Sums of four squares and Waring s Problem Brandon Lukas

Transcription

1 Sums of four squares and Waring s Problem Brandon Lukas Introduction The four-square theorem states that every natural number can be represented as the sum of at most four integer squares. Look at the following examples: There could be multiple solutions: 4 = = = No matter how large, all natural numbers can be written as a sum of four-squares: = The earliest appearances of the theorem was found in Alexandrian Greek mathematician Diophantus s Arithmetica, a book written in the rd century AD. The book was translated in 161 by French mathematician Bachet who stated the theorem in the notes of his translation (for this reason, the theorem is also known as Bachet s conjecture). In 1748, Swiss mathematician Leonhard Euler wrote about the four-square identity in a letter to Goldbach. Even though the identity could be proven with elementary algebra, his identity was important in proving the four-square theorem because it implied that it was sufficient to prove the theorem for prime numbers. Using Euler s four-square identity, the theorem was first proven by Joseph Louis Lagrange in An alternative proof of the theorem utilizes Hurwitz quaternions exists. Quaternions are members of a noncommutative division algebra. The set of quaternions is denoted by H after Irish mathematician Hamilton who discovered the idea for quaternions while walking along the Royal Canal on his way to a meeting in 184. The proof that uses Hurwitz quaternions will be discussed in this paper. Waring s problem is a generalization of Lagrange s four-square theorem proposed by Edward Warring in It states that every rational integer is the sum of a fixed number g(k) of k th powers of positive integers. Lagrange proved that g() = 4 in his four-square theorem. Another example is g() = 9: any rational integer is the sum of at most 9 integers raised to the third power (this was proven by Wieferich in 1909). In the later 1900s, mathematicians proved g(4) = 19 and g(5) = 7. In 177, two years after Lagrange proved Waring s problem for k =, Euler gave a lower bound for g(k) calculated by (/) k k. By the end of the nineteenth century, the existence of g(k) was known for only a finite number of values of k thanks to the work done by J. Liouville, E. Lucas, and many others. In 1909, Dr. Hilbert presented a proof regarding the existence of g(k) for any positive integer k. His proof used complicated analysis and multiple integrals to show that Euler s expression above is in fact g(k) for 6 k Essentially, provided that g(k) = k {( ( ) k ) k } k ( ) k k

2 where {x} = x x. If the inequality above fails, then g(k) = k ( ) k ( ) 4 k θ where θ is either or depending whether equals or exceeds k. ( ) 4 k ( ) k ( ) 4 k ( ) k This paper will only focus on Waring s problem for k = whose solution was proved by Lagrange in 1770 in his four-square theorem. However, as you can see from the formulas above, g() = 4 because ( ) g() = g() =.5 4 g() = 4 g() = 4. This is valid because 4 {( (( ) ) } ( ( ) ) ) 4 (.5 ) ( ) 4 Goal The goal of this paper is to prove Lagrange s four-square theorem which states that any natural number N can be represented as the sum of four squares, or where a, b, c, and d are integers. This is also known as Waring s problem for k =. N = a b c d

3 We will be using Hurwitz quaternions to prove this theorem. Background on Quaternions We can think of quaternions as a number system that is similar to but extends the complex numbers. We define the quaternions to be the matrices [ ] a bi c di. c di a bi More commonly though, quaternions are written in the form a bi cj dk where a, b, c, and d are real numbers and i, j, and k are imaginary units which can be defined as [ ] [ ] [ ] i i 0 i =, j =, k =. 1 0 i 0 0 i The imaginary units satisfy the following relation: i = j = k = ijk = 1. There s not much of a difference between a quaternion and a Hurwitz quaternion. A Hurwitz quaternion is denoted by H = {a bi cj dk H : a, b, c, d Z or a, b, c, d Z 1 }. In other words, now a, b, c, and d are either all integers or all half-integers. Half-integers are simply all numbers that are half of an odd integer (the set Z 1 ). A Hurwitz quaternion α is prime if it is divisible only by the units ±1, ±i, ±j, ±k, and ± 1, ± 1 i, ± 1 j, ± 1 k, and multiples of α with these. A Hurwitz quaternion β divides α if there exists a Hurwitz quaternion φ such that α = βφ or α = φβ. This is also true for ordinary numbers. The norm of a quaternion q is denoted by q. The norm is equivalent to the square root of the quaternion s determinant. Hence, the following relationship is true: ( ) q = a bi cj dk a bi c di = det = a b c d. c di a bi Furthermore, norms have this multiplicative property: q 1 q = q 1 q. The conjugate of any quaternion q = a bi cj dk is q = a bi cj dk and they satisfy the following relations: qq = q q 1 q = q q 1

4 When you multiply a quaternion by its conjugate, the product will be a real number. This is also true with complex numbers. For example, consider a complex number a bi and its conjugate a bi. The product is a b which is a real number. Lemmas Lemmas are short theorems used in proving a larger theorem. Lemma 1: if p is an odd prime p = n 1, n N, then there are l, m Z such that p divides 1 l m. Proof: Let x and y be any two of the numbers 0, 1,,..., n. We can show the squares x and y are incongruent (mod p) by contradiction: x y (mod p) x y 0 (mod p) (x y)(x y) 0 (mod p) x y or x y 0 (mod p) But x y 0 (mod p) because 0 < x y < p (this is because x and y can be at most the value n = p 1, and if x = y = n, then x y = n = p 1 and clearly p 1 < p). Therefore, the n 1 different numbers l = 0, 1,,..., n give n 1 incongruent values of l (mod p). Similarly, the numbers m = 0, 1,,..., n also give n 1 incongruent values of m (mod p), and hence of m (mod p), and hence of 1 m (mod p). But only n 1 incongruent values exist mod p, so by the pigeon-hole principle (if n items are put into m containers, with n > m, then at least one container must contain more than one item) that means for some l and m, we have l 1 m (mod p) 1 l m 0 (mod p) which means p divides 1 l m. Lemma : if a Hurwitz prime divides a product of Hurwitz quaternions αβ, then the prime divides α or β. This is similar to Euclid s lemma which states that if p is prime and p ab, then p a or p b (or both) where a and b are integers. Proof: I will provide a proof of Euclid s lemma using Bézout s identity which says that if x and y are relatively prime integers (meaning they share no common divisors other than 1 ), then there exists integers r and s such that rx sy = 1. We will let a and p be relatively prime and assume that p ab. We will show that p also divides b. rp sa = 1 We will multiply both sides by b, rpb sab = b 4

5 Now we can see that the first term rpb is clearly divisible by p because it is a multiple of p. The second term sab is clearly divisible by ab because it is a multiple of ab. Because we assumed ab is divisible by p, the second term is also divisible by p. Therefore, the sum of these two terms b is also divisible by p which was meant to be proven. Lemma : the product of two numbers, each of which is a sum of four squares, is itself a sum of four squares. This is known as Euler s four-square identity (not the same as Lagrange s four-square theorem). Proof: Euler s four-square identity can be proven using quaternions and the multiplicative property of norms. Let x 1 and x each be the sum of four square integers. x 1 = a 1 b 1 c 1 d 1 x = a b c d By using the norm relationship stated earlier, we can see x 1 = q 1 for some quaternions q 1 and q. x = q Now we can multiply x 1 and x to see x 1 x = q 1 q = ( q 1 q ) = q 1 q = A Bi Cj Dk = A B C D for some A, B, C, and D Z. Alternatively, this identity can be easily proven without quaternions and only using elementary algebra. This is the shortest possible proof of Euler s identity: (a b c d )(w x y z ) = (aw bx cy dz) (bwax dycz) (cwdxay bz) (dw cxbyaz). Lemma 4: any integer greater than 1 can be written as a unique product of prime numbers. This is known as the fundamental theorem of arithmetic (or unique prime factorization theorem). Proof: We will first show existence using Euclid s proof. It will be proven by contradiction. Suppose there were positive integers which could not be written as a product of prime numbers. By the wellordering principle (every non-empty set of positive integers contains a least element), there must be a smallest such number n. Because n > 1 and n is not prime nor a product of prime numbers, n is composite. Thus, n = a b where a and b are positive integers smaller than n. Since we claimed n is the smallest number which cannot be written as the product of primes, both a and b can be written as the product of primes. However, this means that n = a b can also be written as a product of primes which is a contradiction. Therefore, all positive integers can be written as a product of prime numbers. Now, we will show uniqueness using Euclid s lemma (proved in Lemma ). We will use proof by contradiction again. 5

6 Suppose that there were two ways to write the product of prime numbers s > 1. s = p 1 p...p m = q 1 q..q n. We will show that m = n and that q j is simply a rearrangement of p i. By Euclid s lemma, p 1 must divide one of the q j ; let s say that p 1 divides q 1. Since q 1 is prime, the only divisors are itself and 1, so p 1 = q 1. We can divide all parts of the equation by p 1 to get s p 1 = p...p m = q...q n. Using the same reasoning above, we show that p must equal one of the remaining q j (say q ). We divide all parts of the equation by p to get s p 1 p = p...p m = q...q n. We can repeat this for each of the m p i s to show that every p i is a q j and that m n. Then, we can apply the same process in reverse to show that every q j is a p i and that n m, therefore making n = m and every q j is a p i and vice versa. Proving Lagrange s Four-Square Theorem Now that we have some background on quaternions and established the needed lemmas and theorems above, we are ready to prove Lagrange s four-square theorem using non-hurwitz primes. We will assume an odd prime p has a non-trivial Hurwitz factorization p = (a bi cj dk)γ. Then we will take the conjugates of both sides to get p = γ(a bi cj dk). Now we will multiply p with p to get pp = (a bi cj dk)γγ(a bi cj dk) p = (a bi cj dk)(a bi cj dk)γγ p = (a b c d ) γ where both a b c d and γ > 1. But the only positive integer factorization of p is pp, so by unique prime factorization (Lemma 4 ), p = a b c d. If a, b, c, and d are integers, then we are done. But since they are the coefficients of a Hurwitz integer, they could be half-integers. Fortunately, there is a way to re-express the half-integer coefficients to achieve whole integer values. In the case where ψ = a bi cj dk has half-integer co-ordinates, consider an integer ω = ±1±i±j±k. If we choose appropriate signs in ω, then ψ can be written as ψ = ω a b i c j d k where a, b, c, and d 6

7 are even integers (adding ω will result in the half-integer coordinates to either go up or go down a half to the nearest even integer). Therefore, for some integers A, B, C, and D. p = (a bi cj dk)(a bi cj dk) p = (ω a b i c j d k)(ω a b i c j d k) p = (ω a b i c j d k)ω ω(ω a b i c j d k) p = (A Bi Cj Dk)(A Bi Cj Dk) p = A B C D. So, Hurwitz integers can be written as the sum of four-squares even if they have half-integer coefficients. The last case to consider is for any odd prime p. By Lemma 1, we know that p divides 1 l m. By factorization, we see 1 l m = (1 li mj)(1 li mj). If p were a Hurwitz prime, then Lemma says that then p should be able to divide at least one of these factors. However, neither 1 p li p mj p nor 1 p li p mj p are Hurwitz integers. This is a contradiction and shows that p is not a Hurwitz prime. Therefore, we are able to apply our previous conclusion and say that any odd prime p is a sum of four integer squares. Before we can use Euler s four-square identity (Lemma ), we need to show that 1 and can also be written as the sum of four-squares. Obviously 1 = , = We now have shown that 1,, and all odd primes can be written as a sum of four squares. Because all integers are either prime or a product of primes (Lemma 4 ), and Euler s four-square identity says that the product of two numbers, each of which can be written as a sum of four-squares, can also be written as the sum of four-squares, every natural number can thereby be written as a sum of four-squares. Finding Solutions Other than using guess and check, one can find the solution to a Lagrange four-square for any number using a simple algorithm. We will find integers a, b, c, and d that satisfy the following equation for a given p: p = a b c d. In this example, we will use the algorithm to solve for p = 4. 7

8 The first step is to determine the range of a. We know the max value for a is floor the square root of p. a max = Floor( p) = Floor( 4) = Floor(4.899) = 4 The min value for a is the first value where a p 4. p 4 = 4 4 = 6 a 6 Testing values for a, we see 1 6, 6, 6. Since a = is the first value where the expression is true, a min =. Obviously, the range of a is [a min, a max ]. In this example, a is in the range [, 4]. Since a is an integer, a can be only or 4. The second step is to test values for a within the range of a. We will determine the range of b using a similar method above. We need to use a value for a in the range of a. We will start with a =. The max value for b is floor the square root of p a. b max = Floor( p a ) = Floor( 4 ) = Floor( 15) = Floor(.87) = The min value for b is the first value where b (p a ).. (p a ) = (4 ) b 5 = (15) = 5 Testing values for b, we see b = is the first value where the expression is true. So b min =. This means that when a =, b must be (the range for b is [, ]; b = ). The third step is to test values for b within the range of b. Now we will determine the range of c using a similar method above. We will use a = and b = for our values. The max value for c is floor the square root of p a b. c max = Floor( p a b ) = Floor( 4 ) = Floor( 6) = Floor(.45) = The min value for c is the first value where c (p a b ). (p a b ) = (4 ) = (6) = 8

9 . c Testing values for c, we see c = is the first value where the expression is true. So c min =. This means that when a = and b =, c must be (the range for c is [, ], so ccanonlybe). The fourth step is to test values for c within the range of c. Now we will determine the range of d using a similar method above. We will use a =, b =, and c = for our values. The only value for d is the square root of p a b c. If this number is not an integer, then this is not a solution. d = p a b c = 4 = = Because d = and is not an integer, this is not a solution. The fifth step is to repeat the previous steps if necessary. If we had more values for c, we would repeat the fourth step using a different value for c to determine d. Afterwards, if we had more values for b, we would repeat the third step using a different value for b to determine c and then move on to the fourth step to determine d. After that, we would repeat the second step using a different value for a to determine b, then move on to the third step to determine c and the fourth step to determine d. Since we don t have any more values for c and b when a =, we will go back to the second step and test the other value for a. In other words, we will now redetermine the values for b, c, and d using a = 4. We use the same formulas above with a = 4 to see b max = b min = b =. Then we see that when a = 4 and b =, c max = c min = c =. Finally, we see that when a = 4, b =, and c =, d = 4 4 = 0 = 0. Because d = 0 and 0 is an integer, our solution is (a, b, c, d) = (4,,, 0). We can verify our solution easily: 4 0 = = 4. References

10 pdf