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;
(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
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
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
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! = + + 4 + 64 + 67776 +... satisfies the inequality n α k=0 k! < ( n! ) n. (3) 5
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