Table of Contents
What is a quadratic residue modulo n?
From Wikipedia, the free encyclopedia. In number theory, an integer q is called a quadratic residue modulo n if it is congruent to a perfect square modulo n; i.e., if there exists an integer x such that: Otherwise, q is called a quadratic nonresidue modulo n.
What are the quadratic residues mod 7?
Thus 1,2,4 are quadratic residues modulo 7 while 3,5,6 are quadratic nonresidues modulo 7. has exactly two incongruent solutions modulo p. If x2≡a(mod p) has a solution, say x=x′, then −x′ is a solution as well.
Is 31 a quadratic residue of modulo 47?
Since the Legendre symbol evaluates to −1, 31 is not a quadratic residue modulo 47. Page 25 Quadratic Reciprocity, VIII Example: Determine whether 357 is a quadratic residue mod 661. Page 26 Quadratic Reciprocity, VIII Example: Determine whether 357 is a quadratic residue mod 661. We want to find (357 661 ) .
Is a quadratic residue mod p if and only if?
modulo p if and only if p≡1(mod3) p ≡ 1 ( mod 3 ) . Proof. Preliminary to the proof, we remark first that −1 is a quadratic residue modulo p , where p is an odd prime, if and only if p≡1(mod4) p ≡ 1 ( mod 4 ) . (−3p)=(−1p)(3p)=(p3)….result on quadratic residues.
| Title | result on quadratic residues |
|---|---|
| Classification | msc 11-00 |
What is the difference between a quadratic residue and a nonresidue?
In other words, a is a quadratic residue if and only if a(p−1)/2 = 1, and a is a quadratic nonresidue if and only if a(p−1)/2 = −1. Example: We have −1 is a quadratic residue in Zp if and only if p = 1 (mod 4).
How do you find the quadratic residue of an odd prime?
Let p be an odd prime, as the case p = 2 is trivial. Let g be a generator of Z p ∗ . Any a ∈ Z p ∗ can be written as g k for some k ∈ [ 0.. p − 2]. Say k is even. Write k = 2 m. Then ( g m) 2 = a, so a is a quadratic residue.
How do you find the square root of a quadratic residue?
Suppose we have b 2 = a. Then ( − b) 2 = a as well, and since b ≠ − b (since p > 2) every quadratic residue has at least two square roots (in fact, we know from studying polynomials there can be at most two), thus at most half the elements of Z p ∗ are quadratic residues. (Otherwise there are more square roots than elements!)
How many elements of Z p – 2 are quadratic residues?
Exactly half of [ 0.. p − 2] is even (since p is odd), hence at least half of the elements of Z p ∗ are quadratic residues. Suppose we have b 2 = a.