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An Introduction to the Theory of Numbers

Exercise 2.1.51 ( $(p-1)! \equiv p-1 \pmod{ 1 + 2 +\cdots + (p-1)}$ if $p$ is a prime)

Prove that ( p − 1 ) ! ≡ p − 1 ( mod 1 + 2 + ⋯ + ( p − 1 ) ) if p is a prime.

Answers

Proof. If p = 2 , then ( p − 1 ) ! = ( 2 − 1 ) ! = 1 = p − 1 , a fortiori ( p − 1 ) ! is congruent to p − 1 modulo any integer.

Suppose now that p is an odd prime. Then

1 + 2 + ⋯ + ( p − 1 ) = p − 1 2 p .

Since p − 1 2 ∧ p = 1 (because p − 2 ( p − 1 2 ) = 1 ), it suffices to show that

( p − 1 ) ! ≡ p − 1 ( m o d p ) , (1) ( p − 1 ) ! ≡ p − 1 ( m o d ( p − 1 ) ∕ 2 ) . (2)
(1)
By Wilson’s theorem, ( p − 1 ) ! ≡ − 1 ≡ p − 1 ( m o d p ) .

(2)
Since p − 1 2 ∣ p − 1 , ( p − 1 ) ! ≡ 0 ≡ p − 1 ( m o d ( p − 1 ) ∕ 2 ) .

So p ∣ ( p − 1 ) ! − ( p − 1 ) and p − 1 2 ∣ p − 1 ) ! − ( p − 1 ) , where p − 1 2 ∧ p = 1 . Therefore p − 1 2 p ∣ ( p − 1 ) ! − ( p − 1 ) .

For any prime p ,

( p − 1 ) ! ≡ p − 1 ( m o d 1 + 2 + ⋯ + ( p − 1 ) ) ) .

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2024-08-04 16:03
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