0
$\begingroup$

Im a noob and trying to prove: $(m!)^2\equiv(-1)^{m+1}\textrm{(mod p)}$ using Wilson's Theorem. Although it's on Wikipedia, I want to really understand what is happening. Could someone explain how $$\begin{align*}1*2\dotsb*(p-1)=1*(p-1)*2*(p-2)\dotsb m*(p-m)\end{align*}$$ What is m? How is it possible for (p-1)! to have this expansion. Then could someone also explain the property that was used to transform this: $$(-1)^m(m!)^2\equiv-1\textrm{(mod p)}$$ into $$(m!)^2\equiv(-1)^{m+1}\textrm{(mod p)}$$ I understand that $-1^1*-1^m=-1^{m+1}$, but how did the $-1^m$ move from the left side to the right. Thanks for help! Wilson's Theorem Quadratic Residue

$\endgroup$
0
$\begingroup$

$m$ is $(p-1)/2$.

You can multiply a congruence by the same thing on both sides. Also, $(-1)^m\cdot(-1)^m=(-1)^{2m}=1$. Thus the result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.