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Modular Arithmetic
Authors: A~Z, perhaps someone else but not yet (or they've decided to remain hidden like a ninja)

Introduction

Thinking not over the integers as a whole but modulo some integer
$n$
instead can prove quite useful in a number of situation. This chapter attempts to introduce to you the basic concepts of working in such a context.

Congruences

For the following chapter, we will assume
$n$
is a natural integer, and
$a$
and
$b$
are two integers. We say that
$a$
and
$b$
are congruent modulo
$n$
when
$n\mid (b-a)$
, or equivalently when there is an integer
$k$
such that
$a=b+kn$
. We denote this by
$a\equiv b~ [n]$
or
$a \equiv b\mod n$
. I will use the first notation throughout this chapter.
Remark: When
$b\neq0$
, we have
$a\equiv r~[b]$
, where
$r$
is the remainder in the euclidean division of
$a$
by
This relation has a number of useful properties:
$\forall c\in \mathbb Z, a\equiv b~[n] \implies ac \equiv bc ~ [n]$
$\forall c \in \mathbb Z, a\equiv b~[n] \implies a+c\equiv b+c ~[n]$
$\forall c \in \mathbb Z, a \equiv b ~[n] \text{ and } b\equiv c~[n]\implies a\equiv c ~[n]$
$\forall m \in \mathbb N, a\equiv b~[n] \implies a^m\equiv b^m ~[n]$
The proofs are left as an exercise to the reader :p (Hint: go back to the definition)
Seeing as addition and multiplication are well defined, the integers modulo
$n$
form a ring, which we note
$\mathbb Z/n\mathbb Z$
. In sage, you can construct such ring with either of the following
1
Zn = Zmod(5)
2
Zn = Integers(5)
3
Zn = IntegerModRing(5)
4
# Ring of integers modulo 5
5
Zn(7)
6
# 2
7
Zn(8) == Zn(13)
8
# True
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Powering modulo
$n$
is relatively fast, thanks to the double-and-square algorithm, so we needn't worry about it taking too much time when working with high powers
1
pow(2, 564654533, 7) # Output result as member of Z/7Z
2
# 4
3
power_mod(987654321, 987654321, 7) # Output result as simple integer
4
# 6
5
Zmod(7)(84564685)^(2^100) # ^ stands for powering in sage. To get XOR, use ^^.
6
# 5
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As a side note, remember that if an equality holds over the integers, then it holds modulo any natural integer
$n$
. This can be used to prove that a relation is never true by finding a suitable modulus, or to derive conditions on the potential solutions of the equation.
Example: by choosing an appropriate modulus, show that not even god is able to find integers
$a$
and
$b$
such that
$a^2 = 2 + 4b$

Modular Inverse

Since we can multiply, a question arises: can we divide? The answer is yes, under certain conditions. Dividing by an integer
$c$
is the same as multiplying by its inverse; that is we want to find another integer
$d$
such that
$cd\equiv 1~[n]$
. Since
$cd\equiv 1~[n]\iff\exists k\in\mathbb Z, cd = 1 + kn$
, it is clear from Bézout's Identity that such an inverse exists if and only if
$\gcd(c, n) = 1$
. Therefore, the units modulo
$n$
are the integers coprime to
$n$
, lying in a set we call the unit group modulo
$n$
:
$\left(\mathbb Z/n\mathbb Z\right)^\times$
1
Zn = Zmod(10)
2
Zn(7).is_unit()
3
# True
4
Zn(8).is_unit()
5
# False
6
3 == 1/Zn(7) == Zn(7)^(-1) == pow(7,-1,10) # member of Z/10Z
7
# True
8
inverse_mod(7, 10) # simple integer
9
# 3
10
Zn(3)/7
11
# 9
12
Zn(3)/8
13
# ZeroDivisionError: inverse of Mod(8, 10) does not exist
14
Zn.unit_group()
15
# Multiplicative Abelian group isomorphic to C4 (C4 being the cyclic group of order 4)
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Finding the modular inverse of a number is an easy task, thanks to the extended euclidean algorithm (that outputs solutions in
$d$
and
$k$
to the equation
$cd-kn=1$
from above).
1
xgcd(7, 10) # find (gcd(a, b), u, v) in au + bv = gcd(a, b)
2
# (1, 3, -2) <-- (gcd(7, 10), d, -k)
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