2.1: Problem 18: Show that if \(p \equiv 3 \pmod{4}\), then \(\left(\frac{p-1}{2}\right)! \equiv \pm 1 \pmod{p}\)
Proof
The expansion of \((p-1)!\) is
$$
\begin{align*}
1 \cdot 2 \cdot 3 \cdot 4 \cdots (p-1)
\end{align*}
$$
Observe that \(\left(\frac{p-1}{2}\right)!\) is the product of the first \(\frac{p-1}{2}\) elements above. But these elements pair off with the remaining \(\frac{p-1}{2}\) numbers (except for the sign) meaning that
$$
\begin{align*}
(p-1) &\equiv -1 \pmod{p} \\
(p-2) &\equiv -2 \pmod{p} \\
(p-3) &\equiv -3 \pmod{p} \\
&\cdots
\end{align*}
$$
Therefore, we can write
$$
\begin{align*}
(p-1)! \equiv \left(\frac{p-1}{2}\right)! \cdot \left(\frac{p-1}{2}\right)! \cdot (-1)^{\frac{p-1}{2}} \pmod{p}
\end{align*}
$$
But by Wilson’s theorem, we know that
$$
\begin{align*}
(p-1)! \equiv -1 \pmod{p}
\end{align*}
$$
Thus
$$
\begin{align*}
-1 &\equiv \left(\frac{p-1}{2}\right)! \cdot \left(\frac{p-1}{2}\right)! \cdot (-1)^{\frac{p-1}{2}} \pmod{p} \\
-1 &\equiv \left[\left(\frac{p-1}{2}\right)!\right]^2 \cdot (-1)^{\frac{p-1}{2}} \pmod{p}
\end{align*}
$$
Now, recall that
$$
\begin{align*}
p \equiv 3 \pmod{4}
\end{align*}
$$
so \(p - 3 = 4k\) for some \(k\). Thus,
$$
\begin{align*}
\frac{p-1}{2} = \frac{4k+3-1}{2} = \frac{4k+2}{2} = 2k+1
\end{align*}
$$
This means that \(\frac{p-1}{2}\) is odd and not even. Therefore
$$
\begin{align*}
1 &\equiv \left[\left(\frac{p-1}{2}\right)!\right]^2 \pmod{p}
\end{align*}
$$
Since the only square roots of \(1 \pmod{p}\) are \(\pm 1\) we conclude
$$
\begin{align*}
\pm 1 &\equiv \left(\frac{p-1}{2}\right)! \pmod{p}
\end{align*}
$$
as desired. \(\ \blacksquare\)