compute $\sum_{k=0}^{(p-1)/2} \dfrac{(p-1)!}{(k!)^2(p-1-2k)!}\mod p$

Question

Let $p$ be an odd prime. Compute $\sum_{k=0}^{(p-1)/2} \dfrac{(p-1)!}{(k!)^2(p-1-2k)!}\mod p.$

For an odd prime p, let $S(p) = \sum_{k=0}^{(p-1)/2} \dfrac{(p-1)!}{(k!)^2(p-1-2k)!}$. Note that we may assume $p\equiv 1, 5\mod 6,$ since for $p=3$, we have $S(p) = \sum_{k=0}^1 \dfrac{2}{(k!)^2(2-2k)!} = 1+2 = 3\equiv 0\mod p.$ Computing a few more values, we get $S(5) = 4!(1 + 1/(1!^2\cdot 2!) + 1/(2!^2 \cdot 0!)) \equiv 2\mod 5, S(7) = 6! (1 + 1/(1!^2 \cdot 4!) + 1/(2!^2 \cdot 2!) + 1/(3!^2 \cdot 0!)) \equiv 6\mod 7.$ All congruences in this problem will be modulo p for simplicity. First consider the case where $p\equiv 1\mod 6$ and write $p= 6r + 5$ for some $r\ge 0$. Then $(p-1)/2 = 3r+2.$

We have $S(p) = \sum_{k=0}^{(p-1)/2} \dfrac{(p-1)(p-2)\cdots (p-2k)}{(k!)^2} \equiv \sum_{k=0}^{(p-1)/2} \dfrac{(2k)!(-1)^{2k}}{(k!)^2} = \sum_{k=0}^{(p-1)/2} \dfrac{(2k)!}{(k!)^2}$

But I'm not sure how to simplify this result. Would the Binomial Theorem be useful?

Answer 1

$$\begin{aligned} (2k)!&=(2k)(2k-2)\cdots2\cdot(2k-1)(2k-3)\cdots1\\ &=k!2^k(2k-1)(2k-3)\cdots1\\ &=k!(4k-2)(4k-6)\cdots2\\ &\equiv_pk!(4k-2-2p)(4k-6-2p)\cdots(2-2p)\\ &=k!(-4)^k(\frac{p-1}2-k+1)(\frac{p-1}2-k+2)\cdots \frac{p-1}2 \end{aligned}$$ _{The idea of the transformation above comes from this comment.} $$\begin{aligned} &\sum_{k=0}^{(p-1)/2} \dfrac{(p-1)!}{(k!)^2(p-1-2k)!}\\ &\equiv_p\sum_{k=0}^{\frac{p-1}2} \dfrac{(2k)!}{(k!)^2}\\ &=\sum_{k=0}^{\frac{p-1}2}\frac{k!(-4)^k({\frac{p-1}2}-k+1)({\frac{p-1}2}-k+2)\cdots {\frac{p-1}2}}{(k!)^2}\\ &=\sum_{k=0}^{\frac{p-1}2}(-4)^k\frac{({\frac{p-1}2}-k+1)({\frac{p-1}2}-k+2)\cdots {\frac{p-1}2}}{k!}\\ &=\sum_{k=0}^{\frac{p-1}2}(-4)^k{{\frac{p-1}2}\choose k}\\ &=(1+(-4))^{\frac{p-1}2}\\ &=(-3)^{\frac{p-1}2}\\ &=(-1)^{\frac{p-1}2}3^{\frac{p-1}2}\\ &\equiv_p(-1)^{\frac{p-1}2}\left({\frac {3}{p}}\right)\\ &=(-1)^{\frac{p-1}2}\left({\frac {p}{3}}\right)(-1)^{\frac{3-1}2\frac{p-1}2}\\ &=(-1)^{p-1}\left({\frac {p}{3}}\right)\\ &=\left({\frac {p}{3}}\right)\\ &=\begin{cases} 1 &\text{ if } p\equiv_31,\\ 0 &\text{ if } p=3,\\ -1 &\text{ if }p\equiv_3-1\end{cases} \end{aligned}$$

The place where $(1+(-4))^{\frac{p-1}2}$ appears uses the binomial theorem.
$(\frac{3}{p})$ and $(\frac{3}{p})$ are Legendre symbols.
The place where $(-1)^{\frac{3-1}2\frac{p-1}2}$ appears uses the law of quadratic reciprocity.