Given a prime number $p$ , establish the congruence:
$$(p-1)! \equiv (p-1) \pmod{1+2+3+\cdots+(p-1)}$$
I have proceeded like this:
$$\begin{align*}&(p-1)! \equiv (-1) \pmod{p} \quad \quad \quad \text{by Wilson's Theorem}\\
&(p-1)! \equiv 0 \pmod{\frac{p-1}{2}} \end{align*}$$
Then I know that I have to apply Chinese remainder theorem but I don't have a thorough understanding of it.So please give me an elaborate answer to this question with respect to Chinese remainder theorem.
Best Answer
One more way to think of this is that Wilson's Theorem says $$ (p-1)!-(p-1)\equiv0\pmod{p}\tag{1} $$ and because $(p-1)!\equiv p-1\equiv0\quad\left(\!\bmod\frac{p-1}{2}\right)$ $$ (p-1)!-(p-1)\equiv0\quad\left(\!\bmod\frac{p-1}{2}\right)\tag{2} $$ Since $\left(p,\frac{p-1}{2}\right)=1$, the Chinese Remainder Theorem says $$ (p-1)!-(p-1)\equiv0\quad\left(\!\bmod p\cdot\frac{p-1}{2}\right)\tag{3} $$