Exercise 5.14

Use the fact that U ( ℤ ∕ pℤ ) is cyclic to give a direct proof that ( − 3 ∕ p ) = 1 when p ≡ 1 ( mod 3 ) .[Hint : There is a ρ in U ( ℤ ∕ pℤ ) of order 3 . Show that ( 2 ρ + 1 ) 2 = − 3 .]

Answers

Proof. Suppose that p ≡ 1 ( mod 3 ) . Let g a generator of 𝔽 p ∗ . Then g has order p − 1 , thus ρ = g ( p − 1 ) ∕ 3 has order 3 . As ρ 3 = 1 , ρ ≠ 1 , then ρ 2 + ρ + 1 = 0 .

( 2 ρ + 1 ) 2 = 4 ρ 2 + 4 ρ + 1 = 4 ( ρ 2 + ρ + 1 ) − 3 = − 3 .

Thus ( − 3 p ) = 1 . □

The converse is also true for an odd prime p : if ( − 3 p ) = 1 , then there exists a ∈ 𝔽 p ∗ such that − 3 ¯ = a 2 . Then ρ = − 1 + a 2 ( = ( − 1 + a ) 2 − 1 ) has order 3 . Indeed ρ 2 = 1 + a 2 − 2 a 4 = − 2 − 2 a 4 = − 1 − a 2 , so

1 + ρ + ρ 2 = 1 + − 1 + a 2 + − 1 − a 2 = 0

thus ρ ≠ 1 , ρ 3 = 1 . The group 𝔽 p ∗ contains an element of order 3, therefore, by Lagrange’s theorem, 3 ∣ p − 1 , that is p ≡ 1 ( mod 3 ) .

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2022-07-19 00:00
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