Problem Set n°3

University of Toronto – MAT246H1-S – LEC0201/9201

Concepts in Abstract Mathematics

Problem Set n°3

Jean-Baptiste Campesato Due on March 12^th, 2021

You can only use the material covered in class up to lecture 12 (i.e. Chapters 1, 2, 3 and 4 up to section 5 included).

Write your solutions concisely but without skipping important steps.

Make sure that your submission is readable on Crowdmark.

Exercise 1.

Let𝑝be a prime number. Prove that

∀𝑠 ∈ ℕ ⧵ {0}, ∀𝑛 ∈ ℕ ⧵ {0}, ∀𝑥₁, … , 𝑥_𝑛 ∈ ℤ, (

𝑛

∑𝑘=1

𝑥_𝑘 )

𝑝^𝑠

≡

𝑛

∑𝑘=1

𝑥^𝑝_𝑘^𝑠 (mod𝑝)

Exercise 2.

The following questions are independent.

1. For which𝑛 ∈ ℕ, is5^𝑛− 3^𝑛a prime number?

2. For which𝑛 ∈ ℕ, is2^2𝑛+ 5a prime number?

Exercise 3.

Solve for𝑥, 𝑦 ∈ ℕ ⧵ {0},

𝑥

∑𝑘=1

(𝑘!) = 𝑦².

Exercise 4.

Let𝑝be a prime number and𝑛 ∈ ℕsatisfying1 ≤ 𝑛 ≤ 𝑝 − 1.

Prove that(𝑝 − 𝑛)!(𝑛 − 1)! ≡ (−1)^𝑛(mod𝑝).

2 Problem Set 3

Sample solution to Exercise 1.

Method 1:

Let’s prove the statement by induction on𝑠 ≥ 1.

• Base case at𝑠 = 1:

Let𝑛 ∈ ℕ ⧵ {0}and𝑥₁, … , 𝑥_𝑛∈ ℤ.

By Fermat’s theorem we have:

• (

𝑛

∑𝑘=1

𝑥_𝑘 )

𝑝

≡

𝑛

∑𝑘=1

𝑥_𝑘(mod𝑝), and,

• For𝑘 = 1, … , 𝑛, 𝑥^𝑝_𝑘≡ 𝑥_𝑘(mod𝑝).

Thus(

𝑛

∑𝑘=1

𝑥_𝑘 )

𝑝

≡

𝑛

∑𝑘=1

𝑥_𝑘(mod𝑝) ≡

𝑛

∑𝑘=1

𝑥^𝑝_𝑘(mod𝑝)

• Induction step:assume that the statement of the question holds for some𝑠 ≥ 1.

Let𝑛 ∈ ℕ ⧵ {0}and𝑥₁, … , 𝑥_𝑛∈ ℤ.

Then

(

𝑛

∑𝑘=1

𝑥_𝑘 )

𝑝^𝑠+1

=

⎛⎜

⎜⎝(

𝑛

∑𝑘=1

𝑥_𝑘 )

𝑝^𝑠

⎞⎟

⎟⎠

𝑝

≡(

𝑛

∑𝑘=1

𝑥^𝑝_𝑘^𝑠 )

𝑝

(mod𝑝) by induction hypothesis

≡

𝑛

∑𝑘=1(𝑥^𝑝

𝑠

𝑘)

𝑝

(mod𝑝) by the case𝑠 = 1

≡

𝑛

∑𝑘=1

𝑥^𝑝_𝑘^𝑠+1 (mod𝑝)

Method 2:

Lemma.Let’s first prove by induction on𝑠that∀𝑠 ∈ ℕ ⧵ {0}, ∀𝑥 ∈ ℤ, 𝑥^𝑝𝑠 ≡ 𝑥 (mod𝑝).

• Base case at𝑠 = 1: Let𝑥 ∈ ℤthen𝑥^𝑝≡ 𝑥 (mod𝑝)by Fermat’s theorem.

• Induction step:assume that the statement of the question holds for some𝑠 ≥ 1.

Let𝑥 ∈ ℤthen 𝑥^𝑝𝑠+1 = (𝑥^𝑝𝑠)

𝑝

≡ 𝑥^𝑝(mod𝑝) by the inductive hypothesis

≡ 𝑥 (mod𝑝) by Fermat’s theorem Which proves the lemma.

Let’s prove the statement of the question:

Let𝑠 ∈ ℕ ⧵ {0},𝑥₁, … , 𝑥_𝑛 ∈ ℤthen

(

𝑛

∑𝑘=1

𝑥_𝑘 )

𝑝^𝑠

=

𝑛

∑𝑘=1

𝑥_𝑘 by the lemma

=

𝑛

∑𝑘=1

𝑥^𝑝_𝑘^𝑠 by the lemma

MAT246H1-S – LEC0201/9201 – J.-B. Campesato 3

Sample solution to Exercise 2.

1. If𝑛 = 0then5⁰− 3⁰= 0is not prime.

If𝑛 = 1then5¹− 3¹= 2is prime.

If𝑛 > 1then5^𝑛− 3^𝑛≡ 1^𝑛− 1^𝑛(mod2) ≡ 0 (mod2). Thus5^𝑛− 3^𝑛is even but5^𝑛− 3^𝑛> 2, therefore it is not prime.

Conclusion:5^𝑛− 3^𝑛is prime for𝑛 = 1only.

2. If𝑛 = 0then2²⁰+ 5 = 2¹+ 5 = 7is prime.

If𝑛 ≥ 1then2^2𝑛+ 5 ≡ (−1)^2𝑛+ 2 (mod3) ≡ 1 + 2 (mod3) ≡ 0 (mod3)(since2^𝑛is even as𝑛 ≥ 1).

Therefore3|2^2𝑛+ 5but2^2𝑛+ 5 > 3. Thus2^2𝑛+ 5is not prime.

Conclusion:2^2𝑛+ 5is prime for𝑛 = 0only.

Sample solution to Exercise 3.

We first compute𝑦²(mod5)in terms of𝑦 (mod5):

𝑦 (mod5) 0 1 2 3 4 𝑦²(mod5) 0 1 4 4 1 We treat several cases.

1. Let𝑥 = 1then

𝑥

∑𝑘=1

(𝑘!) = 1.

The unique𝑦 ∈ ℕ ⧵ {0}such that𝑦² = 1is𝑦 = 1.

2. Let𝑥 = 2then

𝑥

∑𝑘=1

(𝑘!) = 1! + 2! ≡ 3 (mod5).

So there exists no𝑦 ∈ ℤsuch that

2

∑𝑘=1

(𝑘!) = 𝑦²by the above table.

3. Let𝑥 = 3then

𝑥

∑𝑘=1

(𝑘!) = 1! + 2! + 3! = 9.

The unique𝑦 ∈ ℕ ⧵ {0}such that𝑦² = 9is𝑦 = 3.

4. Let𝑥 ≥ 4.

Note that for𝑘 ≥ 5, we have5|𝑘!.

Thus

𝑥

∑𝑘=1

(𝑘!) ≡ 1! + 2! + 3! + 4! (mod5) ≡ 33 (mod5) ≡ 3 (mod5).

So there exists no𝑦 ∈ ℤsuch that

𝑥

∑𝑘=1

(𝑘!) = 𝑦²when𝑥 ≥ 4, by the above table.

So the solutions are(𝑥, 𝑦) = (1, 1)and(𝑥, 𝑦) = (3, 3).

Sample solution to Exercise 4.

Let𝑝be a prime number and𝑛 ∈ ℕsatisfying1 ≤ 𝑛 ≤ 𝑝 − 1.

Note that

(𝑝 − 1)! = (𝑝 − 𝑛)!(𝑝 − (𝑛 − 1))(𝑝 − (𝑛 − 2)) ⋯ (𝑝 − 1)

≡ (𝑝 − 𝑛)!(−(𝑛 − 1))(−(𝑛 − 2)) ⋯ (−1) (mod𝑝)

≡ (𝑝 − 𝑛)!(−1)^𝑛−1(𝑛 − 1)(𝑛 − 2) ⋯ 1 (mod𝑝)

≡ (𝑝 − 𝑛)!(−1)^𝑛−1(𝑛 − 1)! (mod𝑝)

Since, by Wilson’s theorem,(𝑝 − 1)! ≡ −1 (mod𝑝), we get that(𝑝 − 𝑛)!(−1)^𝑛−1(𝑛 − 1)! ≡ −1 (mod𝑝)and thus, multiplying both side by(−1)^𝑛−1, that(𝑝 − 𝑛)!(𝑛 − 1)! ≡ (−1)^𝑛(mod𝑝).