91Ó°ÊÓ

Wilson's Theorem states that if \(p\) is a prime number, then \((p-1)! \equiv -1 \pmod p\). We will be using this theorem in our proof. Recall also the properties of modular arithmetic: 1. If \(a \equiv b \pmod{m}\) and \(c \equiv d \pmod{m}\), then \(a+c \equiv b+d \pmod{m}\) and \(ac \equiv bd \pmod{m}\). 2. If \(a \equiv b \pmod{m}\), then \(ac \equiv bc \pmod{m}\). Now, let's use these properties to show the given congruence.

Since \(p\) is an odd prime, we have that \((p-1)\) is even. Let \((p-1)=2n\), where \(n\) is an integer greater than or equal to 1. We can rewrite the given expression as \(2(n!)(n-1)!\). By Wilson's theorem, we know that \((p-1)! \equiv -1 \pmod p\). Since \((p-1) = 2n\), we have: $$2n! \equiv -1 \pmod p$$

We want to show that: $$2(n!)(n-1)! \equiv -1 \pmod p$$ Multiply both sides of the previous equation (\(2n! \equiv -1 \pmod p\)) by \((n-1)!\): $$2n!(n-1)! \equiv -1(n-1)! \pmod p$$ Using the fact that \((p-1) = 2n\), we can rewrite the congruence as: $$2(p-3) ! \equiv -1(n-1)! \pmod p$$

Our final goal is to show that \(2(p-3) ! \equiv -1 \pmod p\). Since we have already reached the expression \(2(p-3) ! \equiv -1(n-1)! \pmod p\), we just need to show that \((n-1)! \equiv 1 \pmod p\). Consider the product \((n!)(n-1)!\). By rearranging the factors, we can rewrite it as \([(n-1)!][(n-1)!] = [(n-1)!]^2\) which is a square term. It's equivalent to the given expression \(2(p-3)!\) but modulo \(p\): $$[(n-1)!]^2 \equiv 2(p-3)! \pmod p$$ We already have the congruence \(2(p-3)! \equiv -1 (n-1)! \pmod p\). Comparing these two congruences, one can infer that $$[(n-1)!]^2 \equiv -1(n-1)! \pmod p.$$ Now, to get the desired congruence, we must show that \((n-1)! \equiv 1 \pmod p\). To do this, divide both sides of the above equation by \((n-1)!\). We get: $$(n-1)! \equiv 1 \pmod p$$ Now we can substitute this back into our expression in Step 3: $$2(p-3) ! \equiv -1(1) \pmod p$$ So, we have successfully shown the required congruence: $$2(p-3) ! \equiv -1 \pmod p$$