Exercise 5.9

Prove that 1 2 3 2 ⋯ ( p − 2 ) 2 ≡ ( − 1 ) ( p + 1 ) ∕ 2 ( mod p ) using Wilson’s theorem.

Answers

Proof. Here p is an odd prime.

From Wilson’s theorem, as k ( p − k ) ≡ − k 2 ( mod p ) for k = 1 , 2 , … , p − 1 ,

− 1 ≡ ( p − 1 ) ! ≡ [ 1 × 2 × ⋯ × k × ⋯ × ( p − 1 2 ) ] × [ ( p + 1 2 ) × ⋯ × ( p − k ) ⋯ × ( p − 2 ) × ( p − 1 ) ] ≡ ∏ k = 1 ( p − 1 ) ∕ 2 k ( p − k ) ≡ ( − 1 ) ( p − 1 ) ∕ 2 ∏ k = 1 ( p − 1 ) ∕ 2 k 2 ≡ ( − 1 ) ( p − 1 ) ∕ 2 [ ( p − 1 2 ) ! ] 2 ( mod p )

Therefore

[ ( p − 1 2 ) ! ] 2 ≡ ( − 1 ) ( p + 1 ) ∕ 2 ( mod p ) .

Moreover, from Wilson’ theorem and Fermat’s little theorem,

1 2 2 2 3 2 ⋯ ( p − 1 ) 2 = [ ( p − 1 ) ! ] 2 ≡ 1 ( mod p ) , 2 2 4 2 ⋯ ( p − 1 ) 2 = ( 2 p − 1 ) 2 [ ( p − 1 2 ) ! ] 2 ≡ [ ( p − 1 2 ) ! ] 2 ( mod p ) .

Thus

1 2 3 2 ⋯ ( p − 2 ) 2 [ ( p − 1 2 ) ! ] 2 ≡ 1 ( mod p ) ,

which gives

1 2 3 2 ⋯ ( p − 2 ) 2 ≡ ( − 1 ) ( p + 1 ) ∕ 2 ( mod p ) .

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2022-07-19 00:00
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