Exercise 5.27

Suppose that f is such that b ≡ af ( mod p ) . Show that f 2 ≡ − 1 ( mod p ) , and that 2 ( p − 1 ) ∕ 4 ≡ f ab ∕ 2 ( mod p ) .

Answers

Proof. Let f such as b ≡ af [ p ] .

This is equivalent to f ¯ = b ¯ a ¯ − 1 in 𝔽 p ∗ .

As a ¯ 2 = − b ¯ 2 , f ¯ 2 = − 1 ¯ , so that f 2 ≡ − 1 [ p ] .

We deduce from Ex. 5.26 (d) and (b) that

( 2 ab ) p − 1 4 ≡ ( a + b ) p − 1 2 = ( a + b p ) ≡ ( − 1 ) ( a + b ) 2 − 1 8 ≡ ( f 2 ) ( a + b ) 2 − 1 8 ≡ f ( a + b ) 2 − 1 4 = f a 2 + b 2 − 1 + 2 ab 4 ≡ f p − 1 4 f ab 2 ( mod p )

Since a p − 1 2 = ( a p ) = 1 by Ex. 5.26(a)), then

( ab ) p − 1 4 ≡ ( a 2 f ) p − 1 4 ≡ a p − 1 2 f p − 1 4 ≡ f p − 1 4 [ p ] ,

so

2 p − 1 4 f p − 1 4 ≡ f ab 2 f p − 1 4 [ p ] .

As f p − 1 4 ≢ 0 [ p ] ,

2 p − 1 4 ≡ f ab 2 [ p ] .

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2022-07-19 00:00
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