Wilson’s Theorem. Lemma If p is a prime, then the only solutions to x 2 p 1 are those integers x satisfying x p 1 or x p -1 Proof: x 2 p 1 x 2 - 1 p 0 (x - 1)(x+1) p 0 p | (x - 1)(x+1).
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x2p 1 x2-1p 0 (x-1)(x+1) p 0 p | (x-1)(x+1).
p prime and p | (x-1)(x+1) p | (x-1) or p | (x+1) x p 1 or x p -1
Assume p is prime. Notice that if 1 < a < p-1, then a-1 1 and a-1 p-1, because 1 and p-1 are their own inverses mod p. Thus 1 < a-1 < p-1.
We also know that a2 1 by the previous lemma, and thus a-1 a.
Therefore, we may rearrange the product 23 (p-2) into pairs of the form aa-1 and thus the product evaluates to 1 (modulo p). It then follows that (p-1)! p -1.
Now assume p is not prime. If p = 4, then (p-1)! = 3! = 6 4 2 -1. So we may assume p > 4.
Since p is composite, p = ab with 1 < a b < p. If a = b, then a > 2 since a2 = p > 4. Thus 1 < a < 2a < a2 = p.
But this means that a and 2a appear in the product (p-1)!; since 2aa = 2a2 = 2p p 0, the entire product is congruent to 0 mod p.
We now have a < b < p, so that both a and b appear in the product (p-1)!. Since ab = p, we again have (p-1)! p 0. Thus (p-1)! is not congruent to -1 modulo p.
234 = 23222 35 114 = 44 35 9, which is not congruent to 1 mod 35.
Next we mark with bold face the entries that are relatively prime to 72
There are two facts that stand out:
1. the selected numbers only appear in columns headed by the integers
that are relatively prime to n = 8: 1, 3, 5, and 7
2. Each of these columns contains exactly (m) entries
This pattern suggests that (mn) = (m)(n) when m n and is the basis of a proof of this fact. We omit the proof here.
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