February23 ,2021 FERMAT’SLITTLETHEOREM ConceptsinAbstractMathematics

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MAT246H1-S – LEC0201/9201

Concepts in Abstract Mathematics

FERMAT’S LITTLE THEOREM

February 23^rd, 2021

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Feb 23, 2021 1 / 4

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Binomial coeﬀicients

Given0 ≤ 𝑘 ≤ 𝑛two natural numbers, we denote by(^𝑛𝑘)(read as ”𝑛choose𝑘”) the number of ways to choose an (unordered) subset of𝑘elements from a fixed set of𝑛elements.

Remember that it satisfies(proved in class, see the video):

• (^𝑛𝑘) = 𝑘!(𝑛−𝑘)!^𝑛!

• For0 ≤ 𝑘 < 𝑛, we have(𝑘+1^𝑛+1) = (^𝑛𝑘) + (𝑘+1^𝑛 ) (Pascal’s triangle)

• ∀𝑥, 𝑦 ∈ ℝ, ∀𝑛 ∈ ℕ, (𝑥 + 𝑦)^𝑛=

𝑛

∑𝑘=0(𝑛

𝑘)𝑥^𝑘𝑦^𝑛−𝑘 (Binomial theorem)

Lemma

Let𝑝be a prime number. Then∀𝑘 ∈ {1, … , 𝑝 − 1}, (^𝑝𝑘) ≡ 0 (mod𝑝).

Proof.

Let𝑘 ∈ {1, … , 𝑝 − 1}. Then𝑘(^𝑝𝑘) = 𝑝(^𝑝−1𝑘−1). Hence,𝑝|𝑘(^𝑝𝑘).

Sincegcd(𝑝, 𝑘) = 1, by Gauss’ lemma, we get that𝑝|(𝑘^𝑝). ■

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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^𝑛 by the binomial formula

≡ 𝑎^𝑝+ 1 (mod𝑝) since𝑝|(𝑝

𝑛)for1 ≤ 𝑛 ≤ 𝑝 − 1

≡ 𝑎 + 1 (mod𝑝) by the induction hypothesis Which ends the induction step.

We still need to prove the theorem for𝑎 < 0.

In this case−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). ■

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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).

Sincegcd(𝑎, 𝑝) = 1, by Gauss’ lemma,𝑝|𝑎^𝑝−1− 1.

Thus𝑎^𝑝−1≡ 1 (mod𝑝). ■

Remark

Note that both versions of Fermat’s little theorem are equivalent.