Explore BrainMass

# The Distribution of Quadratic Residues

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

Theory of Numbers
Sums of Squares
Sums of Four Squares

1. Let N1(p) denote the number of pairs of integers in [1, p - 1] in which the first is a quadratic residue
and the second is a quadratic nonresidue modulo p. Prove that
N1(p) = (1/4) (p - ( - 1)^((p - 1)/2))

2. Let N2(p) denote the number of pairs of integers in [1, p - 1] in which the first is a quadratic nonresidue
and the second is a quadratic residue modulo p. Prove that
N2(p) = (1/4) (p - 2 + ( - 1)^((p - 1)/2))

3. Let N3(p) denote the number of pairs of integers in [1, p - 1] in which the first is a quadratic nonresidue
and the second is a quadratic nonresidue modulo p. Prove that
N3(p) = (1/4) (p - 2 + ( - 1)^((p - 1)/2))

4. Use the results of theorem 10-4 and corollary 10-1 to construct solutions of x^2 + y^2 =29.

5. Prove ( without assuming corollary 10-1) that, if p is a prime &#8801; 1 (mod 4), then there exists positive
integers m, x, and y such that x^2 + y^2 =mp, with p &#9532; x, p &#9532; y, 0 < m < p [ Hint: use the proof of
Theorem 11-2].

Theorem 10-4: If p is an odd prime, then &#957;(p) = (1/8)p + Ep where | Ep| < (1/4)(p)^(1/2) +2.

Corollary 10-1: Every prime p &#8801; 1(mod 4) is representable as a sum of two squares.

Theorem 11-2: For each prime p there exist integers A,B and C, not all zero, such that
A^2 + B^2 + C^2 &#8801; 0 (mod p).

See the attached file.

#### Solution Preview

Theory of Numbers
Sums of Squares
Sums of Four Squares

1. Let N1(p) denote the number of pairs of integers in [1, p - ...

#### Solution Summary

This solution is comprised of a detailed explanation of the Distribution of Quadratic Residues
and Sums of Four Squares.

It contains step-by-step explanation for the following problem:

1. Let N1(p) denote the number of pairs of integers in [1, p - 1] in which the first is a quadratic residue
and the second is a quadratic nonresidue modulo p. Prove that
N1(p) = (1/4) (p - ( - 1)^((p - 1)/2))

2. Let N2(p) denote the number of pairs of integers in [1, p - 1] in which the first is a quadratic nonresidue
and the second is a quadratic residue modulo p. Prove that
N2(p) = (1/4) (p - 2 + ( - 1)^((p - 1)/2))

3. Let N3(p) denote the number of pairs of integers in [1, p - 1] in which the first is a quadratic nonresidue
and the second is a quadratic nonresidue modulo p. Prove that
N3(p) = (1/4) (p - 2 + ( - 1)^((p - 1)/2))

4. Use the results of theorem 10-4 and corollary 10-1 to construct solutions of x^2 + y^2 =29.

5. Prove ( without assuming corollary 10-1) that, if p is a prime &#8801; 1 (mod 4), then there exists positive
integers m, x, and y such that x^2 + y^2 =mp, with p &#9532; x, p &#9532; y, 0 < m < p [ Hint: use the proof of
Theorem 11-2].

Theorem 10-4: If p is an odd prime, then &#957;(p) = (1/8)p + Ep where | Ep| < (1/4)(p)^(1/2) +2.

Corollary 10-1: Every prime p &#8801; 1(mod 4) is representable as a sum of two squares.

Theorem 11-2: For each prime p there exist integers A,B and C, not all zero, such that
A^2 + B^2 + C^2 &#8801; 0 (mod p).

Solution contains detailed step-by-step explanation.

\$2.49