Show that $(r!)^2 ≡ (−1)^{r−1} \pmod p,\ r = (p-1)/2$

Question

I need to prove that if p is an odd prime and

$r = (p-1)/2$ then

$(r!)^2 ≡ (−1)^{r−1} \pmod p$

I think it has something to do with gauss's lemma

but I tried a lot and couldn't find a way to break it .

any help or hint will be appreciated.

Special case $,h = (p-1)/2 = r,$ in the linked Wilson Reflection Formula. See also the comments here for the more difficult problem of determining $r!$ (i.e. the sign of the square root). — Bill Dubuque, May 04 '23 at 19:19

score 5 · Accepted Answer · answered Jun 25 '15 at 15:24

Hint: We have $p-1\equiv -1\pmod{p}$, $p-2\equiv -2\pmod{p}$, $p-3\equiv -3\pmod{p}$, and so on up to $\frac{p+1}{2}\equiv -\frac{p-1}{2}\pmod{p}$.

It follows that $$(p-1)!\equiv (-1)^{(p-1)/2}\left(\left(\frac{p-1}{2}\right)!\right)^2\pmod{p}.$$ Now use Wilson's Theorem.

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