PMATH 340 Assignment 6 (Due Monday April 3rd at noon)

Fermat’s Infinite Descent

1. (10 marks) Use Femtat’s method of infinite descent to prove that the Dio- phantine equation x3 + 2y3 = 4z3 has no solutions in positive x, y and z. Hint: Let (x,y,z) be a solution. Explain why x is even, and then write x = 2x0. Deduce that y and z are even as well, so y = 2y0 and z = 2z0. Explain why (x0,y0,z0) is also a solution and why this leads to a contradiction.

Eisenstein Integers Let ω denote the primitive third . That is, √ 2πi −1 + −3 ω = e 3 = . 2 Note that ω satisfies the equation ω2 + ω + 1 = 0. The set

Z[ω] := {a + bω : a,b ∈ Z} is called the of Eisenstein integers. For any Eisenstein α = a + bω, where a,b ∈ Z, the norm map is defined by

N(a + bω) := a2 − ab + b2. (1)

Just like the ring of Gaussian integers, the ring of Eisenstein integers is a Unique Domain.

2. (a) (2 marks) Prove that Z[ω] is a ring by showing that 0,1 ∈ Z[ω], and for all α,β ∈ Z[ω] it is the case that α ± β ∈ Z[ω] and α · β ∈ Z[ω]; (b) (3 marks) Prove that the norm map defined in (1) is multiplicative. That is, for every α,β ∈ Z[ω] it is the case that N(αβ) = N(α)N(β). Explain why N(α) ≥ 0 for every α ∈ Z[ω] and why N(α) = 0 if and only if α = 0;

1 (c) (3 marks) We say that υ ∈ Z[ω] is a if υ | α for every α ∈ Z[ω]. Prove that υ ∈ Z[ω] is a unit if and only if N(υ) = 1. Hint. To prove the sufficient condition, let υ = a+bω. Given the fact that N(υ) = 1, we will show that υ | 1, i.e. there exists c + dω ∈ Z[ω] such that (c + dω)(a + bω) = 1. To see that this is true, determine c and d in terms of a and b by view- ing the above equation as the equation in complex numbers. That is, convert a + bω and c + dω into their complex form, and then equate real and imaginary parts of (c + dω)(a + bω) and 1. You will get a system of two equations with two unknowns c and d. Explain why c and d are integers. Conclude that c + dω ∈ Z[ω], which means that a + bω | 1. Why does this imply that a + bω is a unit? (d) (2 marks) Find all units in Z[ω]. The Diophantine Equation n = x2 − xy + y2 Consider the setup as in Question 1. We say that γ 6= 0 is an Eisenstein prime if the factorization γ = αβ for α,β ∈ Z[ω] implies that either α is a unit or β is a unit.

3. (a) (5 marks) Prove that every rational prime p ≡ 2 (mod 3) is also an Eisenstein prime. Hint: See Example 29.10. (b) (5 marks) Note that 3 = (1 − ω)(1 − ω2), so 3 is not an Eisenstein prime. Also, it can be shown that every rational prime p ≡ 1 (mod 3) is not an Eisenstein prime. Use this fact, as well as the results you proved in Questions 2 (b) and 3 (a), to show that every integer n with the prime factorization

t e1 e2 ek 2 f1 2 f2 2 f` n = 3 p1 p2 ··· pk q1 q2 ···q` ,

where pi ≡ 1 (mod 3) for all i = 1,2,...,k and q j ≡ 2 (mod 3) for all j = 1,2,...,`, admits a solution (x,y) to the Diophantine equation n = x2 − xy + y2. It can be shown that the numbers of the above form are the only numbers that admit solutions, but you do not have to prove that.

Rings With Infinitely Many Units

2 4. (10 marks) Let √ n √ o Z[ 2] := a + b 2: a,b ∈ Z . √ √ We say that υ ∈ Z[ 2] is a unit if √υ | α for every α ∈ Z[ 2]. Prove that there are infinitely many units in Z[ 2]. 2 2 Hint: Consider the Pell equation x − 2y = ±1. Explain why,√ for every (x1,y√1) satisfying this Diophantine equation, the value x1 + y1 2 is a unit in Z[ 2]. Find any solution (x1,y1), and then prove that, for every positive√ integer n√, the integer coefficients xn and yn of the number xn + yn 2 := n 2 2 (x1 + y1 2) also satisfy the equation xn − 2yn = ±1.

The Failure of Unique Factorization Consider the ring √ √ Z[ −13] = {a + b −13: a,b ∈ Z}. √ For every a,b ∈ Z, the norm map on Z[ −13] is defined by √ N(a + b −13) := a2 + 13b2.

You may assume that the√ norm is multiplicative. We will show that the unique factorization fails in Z[ −13]. To solve this problem, you might want to refer to Section 2.3 in Frank Zorzitto, A Taste of Number Theory. √ 5. (a) (3 marks) Prove that the only units of [ −13] are ±1. √ Z √ Hint: Let υ = a + b −13 for a,b√∈ Z. By definition, υ ∈ Z[ −13] is a unit if υ | α for every α ∈ Z[ −13]. Thus, in particular, υ | 1. Explain why this fact implies the equality a2 +13b2 = 1. What are the solutions to this Diophantine equation? √ (b) (5 marks) We say that a non-zero number√ γ ∈ Z[ −13] is prime if the factorization γ = αβ for α,β ∈ Z[ −13] implies that√ either α is a√ unit or β is a unit. Prove√ that the numbers 2,7, 1 + −13 and 1 − −13 are prime in Z[ −13]; (c) (2√ marks) Using Part (b), explain why the unique factorization fails in Z[ −13].

Preperiodic and Periodic Continued Fractions

3 Let α be a real number with the canonical continued fraction expansion

α = [a0,a1,...,an;b1,b2,...,bk,b1,b2,...,bk,b1,...].

In other words, at some point the elements of the continued fraction expansion start to repeat. We indicate this by writing

α = [a0,a1,...,an;b1,b2,...,bk].

A canonical continued fraction expansion of such kind is called preperiodic, and if the terms a0,a1,a2,...,an are missing we say that it is periodic. The smallest number k such that the terms repeat is called the period of a continued fraction. √ 6. (a) (5 marks) Determine canonical continued fraction expansions for 1+ 5 √ 2 and 2. Are they both preperiodic? Are they both periodic? What are the periods of their continued fraction expansions? (b) (5 marks) Prove that if a real number α has a preperiodic canonical continued fraction expansion, then there exist rational integers a,b and c, not all zero, such that

aα2 + bα + c = 0.

Properties of Convergents Let α be a real number with the canonical continued fraction expansion

α = [a0,a1,a2,...] and for a non-negative integer n let the pn/qn := [a0,a1,...,an] denote the n-th convergent of α.

n−1 7. (a) (5 marks) Prove that pnqn−1 − pn−1qn = (−1) . (b) (5 marks) Let p/q be a rational number and let α be a real number. Prove that, if

p 1 α − < , q 2q2

then p/q = pn/qn for some positive integer n. That is, p/q appears as a convergent in the canonical continued fraction expansion of α.

4 Hint: Suppose not and p/q 6= pn/qn for any n ∈ N. Since the denom- inators q1 < q2 < q3 < ... of n-th convergents pn/qn grow, there must exist an integer n such that qn < q < qn+1. Explain why

p pn 1 − ≥ . (2) q qn qqn Then, using the triangle inequality

p pn p pn − ≤ α − + α − , q qn q qn as well as Proposition 31.14 in the lecture notes, prove that

p pn 1 − < , q qn qqn in contradiction to the inequality (2).

The First Transcendental Number Let α be a . We say that α is algebraic if there exists a non- zero polynomial f (x) with rational coefficients such that f (α) = 0. Otherwise we call it transcendental. In 1840’s, the French mathematician Joseph Liouville proved that for each irrational α there exist an integer d ≥ 2 and a real number C > 0 such that

x C α − ≥ y yd for all integers x and y > 0. In other words, every irrational algebraic number cannot be approximated “too well” by a rational number x/y. This property of irrational algebraic numbers allowed him to discover the first transcendental num- ∞ −k! ber ∑k=0 10 , which is now called the Liouville Number. 8. (a) (6 marks) Prove that, for every integer n ≥ 1, the number ∞ 1 1 1 1 1 α := ∑ k! = 1 + + + + + ... k=0 2 2 4 64 16777216 satisfies the inequality

n 1 1 α − < . (3) ∑ k! n! n k=0 2 (2 )

5 Hint: Note that ∞ 1 ∞ 1 ∑ k! < ∑ k . k=n+1 2 k=(n+1)! 2 Use the formula for the infinite geometric series afterwards. (b) (4 marks) Use Liouville’s Theorem and the inequality established in Part (a) to prove that the number α is either rational or transcendental. Hint: Suppose not. Then there exist fixed integers d ≥ 2 and C > 0 such that x C α − ≥ y yd for all integers x and y > 0. Why does this inequality contradict the inequality (3)?

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