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

Zn = Zmod(5)
Zn = Integers(5)
Zn = IntegerModRing(5)
# Ring of integers modulo 5
Zn(7)
# 2
Zn(8) == Zn(13)
# True

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

pow(2, 564654533, 7) # Output result as member of Z/7Z
# 4
power_mod(987654321, 987654321, 7) # Output result as simple integer
# 6
Zmod(7)(84564685)^(2^100) # ^ stands for powering in sage. To get XOR, use ^^.
# 5

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$n$: $\left(\mathbb Z/n\mathbb Z\right)^\times$​

Zn = Zmod(10)
Zn(7).is_unit()
# True
Zn(8).is_unit()
# False
3 == 1/Zn(7) == Zn(7)^(-1) == pow(7,-1,10) # member of Z/10Z
# True
inverse_mod(7, 10) # simple integer
# 3
Zn(3)/7
# 9
Zn(3)/8
# ZeroDivisionError: inverse of Mod(8, 10) does not exist
Zn.unit_group()
# Multiplicative Abelian group isomorphic to C4 (C4 being the cyclic group of order 4)

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).

xgcd(7, 10) # find (gcd(a, b), u, v) in au + bv = gcd(a, b)
# (1, 3, -2) <-- (gcd(7, 10), d, -k)