Science:Math Exam Resources/Courses/MATH312/December 2009/Question 04 (b)/Solution 1

We begin by multiplying the product by $(- 1)^{p - 1} \equiv 1 mod p$ . This gives

$1^{2} \cdot 3^{2} \cdot 5^{2} \cdot . . . \cdot (p - 2)^{2} \equiv 1^{2} \cdot 3^{2} \cdot 5^{2} \cdot . . . \cdot (p - 2)^{2} \cdot (- 1)^{p - 1} mod p$

Now, rewrite each square as $a^{2} = a \cdot a$ and for each value of a, distribute a power of -1. Notice that we have $(p - 1) / 2$ terms in our product. So we need to take a factor of $(- 1)^{(p - 1) / 2}$ to accomplish this. This gives

$\begin{aligned} 1^{2} \cdot & 3^{2} \cdot 5^{2} \cdot . . . \cdot (p - 2)^{2} \cdot (- 1)^{p - 1} \\ \equiv (1) (- 1) (3) (- 3) (5) (- 5) . . . (p - 2) (- (p - 2)) \cdot (- 1)^{(p - 1) / 2} mod p \end{aligned}$

Now for each negative term, we add p to get.

$\begin{aligned} (1) (- 1) & (3) (- 3) (5) (- 5) . . . (p - 2) (- (p - 2)) \cdot (- 1)^{(p - 1) / 2} \\ \equiv (1) (p - 1) (3) (p - 3) (5) (p - 5) . . . (p - 2) (2) \cdot (- 1)^{(p - 1) / 2} mod p \end{aligned}$

After rearranging the right hand side above, we see that

$\begin{aligned} (1) (p - 1) & (3) (p - 3) (5) (p - 5) . . . (p - 2) (2) \cdot (- 1)^{(p - 1) / 2} \\ \equiv (p - 1)! \cdot (- 1)^{(p - 1) / 2} mod p \end{aligned}$

By Wilson's Theorem and combining the above, we have

$\begin{aligned} 1^{2} \cdot & 3^{2} \cdot 5^{2} \cdot . . . \cdot (p - 2)^{2} \\ \equiv (p - 1)! \cdot (- 1)^{(p - 1) / 2} mod p \\ \equiv (- 1) \cdot (- 1)^{(p - 1) / 2} mod p \\ \equiv (- 1)^{(p + 1) / 2} mod p \end{aligned}$

and this was what we wanted to show.