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7/8/2011 ο½ Any natural number π can be represented as π = π 2 + π 2 + π 2 + π2 where π, π, π, and π are integers. Erin Compaan and Cynthia Wu SPWM 2011 ο½ ο½ ο½ ο½ Diophantus β ca. 200 A.D. Bachet β 1621 Fermat β 17th c. Lagrange β 1770 Lagrangeβs Four Square Theorem: Our goal is to prove this theorem using Hurwitz Quaternions. ο½ ο½ Denoted by β Members of a non-commutative division algebra ο½ Form of quaternions: π + ππ + ππ + ππ where π, π, π, and π are real numbers ο½ Fundamental formula of quaternion algebra: π 2 = π 2 = π 2 = πππ = β1 1 7/8/2011 ο½ Not that much of a difference! π― = *π + ππ + ππ + ππ πβ βΆ π, π, π, π π β€ or π, π, π, π π ο½ ο½ ο½ 1 β€+ + 2 So now π, π, π, and π are either all integers or all half integers Half integers: all numbers that are half of an odd 1 integer β the set β€ + . 2 A Hurwitz quaternion πΌ is prime if it divisible only 1 1 by the quaternions ±1, ±π, ±π, ±π, and ± ± π ± 1 π 2 ο½ 1 2 2 2 ± π, and multiples of πΌ with these. A Hurwitz quaternion π½ divides πΌ if there exists a Hurwitz quaternion π such that πΌ = π½π or πΌ = ππ½. β¦ EG: 7/2, -13/2, 8.5 ο½ A Lipschitz quaternion is a quaternion of the form π + ππ + ππ + ππ, with a, b, c, d π β€. β¦ E.g. 1 + 7π β 83π + 12π. Any natural number π can be represented as π = π 2 + π 2 + π 2 + π2 where π, π, π, and π are integers. Prove this using Hurwitz Quaternions: π― = *π + ππ + ππ + ππ πβ βΆ π, π, π, π π β€ or π, π, π, π π β€+ 2 7/8/2011 ο½ ο½ If π is a prime π = 2π + 1, π π β, then there are π, π π β€ such that π divides 1 + π2 + π2. If a Hurwitz prime divides a product of Hurwitz quaternions πΌπ½, then the prime divides πΌ or π½. ο½ ο½ Then π’ = π + ππ + ππ + ππ 2 = πΌ 2 and π£ = π€ + π₯π + π¦π + π§π 2 = π½ 2 , for some Lipschitz quaternions πΌ and π½. Then π’π£ = πΌ π½ 2 = πΌπ½ 2 = π΄ + π΅π + πΆπ + π·π 2= π΄2 + π΅ 2 + πΆ 2 + π·2 for some π΄, π΅, πΆ, π· π β€. ο½ ο½ Base Cases 1 = 12 + 02 + 02 + 02 2 = 12 + 12 + 02 + 02 If two numbers can be written as a sum of four integer squares, then so can their product. Proof: Suppose that π’ = π 2 + π2 + π 2 + π 2 and π£ = π€ 2 + π₯2 + π¦2 + π§2. ο½ ο½ ο½ ο½ ο½ ο½ Suppose π is an odd prime which has a non-trivial Hurwitz factorization π = (π + ππ + ππ + ππ)πΌ. Conjugating: π = π = πΌ(π β ππ β ππ β ππ). Multiplying the equations: π2 = π + ππ + ππ + ππ πΌπΌ π β ππ β ππ β ππ = π2 + π 2 + π 2 + π 2 πΌ 2 Since π is prime, the factors of π2 must both be π. Thus π = π 2 + π 2 + π 2 + π2. If π, π, π, and π are integers , weβre done. If not, we can still show that p is a sum of four integer squares. 3 7/8/2011 ο½ ο½ ο½ Now let π be an odd prime. Then there exist integers π and π such that π divides 1 + π2 + π2. Then π divides (1 + ππ + ππ)(1 β ππ β ππ). By the previously stated lemma, if π were a Hurwitz prime, it must divide one of these factors. 1 π β π π π ο½ 1 π π π But this would imply that + π + π β π π π π π or ο½ ο½ ο½ is a Hurwitz integer, a contradiction. Thus π is not a Hurwitz prime. Fermatβs Two Square Theorem: If a prime π is of the form 4π + 1 for some π π β, then π = π 2 + π 2 for some π, π π β€. ο½ Gaussian integers: Complex numbers with integer coefficients. ο½ Gaussian integer prime: A Gaussian integer z which is divisible only by ±1 or ± π, or products of z with these. ο½ ο½ Lemma: For any prime π of the form 4π + 1, π π β, there exists an integer π such that π divides 1 + π2 . ο½ ο½ Lemma: If a Gaussian integer prime π divides πΌπ½ for some Gaussian integers πΌ and π½, then π divides πΌ or π divides π½. ο½ ο½ ο½ Since π is not a Hurwitz prime, we can apply our previous conclusion and say that π is a sum of four integer squares. We now have that 1, 2, and all odd primes can be written as a sum of four squares. By the Four Squares identity, every natural number can be written as a sum of four squares. Suppose p is a prime of the form 4π + 1 for some π π β. Then p divides 1 + π2 = (1 + ππ)(1 β ππ) for some π π β. Since p divides neither factor of 1 + π2, π is not a Gaussian prime. Then p has a nontrivial factorization in the Gaussian integers π = (π + ππ)(π₯ + π¦π). 4 7/8/2011 ο½ ο½ ο½ ο½ ο½ Conjugating and multiplying equations π2 = π + ππ π₯ + π¦π π + ππ π₯ + π¦π = (π + ππ)(π β ππ)(π₯ + π¦π)(π₯ β π¦π) = π2 + π 2 π₯ 2 + π¦ 2 . Since p is prime and the factorization was nontrivial, the factors π 2 + π 2 and π₯ 2 + π¦ 2 are equal to p. Thus p can be written as a sum of two integer squares. ο½ ο½ ο½ ο½ ο½ Various contributions Calculus of Variations, Lagrange Multipliers, PDEβs Prolific writer Proved four square theorem in 1770 Meticulous and shy Joseph-Louis Lagrange ο½ ο½ ο½ ο½ Born to a Jewish family Number theorist Mostly contributed to number theory and algebras Sickly ο½ ο½ ο½ ο½ Adolf Hurwitz German mathematician Investigated number theory, Bessel functions, PDEβs Work on quadratic mechanics influenced Einstein Also very sickly Rudolf Lipschitz 5 7/8/2011 Any natural number N can be represented as π = π2 + π 2 + π 2 + π 2 where a, b, c, and d are integers ο½ Hurwitz quaternions can be useful in a variety of ways! 6