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

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

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

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