二項係数と剰余類の難問 – 2023年京大 特色入試 数学 第4問

\begin{aligned}
c_2  = & \sum_{1 \leqq j< k \leqq p-1}\prod_{\substack{1 \leqq i \leqq p-1 \\ i \ne j \\i \ne k} } i \\
 
\end{aligned}

\begin{aligned}
 & \prod_{\substack{1 \leqq i \leqq p-2 \\ i \ne j \\i \ne k} } i  \\

 \equiv & \prod_{\substack{1 \leqq i \leqq p-2 \\ i \ne j \\i \ne k} } i \cdot k \cdot k^{-1}  \\
 \equiv & \prod_{\substack{1 \leqq i \leqq p-2 \\ i \ne j } } i \cdot k^{-1}  \\
\equiv &  j^{-1} k^{-1} ( \mod p)
\end{aligned}

\begin{aligned}
c_2  = & \sum_{1 \leqq j< k \leqq p-1}\prod_{\substack{1 \leqq i \leqq p-1 \\ i \ne j \\i \ne k} } i \\
 \equiv & \sum_{1 \leqq j< k \leqq p-1} j^{-1} k^{-1} ( \mod p)
 

\end{aligned}

\begin{aligned}
c_2 \equiv   \sum_{\substack{1 \leqq s< t \leqq p-1 }} st (\mod p)  
\end{aligned}

\begin{aligned}

 &   \sum_{1  \leqq s< t \leqq p-1} st　\\
= & \sum_{s=1}^{p-2} s \sum_{t=s+1}^{p-1}i \\
= & \sum_{s=1}^{p-2} s \sum_{t=1}^{p -s-1}(t +s)\\

= & \sum_{s=1}^{p-2} s \left \{ \frac{(p-s-1)(p-s)}{2} +s(p-s-1) \right \} \\

= & \frac{1}2 \sum_{s=1}^{p-2} s(p-s-1)(p + s) \\
= & \frac{1}2 \sum_{s=1}^{p-2} \{- s^3 -s^2 +p(p-1)s \} \\

= & \frac{(p-2)(p-1)}2  \\
& \times  \left \{ - \frac{(p-2)(p-1)}{4} - \frac{2p-3}{6} + \frac{p(p-1)}2 \right\} \\

= & \frac{(p-2)(p-1)}{24} \\
 & \times \{ -3(p^2-3 p +2) -2(2p -3) +6p(p-1) \} \\
= &  \frac{(p-2)(p-1)( 3p -1)p}{24} 

\end{aligned}

(p-1)(3p-1) =4j(12j +2) = 8j(6j+1)

\begin{aligned}
(p-1)(3p-1) & =(4j+2)(12j +8)\\
 &  = 8(2j+1)(3j+2)

\end{aligned}

\begin{aligned}
c_2 \equiv   \sum_{\substack{1 \leqq s< t \leqq p-1 }} st \equiv 0 (\mod p)  
\end{aligned}

\begin{aligned}
  {}_{p+lp-1} \mathrm{C}_{lp} 
 & = \frac{\sum\limits_{k=1}^{p-1} c_k (lp)^k  }{(p-1)!}  +1\\

\end{aligned}

\begin{aligned}
  {}_{p+lp-1} \mathrm{C}_{lp} = \frac{ \xi p^3 }{(p-1)!}  +1\\

\end{aligned}