Contents 4-Modulararithmetic ConceptsinAbstractMathematics

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4 - Modular arithmetic

Jean-Baptiste Campesato

Modular arithmetic was introduced by Gauss during the beginning of the 19th century. Working modulo a natural number 𝑛 > 0 means that, given an integer 𝑎, we identify it with its remainder 𝑟 for the Euclidean division by𝑛. Basically, it means that weforce𝑎to be equal to 𝑟(of course, not as integers, but equalmodulo n). Informally, we windℤon itself as represented below.

−5 −4 −3 −2 −1 0 1 2 3 4 5

0 1 2 3 4 5 ℤ

… , −12, −6, 0, 6, 12, …

… , −11, −5, 1, 7, 13, …

… , −10, −4, 2, 8, 14, …

… , −9, −3, 3, 9, 15, …

… , −8, −2, 4, 10, 16, … … , −7, −1, 5, 11, 17, …

ℤmodulo6

This extra layer of abstraction allowed Gauss, and subsequently other mathematicians, to obtain simpler proofs of already known results concerning integers but also to prove new theorems, simply by introducing this new eﬀicient notation which has many good properties.

1 Congruences

Definition 1. We say that a binary relationℛon a set𝐸is anequivalence relationif (i) ∀𝑥 ∈ 𝐸, 𝑥ℛ𝑥 (reflexivity)

(ii) ∀𝑥, 𝑦 ∈ 𝐸, 𝑥ℛ𝑦 ⟹ 𝑦ℛ𝑥 (symmetry)

(iii) ∀𝑥, 𝑦, 𝑧 ∈ 𝐸, (𝑥ℛ𝑦and𝑦ℛ𝑧) ⟹ 𝑥ℛ𝑧 (transitivity)

Definition 2. Let𝑛 ∈ ℕ ⧵ {0}and𝑎, 𝑏 ∈ ℤ. We say that𝑎and𝑏arecongruent modulo𝑛if𝑛|𝑎 − 𝑏, which we denote by𝑎 ≡ 𝑏 (mod𝑛).

Proposition 3. Congruence modulo𝑛is an equivalence relation onℤ.

Proof.

• Reflexivity.Let𝑎 ∈ ℤthen𝑛|0 = 𝑎 − 𝑎. Hence𝑎 ≡ 𝑎 (mod𝑛).

• Symmetry.Let𝑎, 𝑏 ∈ ℤbe such that𝑎 ≡ 𝑏 (mod𝑛). Then𝑛|𝑏 − 𝑎 = −(𝑎 − 𝑏)hence𝑏 ≡ 𝑎 (mod𝑛).

• Transitivity. Let𝑎, 𝑏, 𝑐 ∈ ℤbe such that𝑎 ≡ 𝑏 (mod 𝑛)and𝑏 ≡ 𝑐 (mod𝑛). Then𝑛|𝑎 − 𝑏and𝑛|𝑏 − 𝑐.

Hence𝑛|𝑎 − 𝑐 = (𝑎 − 𝑏) + (𝑏 − 𝑐). Thus𝑎 ≡ 𝑐 (mod𝑛).

■ Proposition 4. Let𝑛 ∈ ℕ ⧵ {0}and𝑎, 𝑏 ∈ ℤ. Then𝑎 ≡ 𝑏 (mod𝑛)if and only if𝑎and𝑏have same remainder for the Euclidean division by𝑛.

Proof.

⇒. Assume that𝑎 ≡ 𝑏 (mod𝑛), then𝑏 − 𝑎 = 𝑘𝑛for some𝑘 ∈ ℤ. By Euclidean division,𝑎 = 𝑛𝑞 + 𝑟for𝑞, 𝑟 ∈ ℤ satisfying0 ≤ 𝑟 < 𝑛. Hence𝑏 = 𝑎 + 𝑘𝑛 = 𝑛𝑞 + 𝑟 + 𝑘𝑛 = (𝑞 + 𝑘)𝑛 + 𝑟.

⇐. Assume that𝑎and𝑏have same remainder for the Euclidean division by𝑛, then𝑎 = 𝑛𝑞₁+𝑟and𝑏 = 𝑛𝑞₂+𝑟 where𝑞₁, 𝑞₂, 𝑟 ∈ ℤwith0 ≤ 𝑟 < 𝑛.

Hence𝑎 − 𝑏 = 𝑛𝑞₁+ 𝑟 − (𝑛𝑞₂+ 𝑟) = 𝑛(𝑞₁− 𝑞₂). Thus𝑛|𝑎 − 𝑏, i.e.𝑎 ≡ 𝑏 (mod𝑛). ■ Proposition 5. Let𝑛 ∈ ℕ ⧵ {0}and𝑎 ∈ ℤ. Then𝑎is congruent modulo𝑛to exactly one element of{0, 1, … , 𝑛 − 1}.

Proof. By Euclidean division𝑎 = 𝑛𝑞 + 𝑟where0 ≤ 𝑟 < 𝑛so that𝑎 ≡ 𝑟 (mod𝑛).

Conversely, if𝑎 ≡ 𝑟^′(mod𝑛)where𝑟^′∈ {0, 1, … , 𝑛 − 1}, then𝑎 − 𝑟^′= 𝑛𝑞for some𝑞 ∈ ℤ. So𝑎 = 𝑛𝑞 + 𝑟^′.

By uniqueness of the Euclidean division,𝑟 = 𝑟^′. ■

Proposition 6. Let𝑎, 𝑏, 𝑐, 𝑑 ∈ ℤand𝑛 ∈ ℕ ⧵ {0}. Assume that𝑎 ≡ 𝑏 (mod𝑛)and that𝑐 ≡ 𝑑 (mod𝑛)then

• 𝑎 + 𝑐 ≡ 𝑏 + 𝑑 (mod𝑛)

• 𝑎𝑐 ≡ 𝑏𝑑 (mod𝑛)

Proof. Let𝑎, 𝑏, 𝑐, 𝑑 ∈ ℤand𝑛 ∈ ℕ ⧵ {0}. Assume that𝑎 ≡ 𝑏 (mod𝑛)and that𝑐 ≡ 𝑑 (mod𝑛). Hence𝑎 − 𝑏 = 𝑛𝑘 and𝑐 − 𝑑 = 𝑛𝑙for some𝑘, 𝑙 ∈ ℤ. Then

• (𝑎 + 𝑐) − (𝑏 + 𝑑) = (𝑎 − 𝑏) + (𝑐 − 𝑑) = 𝑛𝑘 + 𝑛𝑙 = 𝑛(𝑘 + 𝑙), hence𝑎 + 𝑐 ≡ 𝑏 + 𝑑 (mod𝑛).

• 𝑎𝑐 − 𝑏𝑑 = (𝑏 + 𝑛𝑘)(𝑑 + 𝑛𝑙) − 𝑏𝑑 = 𝑏𝑛𝑙 + 𝑑𝑛𝑘 + 𝑛²𝑘𝑙 = 𝑛(𝑏𝑙 + 𝑑𝑘 + 𝑛𝑘𝑙), hence𝑎𝑐 ≡ 𝑏𝑑 (mod𝑛).

■ Example 7. 1729 × 16 ≡ 12 × (−1) (mod17) ≡ −12 (mod17) ≡ 5 (mod17)

Corollary 8. Let𝑎, 𝑏 ∈ ℤand𝑛 ∈ ℕ ⧵ {0}. Then∀𝑘 ∈ ℕ, 𝑎 ≡ 𝑏 (mod𝑛) ⟹ 𝑎^𝑘≡ 𝑏^𝑘(mod𝑛).

Proof. We prove the statement by induction on𝑘.

Base case at𝑘 = 0:𝑎⁰= 𝑏⁰ = 1hence𝑎⁰ ≡ 𝑏⁰(mod𝑛).

Induction step:assume that𝑎 ≡ 𝑏 (mod𝑛) ⟹ 𝑎^𝑘≡ 𝑏^𝑘(mod𝑛)for some𝑘 ∈ ℕ.

If𝑎 ≡ 𝑏 (mod𝑛)then by induction hypothesis we also have𝑎^𝑘≡ 𝑏^𝑘(mod𝑛). Hence, combining both previous congruences, we get that𝑎^𝑘𝑎 ≡ 𝑏^𝑘𝑏 (mod𝑛), i.e.𝑎^𝑘+1≡ 𝑏^𝑘+1(mod𝑛). Which proves the induction step. ■ Remark 9. Therefore addition, substraction (which is a special case of addition inℤ), multiplication and exponentiation are compatible with congruences.

Beware: division isnotcompatible with congruences:10 ≡ 4 (mod6)but5 ≢ 2 (mod6).

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Proposition 10. Let𝑎 ∈ ℤand𝑛 ∈ ℕ⧵{0}. Then𝑎has a multiplicative inverse modulo𝑛if and only ifgcd(𝑎, 𝑛) = 1.

Otherwise stated,

∃𝑏 ∈ ℤ, 𝑎𝑏 ≡ 1 (mod𝑛) ⇔gcd(𝑎, 𝑛) = 1

Proof. ∃𝑏 ∈ ℤ, 𝑎𝑏 ≡ 1 (mod𝑛) ⇔ ∃𝑏, 𝑐 ∈ ℤ, 𝑎𝑏 + 𝑛𝑐 = 1 ⇔gcd(𝑎, 𝑛) = 1 ■

Remark 11. Then the multiplicative inverse is unique modulo𝑛. Indeed if𝑎𝑏 ≡ 1 (mod𝑛) ≡ 𝑎𝑏^′ (mod 𝑛) then𝑛|(𝑏 − 𝑏^′)𝑎. Since gcd(𝑎, 𝑛) = 1, using Gauss’ lemma, we get that𝑛|𝑏 − 𝑏^′, i.e.𝑏 ≡ 𝑏^′(mod𝑛).

Remark 12. There is no cancellation law for congruences. For instance,50 ≡ 20 (mod15)but5 ≢ 2 (mod15).

Nonetheless, we have the following proposition.

Proposition 13. Let𝑛 ∈ ℕ⧵{0}and𝑎, 𝑥, 𝑦 ∈ ℤsatisfying𝑎𝑥 ≡ 𝑎𝑦 (mod𝑛)andgcd(𝑎, 𝑛) = 1. Then𝑥 ≡ 𝑦 (mod𝑛).

Proof. Since gcd(𝑎, 𝑛) = 1,𝑎admits an inverse modulo𝑛, i.e. there exists𝑏 ∈ ℤsuch that𝑎𝑏 ≡ 1 (mod𝑛).

Then𝑎𝑥 ≡ 𝑎𝑦 (mod𝑛) ⟹ 𝑏𝑎𝑥 ≡ 𝑏𝑎𝑦 (mod𝑛) ⟹ 𝑥 ≡ 𝑦 (mod𝑛). ■

2 Applications: divisibility criteria

In our everyday life, we usually use a base ten positional notation. It allows use to write all natural numbers using only10digits althoughℕis infinite. The idea is that the position of a digit changes its value.

Indeed, using the well-ordering principle and Euclidean division, it is possible to prove that any𝑛 ∈ ℕ can be uniquely written as𝑛 =

𝑟

∑𝑘=0

𝑎_𝑘10^𝑘where𝑎_𝑘∈ {0, 1, … , 9}and𝑎_𝑟 ≠ 0(see the appendix for a proof).

We usually write𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰ for

𝑟

∑𝑘=0

𝑎_𝑘10^𝑘 but we may omit the line over the digits when there is no possible confusion. For instance,590743 = 5 × 10⁵+ 9 × 10⁴+ 0 × 10³+ 7 × 10²+ 4 × 10¹+ 3 × 10⁰.

Note that we also use other bases: base 2 and base 16 are quite common nowadays in computer sciences.

And other bases were also commonly used by human beings in various places in the past: we still have the influence of a base 60 positional system when describing time (1 hour is 60 minutes), and the influence of a base 20 positional system in several languages (in French96is litteraly pronounced4 × 20 + 16).

In this section, we are going to use modular arithmetic in order to prove some divisibility criteria using our base ten positional notation.

Proposition 14. 3|𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰if and only if3|

𝑟

∑𝑘=0

𝑎_𝑘. Proof. Note that10 ≡ 1 (mod3), hence

𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰=

𝑟

∑𝑘=0

𝑎_𝑘10^𝑘≡

𝑟 𝑘=0∑

𝑎_𝑘1^𝑘(mod3) ≡

𝑟

∑𝑘=0

𝑎_𝑘(mod3)

Thus,

3|𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰⇔ 𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰ ≡ 0 (mod3)

⇔

𝑟

∑𝑘=0

𝑎_𝑘≡ 0 (mod3)

⇔ 3|

𝑟

∑𝑘=0

𝑎_𝑘

■

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Examples 15.

• 91524is divisible by3since9 + 1 + 5 + 2 + 4 = 21 = 7 × 3is.

• Let’s study whether8546921469is a multiple of3or not:

3|8546921469 ⇔ 3|8 + 5 + 4 + 6 + 9 + 2 + 1 + 4 + 6 + 9 = 54

⇔ 3|5 + 4 = 9 But9 = 3 × 3, hence3|8546921469.

Proposition 16. 9|𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰if and only if9|

𝑟

∑𝑘=0

𝑎_𝑘.

Proof. That’s a similar proof since10 ≡ 1 (mod9). ■

Proposition 17. 4|𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰if and only if4|𝑎₁𝑎₀¹⁰.

Proof. Note that10² = 4 × 25hence10^𝑘≡ 0 (mod4)for𝑘 ≥ 2. Hence 4|𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰⇔ 𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰ ≡ 0 (mod4)

⇔

𝑟

∑𝑘=0

𝑎_𝑘10^𝑘≡ 0 (mod4)

⇔ 𝑎₁× 10 + 𝑎₀≡ 0 (mod4)

⇔ 𝑎₁𝑎₀¹⁰ ≡ 0 (mod4)

⇔ 4|𝑎₁𝑎₀¹⁰

■ Examples 18.

• 4 ∤ 856987454251100125since4 ∤ 25.

• 4|98854558715580since4|80 = 4 × 20.

3 Fermat’s little theorem

Lemma 19. Let𝑝be a prime number. Then∀𝑛 ∈ {1, … , 𝑝 − 1}, (^𝑝𝑛) ≡ 0 (mod𝑝).

Proof. Let𝑛 ∈ {1, … , 𝑝 − 1}. Remember that𝑛(^𝑝𝑛) = 𝑝(^𝑝−1𝑛−1). Hence,𝑝|𝑛(^𝑝𝑛).

Since gcd(𝑝, 𝑛) = 1, by Gauss’ lemma, we get that𝑝|(^𝑝𝑛). ■

Theorem 20(Fermat’s little theorem, version 1).

Let𝑝be a prime number and𝑎 ∈ ℤ. Then𝑎^𝑝 ≡ 𝑎 (mod𝑝).

Proof. We first prove the theorem for𝑎 ∈ ℕby induction.

Base case at𝑎 = 0: 0^𝑝= 0 ≡ 0 (mod𝑝).

Induction step:assume that𝑎^𝑝≡ 𝑎 (mod𝑝)for some𝑎 ∈ ℕ. Then (𝑎 + 1)^𝑝 =

𝑝

∑𝑛=0(𝑝

𝑛)𝑎^𝑛by the binomial formula

≡ 𝑎^𝑝+ 1 (mod𝑝) since, by the above lemma,𝑝|(𝑝

𝑛)for1 ≤ 𝑛 ≤ 𝑝 − 1

≡ 𝑎 + 1 (mod𝑝) by the induction hypothesis

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Which ends the induction step.

We still need to prove the theorem for 𝑎 < 0. Then −𝑎 ∈ ℕ, hence, from the first part of the proof, (−𝑎)^𝑝≡ −𝑎 (mod𝑝). Multiplying both sides by(−1)^𝑝we get that𝑎^𝑝≡ (−1)^𝑝+1𝑎 (mod𝑝).

If𝑝 = 2then either𝑎 ≡ 0 (mod2)or𝑎 ≡ 1 (mod2), and the statement holds for both cases.

Otherwise,𝑝is odd, and hence(−1)^𝑝+1= 1. Thus𝑎^𝑝≡ 𝑎 (mod𝑝). ■ Theorem 21(Fermat’s little theorem, version 2).

Let𝑝be a prime number and𝑎 ∈ ℤ. Ifgcd(𝑎, 𝑝) = 1then𝑎^𝑝−1≡ 1 (mod𝑝).

Proof. By the first version of Fermat’s little theorem,𝑎^𝑝≡ 𝑎 (mod𝑝). Hence𝑝|𝑎^𝑝− 𝑎 = 𝑎(𝑎^𝑝−1− 1).

Since gcd(𝑎, 𝑝) = 1, by Gauss’ lemma,𝑝|𝑎^𝑝−1− 1. Thus𝑎^𝑝−1≡ 1 (mod𝑝). ■ Remark 22. Note that both versions of Fermat’s little theorem are equivalent.

4 Wilson’s theorem

Lemma 23. Let𝑝be a prime number. Then

∀𝑎 ∈ ℤ, 𝑎²≡ 1 (mod𝑝) ⟹ (𝑎 ≡ −1 (mod𝑝)or𝑎 ≡ 1 (mod𝑝))

Proof. Let𝑝be a prime number and𝑎 ∈ ℤsatisfying𝑎²≡ 1 (mod𝑝). Then𝑝|𝑎²− 1 = (𝑎 − 1)(𝑎 + 1).

By Euclid’s lemma, either𝑝|𝑎 − 1or𝑝|𝑎 + 1, i.e. 𝑎 ≡ 1 (mod𝑝)or𝑎 ≡ −1 (mod𝑝). ■ Theorem 24(Wilson’s theorem). Let𝑛 ∈ ℕ ⧵ {0, 1}. Then𝑛is prime if and only if(𝑛 − 1)! ≡ −1 (mod𝑛).

Proof. Let𝑛 ∈ ℕ ⧵ {0, 1}.

• Assume that𝑛is a composite number.Then there exists𝑘 ∈ ℕsuch that𝑘|𝑛and1 < 𝑘 < 𝑛.

Assume by contradiction that(𝑛 − 1)! ≡ −1 (mod𝑛)then𝑛|(𝑛 − 1)! + 1and hence𝑘|(𝑛 − 1)! + 1.

But𝑘|(𝑛 − 1)!, thus𝑘|((𝑛 − 1)! + 1 − (𝑛 − 1)!), i.e. 𝑘|1. So𝑘 = 1which leads to a contradiction.

• Assume that𝑛is prime.

Let𝑎 ∈ {1, 2, … , 𝑛 − 1}then gcd(𝑎, 𝑛) = 1. Hence𝑎admits a multiplicative inverse modulo𝑛, so there exists𝑏 ∈ {1, 2, … , 𝑛 − 1}such that𝑎𝑏 ≡ 1 (mod𝑛).

Note that this𝑏is unique by Remark11.

By the above lemma,𝑎 = 1and𝑎 = 𝑛 − 1are the only𝑎as above being their self-multiplicative inverse (i.e. such that𝑎²≡ 1 (mod𝑛)). Otherwise𝑏 ≠ 𝑎.

Thus(𝑛 − 1)! = 1 × 2 × ⋯ × (𝑛 − 1) ≡ 1 × (𝑛 − 1) (mod𝑛) ≡ −1 (mod𝑛).

Indeed, in the previous product each term simplifies with its multiplicative inverse except1and𝑛 − 1.

■

Examples 25.

• Take𝑝 = 17then(17 − 1)! + 1 = 20922789888001 = 17 × 1230752346353.

• Take𝑝 = 15then(15 − 1)! + 1 = 87178291201 = 15 × 5811886080 + 1.

Remark 26. Wilson’s theorem is a very ineﬀicient way to check whether a number is prime or not. Nonethe- less, it has some interesting theoretical applications.

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5 Chinese remainder theorem

Theorem 27(Chinese remainder theorem).

Let𝑛₁, 𝑛₂∈ ℕ ⧵ {0, 1}be such thatgcd(𝑛₁, 𝑛₂) = 1and let𝑎₁, 𝑎₂ ∈ ℤ. Then there exists𝑥 ∈ ℤsatisfying

{

𝑥 ≡ 𝑎₁(mod𝑛₁) 𝑥 ≡ 𝑎₂(mod𝑛₂)

Besides, if𝑥₁, 𝑥₂∈ ℤare two solutions of the above system then𝑥₁≡ 𝑥₂(mod𝑛₁𝑛₂).

Proof.

• Existence. By Bézout’s identity, there exist𝑚₁, 𝑚₂ ∈ ℤsuch that𝑛₁𝑚₁+ 𝑛₂𝑚₂= 1.

Note that𝑛₁𝑚₁ ≡ 0 (mod𝑛₁)and that𝑛₁𝑚₁ ≡ 𝑛₁𝑚₁+ 𝑛₂𝑚₂(mod𝑛₂) ≡ 1 (mod𝑛₂).

Similarly𝑛₂𝑚₂ ≡ 0 (mod𝑛₂)and𝑛₂𝑚₂≡ 1 (mod𝑛₁).

Thus, if we set𝑥 = 𝑎₂𝑛₁𝑚₁+ 𝑎₁𝑛₂𝑚₂then

– 𝑥 ≡ 𝑎₂× 0 + 𝑎₁× 1 (mod𝑛₁) ≡ 𝑎₁(mod𝑛₁), – 𝑥 ≡ 𝑎₂× 1 + 𝑎₁× 0 (mod𝑛₂) ≡ 𝑎₂(mod𝑛₂).

• Uniqueness modulo𝑛₁𝑛₂.Let𝑥₁, 𝑥₂ ∈ ℤbe two solutions.

Then𝑥₁− 𝑥₂≡ 0 (mod𝑛₁)so𝑥₁− 𝑥₂ = 𝑘𝑛₁for some𝑘 ∈ ℤ. Similarly𝑛₂|𝑥₁− 𝑥₂= 𝑘𝑛₁. Since gcd(𝑛₁, 𝑛₂) = 1, by Gauss’ lemma,𝑛₂|𝑘. So there exists𝑙 ∈ ℤsuch that𝑘 = 𝑛₂𝑙.

Thus𝑥₁− 𝑥₂= 𝑙𝑛₁𝑛₂and therefore𝑥₁≡ 𝑥₂(mod𝑛₁𝑛₂).

■

6 Euler’s theorem

Definition 28. Euler’s totient functionis the function𝜑 ∶ ℕ ⧵ {0} → ℕ ⧵ {0}defined by 𝜑(𝑛) ≔ # {𝑘 ∈ ℕ ∶ 1 ≤ 𝑘 ≤ 𝑛and gcd(𝑘, 𝑛) = 1}

Proposition 29. ∀𝑛₁, 𝑛₂ ∈ ℕ ⧵ {0},gcd(𝑛₁, 𝑛₂) = 1 ⟹ 𝜑(𝑛₁𝑛₂) = 𝜑(𝑛₁)𝜑(𝑛₂)

Proof. If𝑛₁ = 1or𝑛₂ = 1then there is nothing to prove. So let’s assume that𝑛₁, 𝑛₂≥ 2.

Define

𝑆_𝑖= {𝑟 ∈ ℕ ∶ 1 ≤ 𝑟 ≤ 𝑛𝑖and gcd(𝑟, 𝑛_𝑖) = 1} , 𝑖 = 1, 2 and

𝑇 = {𝑘 ∈ ℕ ∶ 1 ≤ 𝑘 ≤ 𝑛1𝑛₂and gcd(𝑘, 𝑛₁𝑛₂) = 1}

For𝑘 ∈ 𝑇, write the Euclidean divisions𝑘 = 𝑛₁𝑞₁+ 𝑟₁with0 ≤ 𝑟₁ < 𝑛₁and𝑘 = 𝑛₂𝑞₂+ 𝑟₂where0 ≤ 𝑟₂< 𝑛₂. Let’s prove that𝑟_𝑖∈ 𝑆_𝑖:

• Assume that𝑟_𝑖= 0then𝑛_𝑖|𝑘and𝑛_𝑖|𝑛₁𝑛₂so that𝑛_𝑖|gcd(𝑘, 𝑛₁𝑛₂) = 1: contradiction. So1 ≤ 𝑟_𝑖< 𝑛_𝑖.

• gcd(𝑟_𝑖, 𝑛_𝑖) =gcd(𝑘 − 𝑛_𝑖𝑞_𝑖, 𝑛_𝑖) =gcd(𝑘, 𝑛_𝑖)|gcd(𝑘, 𝑛₁𝑛₂) = 1, hence gcd(𝑟_𝑖, 𝑛_𝑖) = 1.

Therefore we can define𝑓 ∶ 𝑇 → 𝑆₁× 𝑆₂by𝑓 (𝑘) = (𝑟₁, 𝑟₂). Let’s prove that𝑓 is a bijection.

Let(𝑟₁, 𝑟₂) ∈ 𝑆₁× 𝑆₂. Then by the Chinese remainder theorem, there exists a unique𝑘 ∈ {1, 2, … , 𝑛₁𝑛₂}such that𝑘 ≡ 𝑟₁(mod𝑛₁)and𝑘 ≡ 𝑟₂(mod𝑛₂).

Note that gcd(𝑘, 𝑛₁) =gcd(𝑟₁+ 𝑙𝑛₁, 𝑛₁) =gcd(𝑟₁, 𝑛₁) = 1(for some𝑙 ∈ ℤ).

Similarly gcd(𝑘, 𝑛₂) =gcd(𝑟₂, 𝑛₂) = 1.

Then gcd(𝑘, 𝑛₁𝑛₂) = 1by Exercise 3 of Problem Set 2, so that𝑘 ∈ 𝑇.

We proved that∀(𝑟₁, 𝑟₂) ∈ 𝑆₁× 𝑆₂, ∃!𝑘 ∈ 𝑇 , (𝑟₁, 𝑟₂) = 𝑓 (𝑘), i.e. that𝑓 is bijective.

Therefore,#𝑇 = #(𝑆₁× 𝑆₂) = #𝑆₁#𝑆₂, i.e.𝜑(𝑛₁𝑛₂) = 𝜑(𝑛₁)𝜑(𝑛₂). ■

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Proposition 30. Let𝑝₁, … , 𝑝_𝑟be pairwise distinct prime numbers and𝛼₁, … , 𝛼_𝑟 ∈ ℕ ⧵ {0}, then 𝜑(

𝑟

∏𝑖=1

𝑝^𝛼_𝑖^𝑖 )=

𝑟

∏𝑖=1(𝑝

𝛼_𝑖

𝑖 − 𝑝^𝛼_𝑖^𝑖⁻¹) Proof.

• First case:let𝑝be a prime number and𝛼 ∈ ℕ ⧵ {0}. Then gcd(𝑝^𝛼, 𝑚) > 1if and only if𝑝|𝑚.

Hence𝜑(𝑝^𝛼) = # ({1, 2, … , 𝑝^𝛼} ⧵ {1 × 𝑝, 2 × 𝑝, … , 𝑝^𝛼−1× 𝑝}) = 𝑝^𝛼− 𝑝^𝛼−1.

• General case:using Proposition29and the first case, we get that 𝜑(

𝑟

∏𝑖=1

𝑝^𝛼_𝑖^𝑖 )=

𝑟

∏𝑖=1

𝜑 (𝑝^𝛼𝑖^𝑖) =

𝑟

∏𝑖=1(𝑝

𝛼_𝑖

𝑖 − 𝑝^𝛼_𝑖^𝑖⁻¹)

■ Remark 31. Assuming that we have already some knowledge aboutℚ, we can also write for𝑛 =

𝑟

∏𝑖=1

𝑝^𝛼_𝑖^𝑖:

𝜑 (𝑛) = 𝑛

𝑟

∏𝑖=1(1 − 1 𝑝_𝑖)

Theorem 32(Euler’s theorem). Let𝑛 ∈ ℕ ⧵ {0}and𝑎 ∈ ℤsuch thatgcd(𝑎, 𝑛) = 1. Then𝑎^𝜑(𝑛) ≡ 1 (mod𝑛).

Remark 33. Note that Fermat’s little theorem is a special case of Euler’s theorem: indeed, if 𝑝is a prime number then𝜑(𝑝) = 𝑝 − 1.

Proof of Euler’s theorem.

Write𝑆 = {𝑘 ∈ ℕ ∶ 1 ≤ 𝑘 ≤ 𝑛and gcd(𝑘, 𝑛) = 1} = {𝑘1, 𝑘₂, … , 𝑘_𝜑(𝑛)}. We will use the following two facts:

(i) Given𝑘_𝑖∈ 𝑆, there exists𝑘_𝑗 ∈ 𝑆such that𝑎𝑘_𝑖≡ 𝑘_𝑗 (mod𝑛).

Let𝑘_𝑖∈ 𝑆 then gcd(𝑎𝑘_𝑖, 𝑛) = 1by Exercise 3 of Problem Set 2.

Thus𝑎𝑘_𝑖≡ 𝑘_𝑗(mod𝑛)for some𝑘_𝑗 ∈ 𝑆.

(ii) ∀𝑘_𝑖, 𝑘_𝑗 ∈ 𝑆, 𝑎𝑘_𝑖≡ 𝑎𝑘_𝑗(mod𝑛) ⟹ 𝑘_𝑖= 𝑘_𝑗.

Indeed, then𝑛|𝑎(𝑘_𝑖− 𝑘_𝑗)and hence𝑛|𝑘_𝑖− 𝑘_𝑗by Gauss’ lemma.

Thus𝑘_𝑖≡ 𝑘_𝑗(mod𝑛).

Finally,𝑘_𝑖 = 𝑘_𝑗since1 ≤ 𝑘_𝑖, 𝑘_𝑗 ≤ 𝑛.

For𝑖 ∈ {1, 2, … , 𝜑(𝑛)}, there exists a unique𝑙_𝑖∈ {0, 1, … , 𝑛 − 1}such that𝑙_𝑖≡ 𝑎𝑘_𝑖(mod𝑛).

Then,{𝑙₁, 𝑙₂, … , 𝑙_𝜑(𝑛)} = {𝑘₁, 𝑘₂, … , 𝑘_𝜑(𝑛)}.

Indeed, by (i),{𝑙₁, 𝑙₂, … , 𝑙_𝜑(𝑛)} ⊂ {𝑘₁, 𝑘₂, … , 𝑘_𝜑(𝑛)}. And by (ii),#{𝑙₁, 𝑙₂, … , 𝑙_𝜑(𝑛)} = #{𝑘₁, 𝑘₂, … , 𝑘_𝜑(𝑛)}.

Hence

𝜑(𝑛)

∏𝑖=1

𝑘_𝑖=

𝜑(𝑛)

∏𝑖=1

𝑙_𝑖≡

𝜑(𝑛)

∏𝑖=1

𝑎𝑘_𝑖(mod𝑛) ≡ 𝑎^𝜑(𝑛)

𝜑(𝑛)

∏𝑖=1

𝑘_𝑖(mod𝑛).

Therefore𝑛|(𝑎^𝜑(𝑛)− 1)

𝜑(𝑛)

∏𝑖=1

𝑘_𝑖.

Since gcd

⎛⎜

⎜⎝ 𝑛,

𝜑(𝑛)

∏𝑖=1

𝑘_𝑖

⎞⎟

⎟⎠

= 1by Exercise 3 of Problem Set 2, we deduce from Gauss’ lemma that𝑛|𝑎^𝜑(𝑛)− 1,

i.e.𝑎^𝜑(𝑛) ≡ 1 (mod𝑛). ■

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A Positional numeral system with base 𝑏

Theorem 34. Let𝑏 ≥ 2be an natural number. Then any natural number𝑛 ∈ ℕadmits a unique expression 𝑛 =∑

𝑘≥0

𝑎_𝑘𝑏^𝑘

where𝑎_𝑘∈ {0, 1, … , 𝑏 − 1}and𝑎_𝑘= 0for all but finitely many𝑘 ≥ 0. Notation 35. We write𝑎_𝑟𝑎_𝑟−1… 𝑎₁𝑎₀^𝑏for

𝑟

∑𝑘=0

𝑎_𝑘𝑏^𝑘. Proof of Theorem34.

Existence.

We are going to prove by strong induction that for any𝑛 ≥ 0, there exist𝑎_𝑘∈ {0, 1, … , 𝑏 − 1}, 𝑘 ∈ ℕ, all but finitely many equal to0such that𝑛 =∑

𝑘≥0

𝑎_𝑘𝑏^𝑘.

• Base case at𝑛 = 0:0 =

∑𝑘≥0

0𝑏^𝑘.

• Induction step.Assume that0, 1, … , 𝑛admit an expression in base𝑏, for some𝑛 ≥ 0.

By Euclidean division,𝑛 + 1 = 𝑏𝑞 + 𝑟where𝑞, 𝑟 ∈ ℕsatisfy0 ≤ 𝑟 < 𝑏.

Note that if𝑞 ≠ 0then𝑞 < 𝑏𝑞 ≤ 𝑏𝑞 + 𝑟 = 𝑛 + 1. Thus0 ≤ 𝑞 ≤ 𝑛.

Therefore, by the induction hypothesis,𝑞 =∑

𝑘≥0

𝑎_𝑘𝑏^𝑘where𝑎_𝑘 ∈ {0, 1, … , 𝑏 − 1}and𝑎_𝑘 = 0for all but finitely many𝑘 ≥ 0.

Hence,𝑛 + 1 = 𝑏𝑞 + 𝑟 =

∑𝑘≥0

𝑎_𝑘𝑏^𝑘+1+ 𝑟𝑏⁰. Uniqueness.

Write∑

𝑘≥0

𝑎_𝑘𝑏^𝑘=

∑𝑘≥0

𝑎^′_𝑘𝑏^𝑘where𝑎_𝑘, 𝑎^′_𝑘∈ {0, 1, … , 𝑏 − 1}are zero for all but finitely many𝑘 ≥ 0.

Assume by contradiction there exists𝑘 ≥ 0such that𝑎_𝑘≠ 𝑎^′_𝑘.

Since{𝑘 ∈ ℕ ∶ 𝑎_𝑘≠ 𝑎^′_𝑘}is finite and non-empty, it admits a greatest elementℓ.

WLOG, we may assume that𝑎_ℓ< 𝑎^′_ℓ. Then0 =

∑𝑘≥0

𝑎_𝑘𝑏^𝑘−

∑𝑘≥0

𝑎^′_𝑘𝑏^𝑘=

∑𝑘≥0

(𝑎_𝑘− 𝑎^′_𝑘)𝑏^𝑘=

ℓ 𝑘=0∑

(𝑎_𝑘− 𝑎^′_𝑘)𝑏^𝑘. So that(𝑎^′_ℓ− 𝑎_ℓ)𝑏^ℓ =

ℓ−1

∑𝑘=0

(𝑎_𝑘− 𝑎^′_𝑘)𝑏^𝑘.

Therefore(𝑎^′_ℓ− 𝑎_ℓ)𝑏^ℓ ≤

ℓ−1

∑𝑘=0

|𝑎_𝑘− 𝑎^′_𝑘|𝑏^𝑘≤

ℓ−1

∑𝑘=0

(𝑏 − 1)𝑏^𝑘= 𝑏^ℓ− 1 < 𝑏^ℓ≤ (𝑎^′_ℓ− 𝑎_ℓ)𝑏^ℓ.

Hence a contradiction. ■

Remark 36. In order to pass from a base10expression to a base𝑏expression, we can perform successive Euclidean divisions as shown below (to pass from a base𝑏expression to a base10we may simply compute the sum).

Example 37.

42 = 2 × 21 + 0

= 2 × (2 × 10 + 1) + 0

= 2 × (2 × (2 × 5 + 0) + 1) + 0

= 2 × (2 × (2 × (2 × 2 + 1) + 0) + 1) + 0

= 2 × (2 × (2 × (2 × (2 × 1 + 0) + 1) + 0) + 1) + 0

= 1 × 2⁵+ 0 × 2⁴+ 1 × 2³+ 0 × 2²+ 1 × 2¹+ 0 × 2⁰ Hence42¹⁰ = 101010².

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The first known positional numeral system is the Babylonian one (circa 2000BC) whose base is 60 and whose digits are:

0 1 𒁹 2 𒈫 3 𒐈 4 𒃻 5 𒐊 6 𒐋 7 𒐌 8 𒐍 9 𒐎

10 𒌋 11 𒌋 𒁹 12 𒌋 𒈫 13 𒌋 𒐈 14 𒌋 𒃻 15 𒌋 𒐊 16 𒌋 𒐋 17 𒌋 𒐌 18 𒌋 𒐍 19 𒌋 𒐎

20 𒌋𒌋 21 𒌋𒌋 𒁹 22 𒌋𒌋 𒈫 23 𒌋𒌋 𒐈 24 𒌋𒌋 𒃻 25 𒌋𒌋 𒐊 26 𒌋𒌋 𒐋 27 𒌋𒌋 𒐌 28 𒌋𒌋 𒐍 29 𒌋𒌋 𒐎

30 𒌍 31 𒌍 𒁹 32 𒌍 𒈫 33 𒌍 𒐈 34 𒌍 𒃻 35 𒌍 𒐊 36 𒌍 𒐋 37 𒌍 𒐌 38 𒌍 𒐍 39 𒌍 𒐎

40 𒄭 41 𒄭𒁹 42 𒄭𒈫 43 𒄭𒐈 44 𒄭𒃻 45 𒄭𒐊 46 𒄭𒐋 47 𒄭𒐌 48 𒄭𒐍 49 𒄭𒐎

50 𒄴 51 𒄴𒁹 52 𒄴𒈫

53 𒄴𒐈

54 𒄴𒃻

55 𒄴𒐊

56 𒄴𒐋

57 𒄴𒐌

58 𒄴𒐍

59 𒄴𒐎

Let’s say that we want to write13655using Babylonian cuneiform numerals. For that, we perform successive Euclidean divisions by60as follows:

13655 = 60 × 227 + 35 = 60 × (60 × 3 + 47) + 35 = 3 × 60²+ 47 × 60¹+ 35 × 60⁰ Hence it was written: 𒐈 𒄭𒐌 𒌍 𒐊

Originally, there was no positional zero and an empty space was used instead (which can be confusing:

𒌋𒌋 𒐈 𒐊 and 𒌋𒌋 𒐈 𒐊 are not equal). The more convenient symbol𒑊was later used instead of the empty space (but it is not the number0, just a placeholder symbol for the positional numeral system).

See below a problem set submission by a MAT246 student circa 1700BC.

Figure 1: YBC 7289, clay tablet, between 1800BC and 1600BC.

It shows (extremely accurate) approximations of√2 ≃ 1 +24 60+ 51

60² + 10 60³ and of30√2 ≃ 42 + 25

60 + 35

60² (diagonal of the square of side length 30, see above de square) Yale Babylonian Collection,

Original picture fromhttps://commons.wikimedia.org/wiki/File:YBC-7289-OBV-REV.jpg

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B The Chinese Remainder Theorem for more than two equations

You won’t need the following result in MAT246, I’ve just added it because it was asked on Piazza (@82).

Theorem 38(Chinese remainder theorem). Let𝑘 ∈ ℕ ⧵ {0, 1}.

Let𝑛₁, 𝑛₂, … , 𝑛_𝑘∈ ℕ ⧵ {0, 1}be pairwise coprine, i.e. ∀𝑖, 𝑗 ∈ {1, … , 𝑘}, 𝑖 ≠ 𝑗 ⟹ gcd(𝑛_𝑖, 𝑛_𝑗) = 1.

Let𝑎₁, … , 𝑎_𝑘∈ ℤ. Then there exists𝑥 ∈ ℤsatisfying

⎧⎪

⎨⎪

⎩

𝑥 ≡ 𝑎₁(mod𝑛₁) 𝑥 ≡ 𝑎₂(mod𝑛₂)

⋮

𝑥 ≡ 𝑎_𝑘(mod𝑛_𝑘)

The proof follows closely the one of Theorem27but applied to𝑛_𝑖and𝑛₁… 𝑛_𝑖−1𝑛_𝑖+1… 𝑛_𝑘. Proof. Let𝑖 ∈ {1, … , 𝑘}. Then gcd(𝑛_𝑖, 𝑛₁… 𝑛_𝑖−1𝑛_𝑖+1… 𝑛_𝑘) = 1.

So, by Bézout’s identity, there exists𝑢_𝑖, 𝑣_𝑖∈ ℤsuch that𝑢_𝑖𝑛_𝑖+ 𝑣_𝑖𝑛₁… 𝑛_𝑖−1𝑛_𝑖+1… 𝑛_𝑘= 1.

Set𝑒_𝑖= 𝑣_𝑖𝑛₁… 𝑛_𝑖−1𝑛_𝑖+1… 𝑛_𝑘then𝑒_𝑖≡ 1 (mod𝑛_𝑖), and for𝑗 ∈ {1, … , 𝑘} ⧵ {𝑖},𝑒_𝑖≡ 0 (mod𝑛_𝑗).

Therefore𝑥 =

𝑘

∑𝑖=1

𝑎_𝑖𝑒_𝑖is a suitable solution. ■