Concepts in Abstract Mathematics

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Concepts in Abstract Mathematics

Homework questions – Week 4

Jean-Baptiste Campesato

February 8^th, 2021 to February 12^th, 2021

Exercise 1.

Compute gcd(3¹²³− 5, 25).

Exercise 2.

Prove that∀𝑛 ∈ ℤ, 6|𝑛(𝑛 + 1)(𝑛 + 2) Exercise 3.

1. Prove that∀𝑛 ∈ ℤ,gcd(2𝑛, 2𝑛 + 2) = 2.

2. Prove that∀𝑛 ∈ ℤ,gcd(2𝑛 − 1, 2𝑛 + 1) = 1.

3. Prove that for𝑎, 𝑏 ∈ ℤnot both zero, gcd(5𝑎 + 3𝑏, 13𝑎 + 8𝑏) =gcd(𝑎, 𝑏).

Exercise 4.

Find all the integer solutions of (a)𝑥𝑦 = 2𝑥 + 3𝑦 (b) 1

𝑥+ 1 𝑦 = 1

5 (c)𝑥 + 𝑦 = 𝑥𝑦

(d)9𝑥 + 15𝑦 = 11 (e)9𝑥 + 15𝑦 = 18 (f)1665𝑥 + 1035𝑦 = 45 Exercise 5.

1. Prove that if𝑎, 𝑏 ∈ ℤare not both zero then there exist𝑎^′, 𝑏^′∈ ℤsuch that gcd(𝑎^′, 𝑏^′) = 1, and𝑎 = 𝑑𝑎^′ and𝑏 = 𝑑𝑏^′where𝑑 =gcd(𝑎, 𝑏).

2. Prove that∀𝑎, 𝑏, 𝑐 ∈ ℤ ⧵ {0}, 𝑐|𝑎𝑏 ⟹ 𝑐|(gcd(𝑎, 𝑐)gcd(𝑏, 𝑐)) Exercise 6.

1. ProveSophie Germain’s identity:𝑎⁴+ 4𝑏⁴= ((𝑎 + 𝑏)²+ 𝑏²) ((𝑎 − 𝑏)²+ 𝑏²). 2. Prove that3⁴⁴+ 4²⁹is a composite number.

3. Prove that for every natural number𝑛 > 1,𝑛⁴+ 4^𝑛is a composite number.

Hint: study the parity of𝑛.

Exercise 7.

Prove that there are infinitely many integers that can’t be written as the sum of a square with a prime number.

Hint:look at(3𝑘 + 2)²for𝑘 ∈ ℕ ⧵ {0}. Exercise 8.

Prove that∀𝑛 ∈ ℕ, 𝑛|(𝑛 − 1)! + 1 ⟹ 𝑛is prime.

Exercise 9.

Let𝑛 ∈ ℕ ⧵ {0}. Find𝑛consecutive natural numbers such that none of them is a prime number.

Exercise 10.

Prove that the following numbers are not rationals using the prime factorization theorem.

1. log₁₀2 2. √2

For this question you can use what you know aboutℚandℝ.

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Exercise 11.

Prove that∀𝑛 ∈ ℤ, 49 ∤ 𝑛³− 𝑛²− 2𝑛 + 1 Exercise 12.

Prove that there are infinitely many prime numbers of the form𝑝 = 4𝑘 + 3where𝑘 ∈ ℕ.

Exercise 13. Goldbach’s theorem about Fermat numbers

1. Prove that∀𝑛 ∈ ℕ, ∀𝑘 ∈ ℕ ⧵ {0}, 2²^𝑛+𝑘− 1 = (2²^𝑛− 1) ×

𝑘−1

∏𝑖=0(2²

𝑛+𝑖+ 1). 2. Let𝑚, 𝑛 ∈ ℕ. Prove that if𝑚 ≠ 𝑛then2²^𝑛+ 1and2²^𝑚+ 1are coprime.

Exercise 14.

Let𝑎, 𝑛 ≥ 2be two natural numbers.

1. Prove that if𝑎^𝑛− 1is prime then𝑎 = 2and𝑛is prime.

A number of the form𝑀_𝑛= 2^𝑛− 1is called a Mersenne number.

2. Is the converse true?

Exercise 15.

Three brothers inherit𝑛gold pieces weighing1, 2, … , 𝑛.

For what𝑛 ∈ ℕ ⧵ {0}can they be split into three equal heaps?

Exercise 16.

A sea pirate wants to share a treasure with its sailors.

The treasure is made of69diamonds,1150pearls and4140gold coins.

He is able to share fairly the treasure such that everyone (including himself) receive the same amount of each object.

How many sailors are there?

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Sample solutions to Exercise 1.

The positive divisors of25are1,5and25. Hence, gcd(3¹²³− 5, 25)has to be equal to one of these numbers.

Assume by contradiction that gcd(3¹²³− 5, 25) = 5or gcd(3¹²³− 5, 25) = 25. In both cases,5|3¹²³− 5and so 5|3¹²³ = (3¹²³− 5) + 5. Contradiction.

Therefore gcd(3¹²³− 5, 25) = 1.

Let𝑛 ∈ ℤ.

Since𝑛,𝑛 + 1and𝑛 + 2are three consecutive integers, one is divisible by 2 and one is divisible by 3.

Hence𝑛(𝑛 + 1)(𝑛 + 2) = 2𝑘and𝑛(𝑛 + 1)(𝑛 + 2) = 3𝑙for some𝑘, 𝑙 ∈ ℤ.

Then2𝑘 = 3𝑙, so that2|3𝑙. Besides gcd(2, 3) = 1thus2|𝑙by Gauss’ lemma, i.e.𝑙 = 2𝑚for some𝑚 ∈ ℤ.

Therefore𝑛(𝑛 + 1)(𝑛 + 2) = 3𝑙 = 6𝑚and6|𝑛(𝑛 + 1)(𝑛 + 2).

In my solutions I use that gcd(𝑎, 𝑏) =gcd(𝑎 + 𝑘𝑏, 𝑏)(see Proposition 35 of Chapter 2).

1. gcd(2𝑛, 2𝑛 + 2) =gcd(2𝑛, (2𝑛 + 2) − 2𝑛) =gcd(2𝑛, 2) =gcd(2𝑛 − 2 × 𝑛, 2) =gcd(0, 2) = 2.

2. gcd(2𝑛−1, 2𝑛+1) =gcd(2𝑛−1, (2𝑛+1)−(2𝑛−1)) =gcd(2𝑛−1, 2) =gcd((2𝑛−1)−2×𝑛, 2) =gcd(−1, 2) = 1.

3. gcd(5𝑎 + 3𝑏, 13𝑎 + 8𝑏) =gcd(5𝑎 + 3𝑏, (13𝑎 + 8𝑏) − 2 × (5𝑎 + 3𝑏))

=gcd(5𝑎 + 3𝑏, 3𝑎 + 2𝑏) =gcd((5𝑎 + 3𝑏) − (3𝑎 + 2𝑏), 3𝑎 + 2𝑏)

=gcd(2𝑎 + 𝑏, 3𝑎 + 2𝑏) =gcd(2𝑎 + 𝑏, (3𝑎 + 2𝑏) − (2𝑎 + 𝑏))

=gcd(2𝑎 + 𝑏, 𝑎 + 𝑏) =gcd((2𝑎 + 𝑏) − 𝑏, 𝑏)

=gcd(𝑎 + 𝑏, 𝑏) =gcd(𝑎, 𝑏)

(a) Let𝑥, 𝑦 ∈ ℤ. Then

𝑥𝑦 = 2𝑥 + 3𝑦 ⇔ (𝑥 − 3)(𝑦 − 2) = 6

⇔ (𝑥 − 3, 𝑦 − 2) ∈ {(1, 6), (2, 3), (3, 2), (6, 1), (−1, −6), (−2, −3), (−3, −2), (−6, −1)}

⇔ (𝑥, 𝑦) ∈ {(4, 8), (5, 5), (6, 4), (9, 3), (2, −4), (1, −1), (0, 0), (−3, 1)}

(b) Let𝑥, 𝑦 ∈ ℤ ⧵ {0}. Then 1

𝑥 +1 𝑦 = 1

5 ⇔ 5𝑦 + 5𝑥 = 𝑥𝑦

⇔ (𝑥 − 5)(𝑦 − 5) = 25

⇔ (𝑥 − 5, 𝑦 − 5) ∈ {(1, 25), (5, 5), (25, 1), (−1, −25), (−25, −1)} since𝑥, 𝑦 ≠ 0

⇔ (𝑥, 𝑦) ∈ {(6, 30), (10, 10), (30, 6), (4, −20), (−20, 4)}

(c) Let𝑥, 𝑦 ∈ ℤ. Then

𝑥 + 𝑦 = 𝑥𝑦 ⇔ 𝑥 + 𝑦 − 𝑥𝑦 + 1 = 1

⇔ (𝑥 − 1)(𝑦 − 1) = 1

⇔ (𝑥 − 1, 𝑦 − 1) = (−1, −1)or(𝑥 − 1, 𝑦 − 1) = (1, 1)

⇔ (𝑥, 𝑦) = (0, 0)or(𝑥, 𝑦) = (2, 2)

(d) For the next questions, look at the slides from Feb 2 and/or the last part of Chapter 2.

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1. Let𝑎, 𝑏 ∈ ℤnot both zero. Set𝑑 =gcd(𝑎, 𝑏).

Since𝑑|𝑎and𝑑|𝑏, we know that𝑎 = 𝑑𝑎^′and that𝑏 = 𝑑𝑏^′for some𝑎^′, 𝑏^′∈ ℤ.

Then𝑑 =gcd(𝑎, 𝑏) =gcd(𝑑𝑎^′, 𝑑𝑏^′) = 𝑑gcd(𝑎^′, 𝑏^′). Hence gcd(𝑎^′, 𝑏^′) = 1.

2. Method 1:Let𝑎, 𝑏, 𝑐 ∈ ℤ ⧵ {0}be such that𝑐|𝑎𝑏.

Set𝑑 = gcd(𝑎, 𝑐)and𝛿 =gcd(𝑏, 𝑐). Then𝑎 = 𝑑𝑎^′,𝑐 = 𝑑𝑐^′,𝑏 = 𝛿𝑏^″,𝑐 = 𝛿𝑐^″where gcd(𝑎^′, 𝑐^′) = 1and gcd(𝑏^″, 𝑐^″) = 1.

Finally𝑐 = 𝛿𝑐^″|𝛿𝑑|𝑑𝑎^′𝛿𝑏^″= 𝑎𝑏.

Method 2:Let𝑎, 𝑏, 𝑐 ∈ ℤ ⧵ {0}be such that𝑐|𝑎𝑏.

Write𝑎 = 𝑝^𝛼₁¹𝑝^𝛼₂²⋯ 𝑝^𝛼_𝑟^𝑟, 𝑏 = 𝑝^𝛽₁¹𝑝^𝛽₂²⋯ 𝑝^𝛽_𝑟^𝑟 and𝑐 = 𝑝^𝛾₁¹𝑝^𝛾₂²⋯ 𝑝^𝛾_𝑟^𝑟 where the 𝑝_𝑖 are prime numbers and 𝛼_𝑖, 𝛽_𝑖, 𝛾_𝑖∈ ℕ.

Then gcd(𝑎, 𝑐) = 𝑝^min(𝛼₁ ¹^,𝛾¹⁾𝑝^min(𝛼₂ ²^,𝛾²⁾⋯ 𝑝^min(𝛼_𝑟 ^𝑟^,𝛾^𝑟⁾and gcd(𝑏, 𝑐) = 𝑝^min(𝛽₁ ¹^,𝛾¹⁾𝑝^min(𝛽₂ ²^,𝛾²⁾⋯ 𝑝^min(𝛽_𝑟 ^𝑟^,𝛾^𝑟⁾. Hence gcd(𝑎, 𝑐)gcd(𝑏, 𝑐) = 𝑝^min(𝛼₁ ¹^,𝛾¹^)+min(𝛽¹^,𝛾¹⁾𝑝^min(𝛼₂ ²^,𝛾²^)+min(𝛽²^,𝛾²⁾⋯ 𝑝^min(𝛼_𝑟 ^𝑟^,𝛾^𝑟^)+min(𝛽^𝑟^,𝛾^𝑟⁾.

Thus𝑐|gcd(𝑎, 𝑐)gcd(𝑏, 𝑐)if and only if𝛾₁ ≤min(𝛼₁, 𝛾₁) +min(𝛽₁, 𝛾₁), …,𝛾_𝑟≤min(𝛼_𝑟, 𝛾_𝑟) +min(𝛽_𝑟, 𝛾_𝑟).

First case: if min(𝛼_𝑖, 𝛾_𝑖) = 𝛾_𝑖or min(𝛽_𝑖, 𝛾_𝑖) = 𝛾_𝑖then𝛾_𝑖≤min(𝛼_𝑖, 𝛾_𝑖) +min(𝛽_𝑖, 𝛾_𝑖).

Otherwise: min(𝛼_𝑖, 𝛾_𝑖) +min(𝛽_𝑖, 𝛾_𝑖) = 𝛼_𝑖+ 𝛽_𝑖but since𝑐|𝑎𝑏, we know that𝛾₁≤ 𝛼₁+ 𝛽₁, …,𝛾_𝑟 ≤ 𝛼_𝑟+ 𝛽_𝑟.

1. ((𝑎 + 𝑏)²+ 𝑏²) ((𝑎 − 𝑏)²+ 𝑏²) = (𝑎²+ 2𝑏²+ 2𝑎𝑏) (𝑎²+ 2𝑏²− 2𝑎𝑏)

= (𝑎²+ 2𝑏²)²− (2𝑎𝑏)²

= 𝑎⁴+ 4𝑎²𝑏²+ 4𝑏⁴− 4𝑎²𝑏²= 𝑎⁴+ 4𝑏⁴

2. 3⁴⁴+ 4²⁹ = (3¹¹)⁴ + 4 × (4⁷)⁴ = ((3¹¹+ 4⁷)²+ 4¹⁴) ((3¹¹− 4⁷)²+ 4¹⁴)is non-trivial (i.e. none of the factor is±1, check it).

3. If𝑛 = 2𝑘with𝑘 ∈ ℕ ⧵ {0}then𝑛⁴+ 4^𝑛is even and greater than2, so it is composite.

If𝑛 = 2𝑘 + 1with𝑘 ∈ ℕ ⧵ {0}then𝑛⁴+ 4^𝑛= (2𝑘 + 1)⁴+ 4 × (2^𝑘)⁴which has a non-trivial factorization using Germain’s identity (check it), so it is a composite.

Let𝑘 ∈ ℕ ⧵ {0}. Assume by contradiction that(3𝑘 + 2)²= 𝑛²+ 𝑝where𝑛 ∈ ℕand𝑝is a prime number.

Then𝑝 = (3𝑘 + 2)²− 𝑛² = (3𝑘 − 𝑛 + 2)(3𝑘 + 𝑛 + 2).

• If3𝑘 − 𝑛 + 2 = 1then𝑛 = 3𝑘 + 1so𝑝 = 3𝑘 + 𝑛 + 2 = 6𝑘 + 3 = 3(2𝑘 + 1)is not prime, which leads to a contradiction.

• If3𝑘 + 𝑛 + 2 = 1then3𝑘 = −𝑛 − 1 < 0, which is not possible since𝑘 > 0.

Therefore𝑝admits a non-trivial factorization. Which is a contradiction.

Compare with Wilson’s theorem from Chapter 4.

We are going to prove the contrapositive:∀𝑛 ∈ ℕ, 𝑛is not prime ⟹ 𝑛 ∤ (𝑛 − 1)! + 1.

Let𝑛 ∈ ℕ. Assume that𝑛is not prime. Then there exists𝑘 ∈ ℕsuch that1 < 𝑘 < 𝑛and𝑘|𝑛.

Assume by contradiction that𝑛|(𝑛 − 1)! + 1. Then𝑘|(𝑛 − 1)! + 1. But𝑘|(𝑛 − 1)!since1 < 𝑘 < 𝑛.

Thus𝑘|(𝑛 − 1)! + 1 − (𝑛 − 1)! = 1. Which is a contradiction.

Therefore𝑛 ∤ (𝑛 − 1)! + 1.

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Let𝑛 ∈ ℕ ⧵ {0}. Consider the following𝑛consecutive natural numbers

(𝑛 + 1)! + 2, (𝑛 + 1)! + 3, (𝑛 + 1)! + 4, … , (𝑛 + 1)! + (𝑛 + 1)

Take(𝑛 + 1)! + 𝑘in the previous list (i.e.𝑘 = 2, … , (𝑛 + 1)). Then𝑘|(𝑛 + 1)! + 𝑘but1 < 𝑘 < (𝑛 + 1)! + 𝑘.

Therefore(𝑛 + 1)! + 𝑘has a non-trivial divisor.

1. Assume by contradiction that log₁₀2 = ^𝑎

𝑏 ∈ ℚ. Then log2

log10 = 𝑎

𝑏 ⇔ 𝑏log2 = 𝑎log10 ⇔log(2^𝑎) =log(10^𝑏) ⇔ 2^𝑎 = 10^𝑏⇔ 2^𝑎= 2^𝑏5^𝑏

By uniqueness of the prime factorization,𝑎 = 𝑏 = 0. Contradiction.

2. Assume by contradiction that√2 = ^𝑎

𝑏 ∈ ℚ. Then2𝑏²= 𝑎².

The prime factorization of the LHS has an odd number of primes (counted with exponents) whereas the RHS has an even number of primes (counted with exponents). Which is impossible since the prime factorization is unique up to order.

Assume by contradiction that49|𝑛³− 𝑛²− 2𝑛 + 1for some𝑛 ∈ ℤ.

Note that𝑛³− 𝑛²− 2𝑛 + 1 = (𝑛 + 2)³− 7𝑛²− 14𝑛 − 7.

Therefore, there exists𝑘 ∈ ℤsuch that𝑛 = 7𝑘 − 2.

Then𝑛³− 𝑛²− 2𝑛 + 1 = 49(7𝑘³− 7𝑘²+ 2𝑘) − 7.

Therefore49|49(7𝑘³− 7𝑘²+ 2𝑘) − (𝑛³− 𝑛²− 2𝑛 + 1) = 7. Which is a contradiction.

It is a special case of Dirichlet’s theorem on arithmetic progressions.

Assume that there are only finitely many primes3 = 𝑝₁< 𝑝₂< ⋯ < 𝑝_𝑟such that𝑝_𝑚= 4𝑘_𝑚+ 3with𝑘_𝑚∈ ℝ.

Set𝑛 = 4𝑝₁𝑝₂⋯ 𝑝_𝑟− 1. Then𝑛 = 4(𝑝₁𝑝₂⋯ 𝑝_𝑟− 1) + 3 Write𝑛 = ∏^𝑠_𝑖=1𝑞_𝑖as a product of prime numbers.

Note that each𝑞_𝑖is not one of the𝑝_𝑚nor2(otherwise𝑞_𝑖|1or2|1).

Therefore𝑞_𝑖 = 4𝑟_𝑖+ 1(the only possible remainder is1).

Hence𝑛 = ∏^𝑠_𝑖=1(4𝑟_𝑖+ 1) = 4𝛼 + 1for some𝛼 ∈ ℕ.

We obtain a contradiction with the uniqueness of the Euclidean division (the remainder of the Euclidean division of𝑛by4can’t3and1).

1. Let𝑛 ∈ ℕ. We are going to prove by induction on𝑘 ≥ 1that

∀𝑘 ∈ ℕ ⧵ {0}, 2²^𝑛+𝑘− 1 = (2²^𝑛− 1) ×

𝑘−1

∏𝑖=0(2²

𝑛+𝑖+ 1)

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Base case at𝑘 = 1:(2²

𝑛− 1) ×∏

𝑖=0(2²

𝑛+𝑖+ 1) = (2²^𝑛− 1) × (2²^𝑛+ 1) = (2²^𝑛)

2− 1²= 2²^𝑛+1− 1.

Induction step.Assume that2²^𝑛+𝑘− 1 = (2²^𝑛− 1) ×

𝑘−1

∏𝑖=0(2²

𝑛+𝑖+ 1)holds for some𝑘 ≥ 1. Then

(2²

𝑛− 1) ×

(𝑘+1)−1

∏𝑖=0 (2²

𝑛+𝑖+ 1) = (2²^𝑛− 1) ×

𝑘

∏𝑖=0(2²

𝑛+𝑖+ 1)

= (2²^𝑛− 1) ×

⎛⎜

⎜⎝

𝑘−1

∏𝑖=0(2²

𝑛+𝑖+ 1)

⎞⎟

⎟⎠

× (2²^𝑛+𝑘+ 1)

= (2²^𝑛+𝑘− 1) × (2²^𝑛+𝑘+ 1) by the induction hypothesis

=((2²^𝑛+𝑘)

2− 1² )

= (2²^𝑛+𝑘+1− 1) Which ends the induction step.

2. We may assume without lost of generality that𝑛 < 𝑚, i.e. 𝑚 = 𝑛 + 𝑘for some𝑘 ∈ ℕ ⧵ {0}.

Write𝐹_𝑖= 2²^𝑖+ 1then from the first question we get that

𝐹_𝑚+ 2 = (2²^𝑛− 1) ×

𝑘−1

∏𝑖=0

𝐹_𝑛+𝑖

Let𝑔 =gcd(𝐹_𝑚, 𝐹_𝑛). Then𝑑divides𝐹_𝑚and𝐹_𝑛thus𝑑divides2 = (2²^𝑛− 1) × (∏^𝑘−1^𝑖=0 𝐹_𝑛+𝑖) − 𝐹^𝑚. So either𝑑 = 2or𝑑 = 1. Since𝐹_𝑚is even, we get that𝑑 = 1.

Therefore gcd(𝐹_𝑚, 𝐹_𝑛) = 1.

1. Assume that𝑎^𝑛− 1is prime.

Note that𝑎^𝑛− 1 = (𝑎 − 1) (𝑎^𝑛−1+ 𝑎^𝑛−2+ ⋯ + ⋯ + 𝑎 + 1).

Since𝑎^𝑛− 1is prime, then it has no trivial divisor, therefore either𝑎 − 1 = 1or𝑎 − 1 = 𝑎^𝑛− 1.

The latter is not possible since𝑎, 𝑛 ≥ 2, thus𝑎 − 1 = 1, i.e. 𝑎 = 2.

Assume that𝑛 = 𝑝𝑞with𝑝, 𝑞 ∈ ℕ. Then2^𝑛− 1 = 2^𝑝𝑞− 1 = (2^𝑝− 1) ((2^𝑝)^𝑞−1+ (2^𝑝)^𝑞−2+ ⋯ + 2^𝑝+ 1). Since2^𝑛− 1is a prime number, then either2^𝑝− 1 = 1or2^𝑝− 1 = 2^𝑝𝑞− 1.

In the first case𝑝 = 1and in the other case𝑝 = 𝑝𝑞 = 𝑛.

Hence the only positive divisors of𝑛are1and itself, i.e. 𝑛is a prime number.

2. No,2¹¹− 1 = 2047 = 23 × 89.

Since1 + 2 + 3 + ⋯ + 𝑛 = ^𝑛(𝑛+1)

2 must be divisible by 3, either3|𝑛or3|𝑛 + 1. It is easy to check that this necessary condition is also suﬀicient when𝑛 > 3.

Note that69 = 3 × 23,1150 = 2 × 5²× 23 and4140 = 2² × 3²× 5 × 23. Note that only positive common divisors are1and23. Assuming the pirate is not alone, the treasure is shared between23people so there are22sailors.