AoPSWiki
Try our innovative online adaptive learning system, Alcumus.
Over 1100 problems and 60+ video lessons. FREE!

User:Temperal/The Problem Solver's Resource9

From AoPSWiki

Introduction | Other Tips and Tricks | Methods of Proof | You are currently viewing page 9.

Advanced Number Theory

Back to page 8 | Continue to page 10

Quadratic Reciprocity

A number m is a quadratic residue mod n if and only there exists a number a such that a^2 \equiv m \pmod{n} and 0 \le m < n.. For example, since 5^2 \equiv 4 \pmod{7}, 4 is a quadratic residue mod 7.

Let p be an odd prime. There are some theorems regarding quadratic residues mod p. Usually, 0 is a "special case." So 0 is not a residue, but it is also not a non-residue.

Adopting these definitions, we prove some theorems.

Theorem 9.1. There are an equal number of non-residues and residues mod p.

Proof: First, we prove that 1^2, 2^2, ... (\frac{p-1}{2})^2 are distinct modulo p. Assume a^2 \equiv b^2 \pmod{p}, where a and b are distinct integers from 1 to \frac{p-1}{2}. Then, a^2-b^2 would have to be a multiple of p. a^2-b^2 can be factored as (a-b)(a+b). However, since a and b are distinct integers from 1 to \frac{p-1}{2}, a-b can't be a multiple of p. Therefore, a+b is a multiple of p. We know that \frac{p-1}{2} \ge a,b \ge 1, so p-1 \ge a+b \ge 2, so a+b also can't be a multiple of p, a contradiction.

Our Precalculus course starts on Dec. 4. Master trig, complex numbers, and vectors and matrices in 2 and 3 dimensions. Click here to enroll today!
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us