Exercise 8.19

Find a formula for the number of solutions to x 1 2 + x 2 2 + ⋯ + x r 2 = 0 in 𝔽 p .

Answers

Proof. Let χ be the Legendre character. Then

N ( x 1 2 + x 2 2 + ⋯ + x r 2 = 0 ) = ∑ a 1 + a 2 + ⋯ + a r = 0 N ( x 1 2 = a 1 ) N ( x 2 2 = a 2 ) ⋯ N ( x r 2 = a r ) = ∑ a 1 + a 2 + ⋯ + a r = 0 ( 1 + χ ( a 1 ) ) ( 1 + χ ( a 2 ) ) ⋯ ( 1 + χ ( a r ) ) = p r − 1 + J 0 ( χ , χ , ⋯ , χ )

(We used Proposition 8.5.1). For all k , χ 2 k = 𝜀 , χ 2 k + 1 = χ .

∙ If r is odd, χ r ≠ 𝜀 , so J 0 ( χ , χ , ⋯ , χ ) = 0 (Proposition 8.5.1).

N ( x 1 2 + x 2 2 + ⋯ + x r 2 = 0 ) = p r − 1 .

∙ If r is even, χ r = 𝜀 , so J 0 ( χ , χ , ⋯ , χ ) = χ ( − 1 ) ( p − 1 ) J ( χ , χ , ⋯ , χ ) , where there are r − 1 components in the Jacobi sum ( Proposition 8.5.1).

By Theorem 3, J ( χ , χ , ⋯ , χ ) g ( χ r − 1 ) = g ( χ ) r − 1 , and g ( χ r − 1 ) = g ( χ ) , so

J ( χ , χ , ⋯ , χ ) = g ( χ ) r − 2 .

g ( χ ) 2 = χ ( − 1 ) p , therefore g ( χ ) r − 2 = χ ( − 1 ) r ∕ 2 − 1 p r ∕ 2 − 1 = ( − 1 ) ( ( p − 1 ) ∕ 2 ) ( r ∕ 2 − 1 ) p ( r ∕ 2 − 1 ) . So

N ( x 1 2 + x 2 2 + ⋯ + x r 2 = 0 ) = p r − 1 + ( − 1 ) p − 1 2 r 2 ( p − 1 ) p r 2 − 1 .

(Verified in C++ with small values of p and r .)

Conclusion :

{ N ( x 1 2 + x 2 2 + ⋯ + x r 2 = 0 ) = p r − 1 if r is odd , = p r − 1 + ( − 1 ) p − 1 2 r 2 ( p − 1 ) p r 2 − 1 if r is even .

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2022-07-19 00:00
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