Exercise 5.16

Using quadratic reciprocity find the primes for which 7 is quadratic residue. Do the same for 15 .

Answers

Proof. 7 is a quadratic residue for 2 and for the odd primes such that ( 7 p ) = 1 .

From the law of quadratic reciprocity,

( 7 p ) = 1 ⟺ ( − 1 ) ( p − 1 ) ∕ 2 ( p 7 ) = 1

iff either p ≡ 1 ( mod 4 ) and ( p 7 ) = 1 , or p ≡ − 1 ( mod 4 ) and ( p 7 ) = − 1 .

In the first case , p ≡ 1 ( mod 4 ) , p ≡ 1 , 4 , 2 ( mod 7 ) , which gives p ≡ 1 , − 3 , 9 ( mod 28 ) .

In the second case, p ≡ − 1 ( mod 4 ) , p ≡ − 1 , − 4 , − 2 ( mod 7 ) , which gives p ≡ − 1 , 3 , − 9 ( mod 28 ) .

Conclusion : the primes for which 7 is a quadratic residue are 2 and the odd primes p such that

( 7 p ) = 1 ⟺ p ≡ ± 1 , ± 3 , ± 9 ( mod 28 ) .

□

15 is a quadratic residue for 2 and for the odd primes such that ( 15 p ) = 1 .

( 15 p ) = 1 ⟺ ( 3 p ) = ( 5 p ) = 1 or ( 3 p ) = ( 5 p ) = − 1

From the examples of theorem 2, we know that

( 3 p ) = 1 ⟺ p ≡ 1 , − 1 ( mod 12 ) , ( 3 p ) = − 1 ⟺ p ≡ 5 , − 5 ( mod 12 ) ,

( 5 p ) = 1 ⟺ p ≡ 1 , − 1 ( mod 5 ) , ( 5 p ) = − 1 ⟺ p ≡ 2 , − 2 ( mod 5 ) .

As 5 ∧ 12 = 1 , there exist 8 cases, all possible, which give

( 15 p ) = 1 ⟺ p ≡ ± 1 , ± 7 , ± 11 , ± 17 ( mod 60 ) .

For instance, the primes 2 , 7 , 11 , 17 , 43 , 53 , 59 , 61 , 67 , 137 , … are suitable.

User profile picture
2022-07-19 00:00
Comments