Exercise 8.20

Generalize Proposition 8.6.1 by finding an explicit formula for the number of solutions to a 1 x 1 2 + a 2 x 2 2 + ⋯ + a r x r 2 = 1 in 𝔽 p .

Answers

Proof. Write χ the Legendre character.

N ( a 1 x 1 2 + ⋯ + a r x r 2 = 1 ) = ∑ a 1 u 1 + ⋯ + a r u r = 1 N ( x 1 2 = u 1 ) ⋯ N ( x r 2 = u r ) = ∑ a 1 u 1 + ⋯ + a r u r = 1 ( 1 + χ ( u 1 ) ) ⋯ ( 1 + χ ( u r ) ) ( v i = a i u i ) = ∑ v 1 + ⋯ + v r = 1 ( 1 + χ ( a 1 ) − 1 χ ( v 1 ) ) ⋯ ( 1 + χ ( a r − 1 ) χ ( v r ) ) = p r − 1 + χ ( a 1 − 1 ) ⋯ χ ( a r − 1 ) J ( χ , χ , ⋯ , χ ) .

χ ( a i − 1 ) = χ ( a i ) ¯ = χ ( a i ) = ( a i p ) .

J ( χ , χ , ⋯ , χ ) is computed in Chapter 8 Section 6. We obtain

{ N ( a 1 x 1 2 + ⋯ + a r x r 2 = 1 ) = p r − 1 + ( a 1 p ) ⋯ ( a r p ) ( − 1 ) r − 1 2 p − 1 2 p r − 1 2 if r is odd , = p r − 1 − ( a 1 p ) ⋯ ( a r p ) ( − 1 ) r 2 p − 1 2 p r 2 − 1 if r is even .

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2022-07-19 00:00
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