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

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1 Fermat s Infinite Descent PMATH 340 Assignment 6 (Due Monday April 3rd at noon). (0 marks) Use Femtat s method of infinite descent to prove that the Diophantine equation x 3 + y 3 = 4z 3 has no solutions in positive integers x, y and z. Hint: Let (x,y,z) be a solution. Explain why x is even, and then write x = x. Deduce that y and z are even as well, so y = y and z = z. Explain why (x,y,z ) is also a solution and why this leads to a contradiction. Eisenstein Integers Let ω denote the primitive third root of unity. That is, ω = e πi 3 = + 3. Note that ω satisfies the equation ω + ω + = 0. The set Z[ω] := {a + bω : a,b Z} is called the ring of Eisenstein integers. For any Eisenstein integer α = a + bω, where a,b Z, the norm map is defined by N(a + bω) := a ab + b. () Just like the ring of Gaussian integers, the ring of Eisenstein integers is a Unique Factorization Domain.. (a) ( marks) Prove that Z[ω] is a ring by showing that 0, 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 () 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;

2 (c) (3 marks) We say that υ Z[ω] is a unit if υ α for every α Z[ω]. Prove that υ Z[ω] is a unit if and only if N(υ) =. Hint. To prove the sufficient condition, let υ = a+bω. Given the fact that N(υ) =, we will show that υ, i.e. there exists c + dω Z[ω] such that (c + dω)(a + bω) =. To see that this is true, determine c and d in terms of a and b by viewing 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. 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. Why does this imply that a + bω is a unit? (d) ( marks) Find all units in Z[ω]. The Diophantine Equation n = x xy + y Consider the setup as in Question. We say that γ 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 (mod 3) is also an Eisenstein prime. Hint: See Example 9.0. (b) (5 marks) Note that 3 = ( ω)( ω ), so 3 is not an Eisenstein prime. Also, it can be shown that every rational prime p (mod 3) is not an Eisenstein prime. Use this fact, as well as the results you proved in Questions (b) and 3 (a), to show that every integer n with the prime factorization n = 3 t p e pe pe k k q f q f q f l l, where p i (mod 3) for all i =,,...,k and q j (mod 3) for all j =,,...,l, admits a solution (x,y) to the Diophantine equation n = x xy + y. 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

3 4. (0 marks) Let Z[ ] := { a + b } : a,b Z. We say that υ Z[ ] is a unit if υ α for every α Z[ ]. Prove that there are infinitely many units in Z[ ]. Hint: Consider the Pell equation x y = ±. Explain why, for every (x,y ) satisfying this Diophantine equation, the value x + y is a unit in Z[ ]. Find any solution (x,y ), and then prove that, for every positive integer n, the integer coefficients x n and y n of the number x n + y n := (x + y ) n also satisfy the equation x n y n = ±. The Failure of Unique Factorization Consider the ring Z[ 3] = {a + b 3: a,b Z}. For every a,b Z, the norm map on Z[ 3] is defined by N(a + b 3) := a + 3b. You may assume that the norm is multiplicative. We will show that the unique factorization fails in Z[ 3]. To solve this problem, you might want to refer to Section.3 in Frank Zorzitto, A Taste of Number Theory. 5. (a) (3 marks) Prove that the only units of Z[ 3] are ±. Hint: Let υ = a + b 3 for a,b Z. By definition, υ Z[ 3] is a unit if υ α for every α Z[ 3]. Thus, in particular, υ. Explain why this fact implies the equality a +3b =. What are the solutions to this Diophantine equation? (b) (5 marks) We say that a non-zero number γ Z[ 3] is prime if the factorization γ = αβ for α,β Z[ 3] implies that either α is a unit or β is a unit. Prove that the numbers,7, + 3 and 3 are prime in Z[ 3]; (c) ( marks) Using Part (b), explain why the unique factorization fails in Z[ 3]. Preperiodic and Periodic Continued Fractions 3

4 Let α be a real number with the canonical continued fraction expansion α = [a 0,a,...,a n ;b,b,...,b k,b,b,...,b k,b,...]. In other words, at some point the elements of the continued fraction expansion start to repeat. We indicate this by writing α = [a 0,a,...,a n ;b,b,...,b k ]. A canonical continued fraction expansion of such kind is called preperiodic, and if the terms a 0,a,a,...,a n 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 + 5 and. 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 aα + bα + c = 0. Properties of Convergents Let α be a real number with the canonical continued fraction expansion α = [a 0,a,a,...] and for a non-negative integer n let the rational number p n / := [a 0,a,...,a n ] denote the n-th convergent of α. 7. (a) (5 marks) Prove that p n p n = ( ) n. (b) (5 marks) Let p/q be a rational number and let α be a real number. Prove that, if α p q < q, then p/q = p n / for some positive integer n. That is, p/q appears as a convergent in the canonical continued fraction expansion of α. 4

5 Hint: Suppose not and p/q p n / for any n N. Since the denominators q < q < q 3 <... of n-th convergents p n / grow, there must exist an integer n such that < q < +. Explain why p q p n. () q Then, using the triangle inequality p q p n α p q + α p n, as well as Proposition 3.4 in the lecture notes, prove that p q p n <, q in contradiction to the inequality (). The First Transcendental Number Let α be a complex number. We say that α is algebraic if there exists a nonzero polynomial f (x) with rational coefficients such that f (α) = 0. Otherwise we call it transcendental. In 840 s, the French mathematician Joseph Liouville proved that for each irrational algebraic number α there exist an integer d and a real number C > 0 such that α x y C y d 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 number k=0 0 k!, which is now called the Liouville Number. 8. (a) (6 marks) Prove that, for every integer n, the number α := k=0 k! = satisfies the inequality n α k=0 k! < ( n! ) n. (3) 5

6 Hint: Note that k=n+ k! < k=(n+)! k. 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 and C > 0 such that α x y C y d for all integers x and y > 0. Why does this inequality contradict the inequality (3)? 6