On $p$-adic congruences involving $\sqrt d$
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arXiv
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| Format: | Preprint |
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2025
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| _version_ | 1866913796584374272 |
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| author | Jiang, Bo Sun, Zhi-Wei |
| author_facet | Jiang, Bo Sun, Zhi-Wei |
| contents | Let $p$ be an odd prime and let $d$ be an integer not divisible by $p$. We prove that $$ \prod_{1\le m,n\le p-1\atop p\nmid m^2-dn^2}\ (x-(m+n\sqrt{d})) \equiv \begin{cases}\sum_{k=1}^{p-2}\frac{k(k+1)}2x^{(k-1)(p-1)}\pmod p &\text{if}\ (\frac dp)=1,\\\sum_{k=0}^{(p-1)/2}x^{2k(p-1)} \pmod p&\text {if}\ (\frac dp)=-1, \end{cases}$$ where $(\frac dp)$ denotes the Legendre symbol. This extends a recent conjecture of N. Kalinin. We also obtain the Wolstenholme-type congruence $$\sum_{1\le m,n\le p-1\atop p\nmid m^2-dn^2}\ \ \frac1{m+n\sqrt d}\equiv0\pmod{p^2}.$$ |
| format | Preprint |
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arxiv_https___arxiv_org_abs_2504_12242 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | On $p$-adic congruences involving $\sqrt d$ Jiang, Bo Sun, Zhi-Wei Number Theory 11A07, 11R11, 11T07 Let $p$ be an odd prime and let $d$ be an integer not divisible by $p$. We prove that $$ \prod_{1\le m,n\le p-1\atop p\nmid m^2-dn^2}\ (x-(m+n\sqrt{d})) \equiv \begin{cases}\sum_{k=1}^{p-2}\frac{k(k+1)}2x^{(k-1)(p-1)}\pmod p &\text{if}\ (\frac dp)=1,\\\sum_{k=0}^{(p-1)/2}x^{2k(p-1)} \pmod p&\text {if}\ (\frac dp)=-1, \end{cases}$$ where $(\frac dp)$ denotes the Legendre symbol. This extends a recent conjecture of N. Kalinin. We also obtain the Wolstenholme-type congruence $$\sum_{1\le m,n\le p-1\atop p\nmid m^2-dn^2}\ \ \frac1{m+n\sqrt d}\equiv0\pmod{p^2}.$$ |
| title | On $p$-adic congruences involving $\sqrt d$ |
| topic | Number Theory 11A07, 11R11, 11T07 |
| url | https://arxiv.org/abs/2504.12242 |