Quadratic residue by prime modulus [TUTORIAL]

Revision en11, by EJIC_B_KEDAX, 2024-07-30 15:37:55

Hello, Codeforces!

Today I'm going to talk about an unpopular technique in number theory.

Definition and elementary properties

Def. Let $$$p > 2$$$ be a prime number, then we will call quadratic residues all $$$1 \le x \le p - 1$$$ modulo $$$p$$$ such that the equation $$$a^2 \equiv x \pmod{p}$$$ has solutions and quadratic nonresidues otherwise. Note that $$$0$$$ is neither quadratic residue nor quadratic nonresidue.

Theorem: quadratic residues and quadratic nonresidues are equally divided.

Proof

Theorem: Denote by $$$R$$$ a quadratic residue and by $$$N$$$ a quadratic nonresidue, then:

$$$R \cdot R = R$$$
$$$N \cdot R = N$$$
$$$N \cdot N = R$$$

.

Proof

With these two theorems, we can already solve 103428K - Tiny Stars.

How to check if the number is residue or nonresidue?

There are several ways to check if a number is quadratic residue. In this blog, we will look at just one of them, you can read about Gauss's lemma and law of quadratic reciprocity.

Euler's criterion

$$$a$$$ is a quadratic residue modulo $$$p$$$ if and only if $$$a^{{\frac{p - 1}{2}}} \equiv 1 \pmod{p}$$$, and a quadratic nonresidue if and only if $$$a^{\frac{p - 1}{2}} \equiv -1 \pmod{p}$$$.

Proof

Consequence: $$$-1$$$ is quadratic residue if and only if $$$p = 4k + 1$$$, for some natural $$$k$$$, and quadratic nonresidue if and only if $$$p = 4k + 3$$$, for some natural $$$k$$$.

Clearly, the complexity of the number check is $$$O(\log_2p)$$$.

Code

Finding $$$i$$$ modulo $$$p$$$

Def. $$$i$$$ is such a number that $$$i^2 = -1$$$ $$$\implies$$$ $$$i$$$ modulo $$$p$$$ let us call such a number that $$$i ^ 2 \equiv -1 \pmod{p}$$$.

Algorithm: If $$$p = 4k + 3$$$ for some natural $$$k$$$, then there is no such $$$i$$$. If $$$p = 2$$$, then $$$i = 1$$$. Otherwise consider the quadratic nonresidue of $$$a$$$, by Euler's criterion we know that $$$a^{\frac{p - 1}{2}} \equiv -1 \pmod{p}$$$, then $$$a^{\frac{p - 1}{4}} \equiv i \pmod{p}$$$. All that remains is to find a quadratic nonresidue, for this we take a random $$$1 \le a \le p - 1$$$ and check it for $$$O(\log_2p)$$$, if it is a quadratic residue then take another random $$$a$$$. Since the number of residues and nonresidue are equal, we need $$$O(1)$$$ of such checks, so the expected running time of the algorithm is $$$O(\log_2p)$$$.

Code
Tags math, number theory

History

 
 
 
 
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en11 English EJIC_B_KEDAX 2024-07-30 15:37:55 0 (published)
ru16 Russian EJIC_B_KEDAX 2024-07-30 15:37:29 0 (опубликовано)
en10 English EJIC_B_KEDAX 2024-07-30 15:21:42 8
ru15 Russian EJIC_B_KEDAX 2024-07-30 15:14:38 11
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en8 English EJIC_B_KEDAX 2024-07-29 01:34:33 29 Tiny change: 's, so the total complexity of the al' -> 's, so the expected running time of the al'
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en6 English EJIC_B_KEDAX 2024-07-29 01:30:19 1 Tiny change: 'y="Proof">.\nFirst, l' -> 'y="Proof">\nFirst, l'
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en1 English EJIC_B_KEDAX 2024-07-29 00:54:10 6734 Initial revision for English translation (saved to drafts)
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