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Proof of Two Congruences Concerning Legendre Polynomials
Results in Mathematics ( IF 1.1 ) Pub Date : 2021-04-12 , DOI: 10.1007/s00025-021-01389-3
Chen Wang , Wei Xia

The Legendre polynomials \(P_n(x)\) are defined by

$$\begin{aligned} P_n(x)=\sum _{k=0}^n\left( {\begin{array}{c}n+k\\ k\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( \frac{x-1}{2}\right) ^k\quad (n=0,1,2,\ldots ). \end{aligned}$$

In this paper, we prove two congruences concerning Legendre polynomials. For any prime \(p>3\), by using the symbolic summation package Sigma, we show that

$$\begin{aligned} \sum _{k=0}^{p-1}(2k+1)P_k(-5)^3\equiv p-\frac{10}{3}p^2q_p(2)\pmod {p^3}, \end{aligned}$$

where \(q_p(2)=(2^{p-1}-1)/p\) is the Fermat quotient. This confirms a conjecture of Z.-W. Sun. Furthermore, we prove the following congruence which was conjectured by V.J.W. Guo

$$\begin{aligned}&\sum _{k=0}^{p-1}(-1)^k(2k+1)P_k(2x+1)^4\\ \equiv&p\sum _{k=0}^{(p-1)/2}(-1)^k\left( {\begin{array}{c}2k\\ k\end{array}}\right) ^2(x^2+x)^k(2x+1)^{2k}\pmod {p^3},\end{aligned}$$

where p is an odd prime and x is an integer.



中文翻译:

有关勒让德多项式的两个同余证明

勒让德多项式\(P_n(x)\)由下式定义

$$ \ begin {aligned} P_n(x)= \ sum _ {k = 0} ^ n \ left({\ begin {array} {c} n + k \\ k \ end {array}} \ right)\左({\ begin {array} {c} n \\ k \ end {array}} \ right)\ left(\ frac {x-1} {2} \ right)^ k \ quad(n = 0,1 ,2,\ ldots)。\ end {aligned} $$

在本文中,我们证明了勒让德多项式的两个等价性。对于任何素数\(p> 3 \),通过使用符号求和包Sigma,我们得出

$$ \ begin {aligned} \ sum _ {k = 0} ^ {p-1}(2k + 1)P_k(-5)^ 3 \ equiv p- \ frac {10} {3} p ^ 2q_p(2 )\ pmod {p ^ 3},\ end {aligned} $$

其中\(q_p(2)=(2 ^ {p-1} -1)/ p \)是Fermat商。这证实了Z.-W.的猜想。太阳。此外,我们证明了以下由VJW Guo猜想的全等

$$ \ begin {aligned}&\ sum _ {k = 0} ^ {p-1}(-1)^ k(2k + 1)P_k(2x + 1)^ 4 \\ \ equiv&p \ sum _ {k = 0} ^ {(p-1)/ 2}(-1)^ k \ left({\ begin {array} {c} 2k \\ k \ end {array}} \ right)^ 2(x ^ 2 + x)^ k(2x + 1)^ {2k} \ pmod {p ^ 3},\ end {aligned} $$

其中p是奇数质数,x是整数。

更新日期:2021-04-12
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