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 integernninstead 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 assumennis a natural integer, andaaandbbare two integers. We say thataaandbbare congruent modulonnwhenn(ba)n\mid (b-a), or equivalently when there is an integerkksuch thata=b+kna=b+kn. We denote this byab [n]a\equiv b~ [n]or abmodna \equiv b\mod n. I will use the first notation throughout this chapter.

Remark: Whenb0b\neq0, we havear [b]a\equiv r~[b], whererris the remainder in the euclidean division ofaaby

This relation has a number of useful properties:

  • cZ,ab [n]acbc [n]\forall c\in \mathbb Z, a\equiv b~[n] \implies ac \equiv bc ~ [n]

  • cZ,ab [n]a+cb+c [n]\forall c \in \mathbb Z, a\equiv b~[n] \implies a+c\equiv b+c ~[n]

  • cZ,ab [n] and bc [n]ac [n]\forall c \in \mathbb Z, a \equiv b ~[n] \text{ and } b\equiv c~[n]\implies a\equiv c ~[n]

  • mN,ab [n]ambm [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 modulonnform a ring, which we noteZ/nZ\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 modulonnis 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 integernn. 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 integersaaandbbsuch thata2=2+4ba^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 integerccis the same as multiplying by its inverse; that is we want to find another integerddsuch thatcd1 [n]cd\equiv 1~[n]. Sincecd1 [n]kZ,cd=1+kncd\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 ifgcd(c,n)=1\gcd(c, n) = 1. Therefore, the units modulonnare the integers coprime tonn, lying in a set we call the unit group modulonn: (Z/nZ)×\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 inddandkkto the equationcdkn=1cd-kn=1from above).

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