## § Wilson's theorem

• We get $p \equiv 1$ (mod $4$) implies $((p-1)/2)!$ is a square root of -1.
• It turns that this is because from Wilson's theorem, $(p-1)! = -1$.
• Pick $p = 13$. Then in the calculation of $(p-1)!$, we can pair off $6$ with $-6=7$, $5$ with $-5=8$ and so on.
• So we get $(p-1)/2 \times (p-1)/2 = (p-1)!$.
• This means that $(p-1)/2 = \sqrt{-1}$.
• The condition $(p-1)/2$ is even is the same as saying that $p-1$ is congruent to $0$ mod $4$, or that $p$ is congruent to $1$ mod $4$.
• It's really nice to be able to see where this condition comes from!