Prove the following using the Wilson's Theorem

Question

Given a prime number, $p$, prove that $(p-1)!\equiv p-1\,\,\left(\text{mod }\frac{p(p-1)}2\right)$

How do we modify the Wilson's theorem into modulo $p(p-1)/2$ ? I can't get any clue

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Answer 1

We can divide by $p-1$. So, we only need to prove that $(p-2)!\equiv1 \pmod{p/2}$.

And multiply by $2$, for $p>2$. So, we only need to prove that $2(p-2)!\equiv 2 \pmod p$.

By Wilson $(p-1)!\equiv -1$. Then $-(p-2)!\equiv (p-1)(p-2)!\equiv -1$.

Hence $(p-2)!\equiv 1$.

Answer 2

Start with Wilson's Theorem $$ (p-1)!\equiv-1\pmod{p} $$ Since $(p,p-1)=1$, we can divide by $p-1\equiv-1\pmod{p}$ to get $$ (p-2)!\equiv1\pmod{p} $$ Multiply by $p-1$ to get $$ (p-1)!\equiv p-1\pmod{p(p-1)} $$ which implies that $$ (p-1)!\equiv p-1\,\,\,\left(\text{mod }\frac{p(p-1)}2\right) $$

Answer 3

$(p\!-\!1)!-(p\!-\!1)\, $ is divisible by coprimes $\,p\!-\!1\,$ and $\,p\,$ (by Wilson), so is divisible by $\,(p\!-\!1)p$