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On $p$-adic congruences involving $\sqrt d$

Number Theory 2025-04-17 v1

Abstract

Let pp be an odd prime and let dd be an integer not divisible by pp. We prove that 1m,np1pm2dn2 (x(m+nd)){k=1p2k(k+1)2x(k1)(p1)(modp)if (dp)=1,k=0(p1)/2x2k(p1)(modp)if (dp)=1, \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 (dp)(\frac dp) denotes the Legendre symbol. This extends a recent conjecture of N. Kalinin. We also obtain the Wolstenholme-type congruence 1m,np1pm2dn2  1m+nd0(modp2).\sum_{1\le m,n\le p-1\atop p\nmid m^2-dn^2}\ \ \frac1{m+n\sqrt d}\equiv0\pmod{p^2}.

Keywords

Cite

@article{arxiv.2504.12242,
  title  = {On $p$-adic congruences involving $\sqrt d$},
  author = {Bo Jiang and Zhi-Wei Sun},
  journal= {arXiv preprint arXiv:2504.12242},
  year   = {2025}
}

Comments

6 pages

R2 v1 2026-06-28T23:00:48.271Z