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§ Wilson's Theorem

created 2022-02-03
  • We get p≡1p \equiv 1p≡1 (mod 444) implies ((p−1)/2)!((p-1)/2)!((p−1)/2)! is a square root of -1.
  • It turns that this is because from Wilson's theorem, (p−1)!=−1(p-1)! = -1(p−1)!=−1.
  • Pick p=13p = 13p=13. Then in the calculation of (p−1)!(p-1)!(p−1)!, we can pair off 666 with −6=7-6=7−6=7, 555 with −5=8-5=8−5=8 and so on.
  • So we get (p−1)/2×(p−1)/2=(p−1)!(p-1)/2 \times (p-1)/2 = (p-1)!(p−1)/2×(p−1)/2=(p−1)!.
  • This means that (p−1)/2=−1(p-1)/2 = \sqrt{-1}(p−1)/2=−1​.
  • The condition (p−1)/2(p-1)/2(p−1)/2 is even is the same as saying that p−1p-1p−1 is congruent to 000 mod 444, or that ppp is congruent to 111 mod 444.
  • It's really nice to be able to see where this condition comes from!
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