MATH312 December 2009
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Question 04 (b)

Suppose that p is an odd prime. Then show that
$\displaystyle 1^{2}\cdot 3^{2}\cdot 5^{2}\cdot ...\cdot (p2)^{2}\equiv (1)^{(p+1)/2}\mod {p}$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1

Try to find a way to make use of Wilson's Theorem.

Hint 2

Multiply the product by $\displaystyle (1)^{p1}$. Write each square like
$\displaystyle 1^{2}\equiv (1)(1)\mod {p}$.
making use of all the negative one powers you included. Then add p to all the negative terms and examine what you get.

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 If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
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Solution

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We begin by multiplying the product by $\displaystyle (1)^{p1}\equiv 1\mod {p}$. This gives
$\displaystyle 1^{2}\cdot 3^{2}\cdot 5^{2}\cdot ...\cdot (p2)^{2}\equiv 1^{2}\cdot 3^{2}\cdot 5^{2}\cdot ...\cdot (p2)^{2}\cdot (1)^{p1}\mod {p}$
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 $\displaystyle (p1)/2$ terms in our product. So we need to take a factor of $\displaystyle (1)^{(p1)/2}$ to accomplish this. This gives
${\begin{aligned}1^{2}\cdot &3^{2}\cdot 5^{2}\cdot ...\cdot (p2)^{2}\cdot (1)^{p1}\\&\equiv (1)(1)(3)(3)(5)(5)...(p2)((p2))\cdot (1)^{(p1)/2}\mod {p}\end{aligned}}$
Now for each negative term, we add p to get.
${\begin{aligned}(1)(1)&(3)(3)(5)(5)...(p2)((p2))\cdot (1)^{(p1)/2}\\&\equiv (1)(p1)(3)(p3)(5)(p5)...(p2)(2)\cdot (1)^{(p1)/2}\mod {p}\end{aligned}}$
After rearranging the right hand side above, we see that
${\begin{aligned}(1)(p1)&(3)(p3)(5)(p5)...(p2)(2)\cdot (1)^{(p1)/2}\\&\equiv (p1)!\cdot (1)^{(p1)/2}\mod {p}\end{aligned}}$
By Wilson's Theorem and combining the above, we have
${\begin{aligned}1^{2}\cdot &3^{2}\cdot 5^{2}\cdot ...\cdot (p2)^{2}\\&\equiv (p1)!\cdot (1)^{(p1)/2}\mod {p}\\&\equiv (1)\cdot (1)^{(p1)/2}\mod {p}\\&\equiv (1)^{(p+1)/2}\mod {p}\end{aligned}}$
and this was what we wanted to show.

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