Exercise 2.1.43 (Consequences of Wilson's theorem)

Question

If $p$ is an odd prime, prove that:
$$1^2\cdot 3^2\cdot5^2\cdots (p-2)^2 \equiv (-1)^{(p+1)/2} \pmod p.$$
and 
$$2^2\cdot 4^2 \cdot 6^2\cdots (p-1)^2 \equiv (-1)^{(p+1)/2} \pmod p.$$

Richard Ganaye · Accepted Answer

\begin{proof} Recall from the text (proof of theorem 2.12) that, by Wilson's theorem, for any odd prime $p$,
$$\left(1\cdot 2 \cdot3 \cdots \cdot\left(\frac{p-1}{2}ight)ight)\left(\left(\frac{p+1}{2}ight)\cdots (p-3)(p-2)(p-1)ight)\equiv -1 \pmod p,$$
so that
$$\prod_{j=1}^{(p-1)/2} j(p-j) \equiv -1 \pmod p,$$
therefore
$$(-1)^{\frac{p-1}{2}} \prod _{j=1}^{(p-1)/2}j^2 \equiv -1 \pmod p,$$
which gives
\begin{align}
\prod _{j=1}^{(p-1)/2} j^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p.
\end{align}

\bigskip
Write
\begin{align*}
P &= 2^2\cdot 4^2 \cdot 6^2\cdots (p-1)^2,\
Q&= 1^2\cdot 3^2\cdot5^2\cdots (p-2)^2.
\end{align*}
Then
\begin{align*}
P&= \prod_{k=1}^{(p-1)/2} (2k)^2\
&=2^{p-1}\prod_{k=1}^{(p-1)/2} k^2\
&\equiv \prod_{k=1}^{(p-1)/2} k^2\qquad\qquad \qquad \qquad 	ext{(by Fermat's theorem)}\
&\equiv(-1)^{\frac{p+1}{2}} \pmod p\qquad \qquad 	ext{(by (1))}.
\end{align*}
Now, by Wilson's theorem,
\begin{align*}
PQ &=(p-1)!^2\
&=1.
\end{align*}
Therefore,
$$(-1)^{\frac{p+1}{2}}  Q \equiv 1\pmod p,$$
so
$$Q \equiv (-1)^{\frac{p+1}{2}}\pmod p .$$
The conclusion is
\begin{align*}
 2^2\cdot 4^2 \cdot 6^2\cdots (p-1)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p,\
1^2\cdot 3^2\cdot5^2\cdots (p-2)^2\equiv (-1)^{\frac{p+1}{2}} \pmod p.
\end{align*}
\end{proof}

Exercise 2.1.43 (Consequences of Wilson's theorem)

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Exercise 2.1.43 (Consequences of Wilson's theorem)

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