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

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.

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 5Zn(7)# 2Zn(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# 4power_mod(987654321, 987654321, 7) # Output result as simple integer# 6Zmod(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$a^2 = 2 + 4b$

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()# TrueZn(8).is_unit()# False3 == 1/Zn(7) == Zn(7)^(-1) == pow(7,-1,10) # member of Z/10Z# Trueinverse_mod(7, 10) # simple integer# 3Zn(3)/7# 9Zn(3)/8# ZeroDivisionError: inverse of Mod(8, 10) does not existZn.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)