Homepage › Solution manuals › Ivan Niven › An Introduction to the Theory of Numbers › Exercise 2.1.18 (if $p\equiv 3 \pmod 4$, then $\left(\frac{p-1}{2} \right)! \equiv \pm 1 \pmod p$)

An Introduction to the Theory of Numbers

Exercise 2.1.18 (if $p\equiv 3 \pmod 4$, then $\left(\frac{p-1}{2} \right)! \equiv \pm 1 \pmod p$)

Show that if p ≡ 3 ( mod 4 ) , then ( p − 1 2 ) ! ≡ ± 1 ( mod p ) .

Answers

Proof. In the proof of Theorem 2.12, we saw that for odd primes p , Wilson’ theorem gives the following congruences modulo p :

− 1 ≡ ( p − 1 ) ! ≡ ∏ j = 1 ( p − 1 ) ∕ 2 j ∏ j = ( p + 1 ) ∕ 2 p − 1 j ≡ ∏ j = 1 ( p − 1 ) ∕ 2 j ∏ i = 1 ( p − 1 ) ∕ 2 ( p − i ) ( j = p − i ) ≡ ∏ j = 1 ( p − 1 ) ∕ 2 j ∏ j = 1 ( p − 1 ) ∕ 2 ( p − j ) ≡ ∏ j = 1 ( p − 1 ) ∕ 2 j ( p − j ) ≡ ( − 1 ) ( p − 1 ) ∕ 2 ∏ j = 1 ( p − 1 ) ∕ 2 j 2 ≡ ( − 1 ) ( p − 1 ) ∕ 2 [ ( p − 1 2 ) ! ] 2 ( mod p ) .

Therefore

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

If p ≡ 3 ( mod 4 ) , p = 3 + 4 k , k ∈ , thus p + 1 2 = 2 ( k + 1 ) is even, so ( − 1 ) ( p + 1 ) ∕ 2 = 1 . in this case,

[ ( p − 1 2 ) ! ] 2 ≡ 1 ( mod p ) .

Then

p ∣ [ ( p − 1 2 ) ! − 1 ] [ ( p − 1 2 ) ! + 1 ] .

Since p is a prime number,

p ∣ ( p − 1 2 ) ! − 1  or  p ∣ ( p − 1 2 ) ! + 1 .

This show that

( p − 1 2 ) ! ≡ ± 1 ( mod p ) .

□

User profile picture
2024-08-21 10:35
Comments