</mi> <mo stretchy="false">)</mo> </mrow> </mrow> <mi>p</mi> </mfrac> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq12"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="41980_2025_982_Article_IEq12.gif" Format="GIF" Height="29" Rendition="HTML" Resolution="72" Type="Linedraw" Width="66" /> </InlineMediaObject> <EquationSource Format="TEX">$(\frac{S_{p-4}(d,p)}{p})$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">(</mo> <mfrac> <mrow> <msub> <mi>S</mi> <mrow> <mi>p</mi> <mo>-</mo> <mn>4</mn> </mrow> </msub> <mrow> <mo stretchy="false">(</mo> <mi>d</mi> <mo>,</mo> <mi>p</mi> <mo stretchy="false">)</mo> </mrow> </mrow> <mi>p</mi> </mfrac> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> in some special cases.

On Some Determinants Arising from Quadratic Residues

期刊论文
发表日期：

2025年7月23日

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Chen-Kai Ren,
Zhi-Wei Sun

摘要

Let $p>3$ $p > 3$ be a prime, and let $d\in {\mathbb {Z}}$ $d \in Z$ with $p\not \mid d$ $p ∤ d$ . For $m\in {\mathbb {Z}}$ $m \in Z$ with $(p-1)/2\leqslant m\leqslant p-1$ $(p - 1) / 2 ⩽ m ⩽ p - 1$ , Sun considered the determinant $\begin{aligned} S_m(d,p)=\det \left[ (i^2+dj^2)^{m}\right] _{1\leqslant i,j \leqslant (p-1)/2}, \end{aligned}$ $\begin{matrix} S_{m} (d, p) = det {[{(i^{2} + d j^{2})}^{m}]}_{1 ⩽ i, j ⩽ (p - 1) / 2}, \end{matrix}$ and determined $S_m(d,p)$ $S_{m} (d, p)$ modulo p when $m\in \{p-2,p-3\}$ $m \in {p - 2, p - 3}$ and $(\frac{-d}{p})=-1$ $(\frac{- d}{p}) = - 1$ . In this paper, we obtain $S_{p-2}(d,p)$ $S_{p - 2} (d, p)$ modulo p in the remaining case $(\frac{-d}{p})=1$ $(\frac{- d}{p}) = 1$ , and determine the Legendre symbols $(\frac{S_{p-3}(d,p)}{p})$ $(\frac{S_{p - 3} (d, p)}{p})$ and $(\frac{S_{p-4}(d,p)}{p})$ $(\frac{S_{p - 4} (d, p)}{p})$ in some special cases.