Number Theory and Graph Theory. Arithmetic functions and roots of unity

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1 1 Number Theory and Graph Theory Chapter 3 Arithmetic functions and roots of unity By A. Satyanarayana Reddy Department of Mathematics Shiv Nadar University Uttar Pradesh, India satya8118@gmail.com

2 2 Module-1: Introduction to arithmetic functions Objectives Introduction to arithmetic functions. Multiplicative function and completely multiplicative function. Properties of few well known multiplicative functions. Perfect numbers. An arithmetic function takes positive integers as inputs and produces real or complex numbers as outputs. Definition 1. A function f : N C is called a number-theoretic (or arithmetic) function. Although, the co-domain of a number-theoretic function is not required to be an integer, but most of the arithmetic functions that we will encounter are integer-valued. We now give a few examples of arithmetic functions and study their properties. Definition 2. Given a positive integer n, the number of positive divisors of n, denoted τ(n), is given by τ(n) = 1. d n n τ(n) the sum of the divisors of n, denoted σ(n), is given by σ(n) = d. d n n σ(n) and a positive integer k, we define σ k (n) = d k. Note that σ 0 = τ and σ 1 = σ. d n

3 3 U(n) = 1. N(n) = n. 1 i f n = 1 I(n) = 0 i f n 1. Möbius function 0, if n is not square free, µ(n) = 1, if n is square free and has even number of prime factors, 1, if n is square free and has odd number of prime factors. n µ(n) Definition 3. Fix a positive integer n and an arithmetic function f. Then, the divisor sum of f at n, denoted (D f )(n), equals the sum of the values of f at the positive divisors of n. Notice that the divisor sum is a function which takes an arithmetic function as input and produces an arithmetic function as output. Mathematically, Theorem 4. (D f )(n) = f (d). d n 1. Let f : N Z + be a function defined by f (n) = n k, for some k Z +. Then, f is an arithmetic function and note that its divisor sum D f = σ k. 2. Similarly, verify that DU = τ and DN = σ. 3. Dµ = I. Proof. Proof of (1) and (2) directly follows from the definition of the respective arithmetic functions. Now, let us prove (3). Note that the result holds true for n = 1. That is, (Dµ)(1) = µ(1) = 1.

4 4 So, let n > 1. Then, by the fundamental theorem of arithmetic n = p a 1 1 pa 2 2 pa k k, where p 1, p 2,..., p k are distinct primes and a i 1 for all i. As µ(d) = 0, whenever d is not square free, we have (Dµ)(n) = µ(1) + k i=1 µ(p i ) + µ(p i p j ) + + µ(p 1 p 2 p k ). i j Thus, using the binomial theorem (a + b) n = n ( n i) a n i b i, where ( n) n! i = i=0 i!(n i)! ) (Dµ)(n) = 1 + ( ) k ( 1) + 1 ( ) k ( 1) ( k 3 ( 1) 3 + +, we have ( ) k ( 1) k = (1 1) k = 0. k Definition An arithmetic function f is said to be a multiplicative function if f (mn) = f (m) f (n), whenever gcd(m,n) = An arithmetic function f is said to be a completely multiplicative function if f (mn) = f (m) f (n). By definition, every completely multiplicative function is multiplicative. For example, U and N are completely multiplicative functions and hence multiplicative. If f is multiplicative and n 1,n 2,...,n k are pairwise relatively prime positive integers then, f (n 1 n 2 n k ) = f (n 1 ) f (n 2 ) f (n k ). Consequently, if f is multiplicative, then in order to evaluate f (n) for any positive integer n 2, it is sufficient to evaluate f (p k 1 1 ), f (pk 2 2 ),..., f (pk r r ), whenever n = p k 1 1 pk 2 2 pk r r is the prime factorization of n into distinct primes. Further, if f is completely multiplicative then, f (n) = f (p 1 ) k 1 f (p 2 ) k 2 f (p r ) k r, i.e., if we know the values of f at prime numbers then, we know its value at every positive integer. If f is multiplicative and not identically zero, then f (1) = 1.

5 5 Proof. Since f is not identically zero, there exists m N such that f (m) 0. But, f (m) = f (m 1) = f (m) f (1) and hence f (1) = 1. Theorem 6. Let f : N C be a multiplicative function. Then D f is also multiplicative. Proof. Let gcd(a,b) = 1 and let x and y be any positive divisors of a and b, respectively. Then, xy is a positive divisor of ab. Conversely, if m ab then, there exists x,y Z such that m = xy and x a and y b. Hence, (D f )(ab) = m ab = = x a y b f (m) = f (xy) x a y b ( f (x) x a f (x) f (y) )( ) f (y) y b = (D f )(a) (D f )(b). Since the functions U,N and I are multiplicative, using Theorems 4 and 6, we observe that τ,σ and µ are multiplicative functions. But, we give an independent proof of all of these results as they also give some extra information about the actual functions. Theorem 7. The Möbius function µ is multiplicative. Proof. Let a,b be relatively prime positive integers and let p be a prime such that p 2 a or p 2 b. Then, p 2 ab and hence µ(ab) = 0 = µ(a)µ(b). So, let us assume that both a and b are square free integers. Then, a = p 1 p 2 p r and b = q 1 q 2 q s, where the primes p 1,..., p r,q 1,...,q s are distinct as gcd(a,b) = 1. Now, by definition µ(ab) = µ(p 1 p 2 p r q 1 q 2 q s ) = ( 1) r+s = ( 1) r ( 1) s = µ(a)µ(b).

6 6 Properties of τ(n) and σ(n) Theorem 8. Let n = p k 1 1 pk 2 2 pk r r be the prime factorization of n 2 into distinct primes. Then, 1. τ(n) = (k 1 + 1)(k 2 + 1) (k r + 1). 1 p r σ(n) = pk pk p 1 1 p 2 1 pk r+1 r Proof. Proof of 1 From Theorem 14 of Module 1 of Chapter 2, any divisor d of n is of the form d = p a 1 1 pa r r, where 0 a i k i. So, for each i,1 i r, a i has k i + 1 possible choices. Also, the choices for a i is independent of the choice of a j, whenever i j, the number of possible divisors of n are Proof of 2 (k 1 + 1)(k 2 + 1) (k r + 1). Observe that each positive divisor of n occurs exactly once as a term in the expansion of the product In other words, (1 + p p k 1 1 )(1 + p p k 2 2 ) (1 + p r + + p k r r ). σ(n) = (1 + p p k 1 1 )(1 + p p k 2 2 ) (1 + p r + + p k r r ) = pk p 1 1 pk p 2 1 pkr+1 r 1 p r 1. Theorem 9. The arithmetic functions τ and σ are multiplicative. Proof. Let m and n be relatively prime positive integers. Then, the result is clearly true when m or n is 1. So, assume that n,m 2. Then, by the fundamental theorem of arithmetic, there exists distinct primes p 1,..., p r,q 1,...,q s and positive integers a 1,...,a r,b 1,...,b r such that m = p a 1 1 pa r r,n = q b 1 1 qb s s and mn = p a 1 1 pa r r q b 1 1 qb s s.

7 7 Thus, Similarly, τ(mn) = (a 1 + 1)(a 2 + 1) (a r + 1) (b 1 + 1)(b 2 + 1) (b s + 1) = τ(m)τ(n). σ(mn) = [ p a p 1 1 par+1 r 1 p r 1 ][ q b q 1 1 ] qbs+1 s 1 = σ(m)σ(n). q s 1 Example 10. Let n 2 be a positive integer. If the prime factorization of n into distinct primes is given by n = p k 1 1 pk r r 2 then, τ(n) is odd if and only if n is a perfect square. Proof. The result follows from τ(n) = (k 1 + 1) (k r + 1) as n is a perfect square if and only if 2 divides each k i,1 i r. Definition 11. A positive number n is said to be a Perfect number if σ(n) = 2n. That is, a number M is a perfect number if the sum of all its divisors is 2M. The smallest perfect number is 6 as = 12 = 2 6. The following perfect number eye chart is taken from Ron Ruemmler s letter, Mathematics Teacher, November

8 8 Theorem 12. (Euclid) Let n 2 be a positive integer such that 2 n 1 is a prime. Then, 2 n 1 (2 n 1) is a perfect number. Proof. Let M = 2 n 1 (2 n 1). Since 2 n 1 is a prime, σ(2 n 1) = 1 + (2 n 1) = 2 n. As σ is a multiplicative function and gcd(2 n 1,(2 n 1)) = 1, we have σ(m) = σ(2 n 1 ) σ(2 n 1) = ( n 1 )2 n = 2n n = 2M. Euler proved the converse of the above result which is stated next. Theorem 13. Let M be an even perfect number. Then M is of the form 2 n 1 (2 n 1), where n 2 and 2 n 1 is a prime number. Proof. Let M be an even perfect number, say M = 2 t q, with q as an odd positive integer. We claim that q is prime and is of the form 2 t+1 1. Since M is perfect number, 2 t+1 q = 2M = σ(m) = (1 = t )σ(q) = (2 t+1 1)σ(q). (1) Since gcd(2 t+1 1,2 t+1 ) = 1, 2 t+1 divides σ(q) and hence, σ(q) = 2 t+1 s, for some odd positive integer s. Thus, (1) reduces to q = (2 t+1 1)s. Now the only thing left to show is s = 1 and q is prime. Since σ(q) = 2 t+1 s, we observe that q = (2 t+1 1)s = σ(q) s. Or equivalently, σ(q) = q +s. Thus, we see that the factors of q should only be q and s. And, for this to happen, s must equal 1 and 2 t+1 1 must be a prime. The following two results are taken from the article The greatest two-point answer ever by William Dunham and Hank Sneddon, Math Horizons, Theorem 14. Let M be an even perfect number. Then σ(σ(m)) > 4M.

9 9 Proof. Let M be an even perfect number. Then, from Theorem 13, M is of the form 2 n 1 (2 n 1), for some positive integer n such that 2 n 1 is a prime. Since σ(m) = 2M, we have σ(σ(m)) = σ(2m) = σ(2 n (2 n 1)) = σ(2 n ) σ(2 n 1) = ( n )[(2 n 1) + 1] = (2 n+1 1)2 n > (2 n+1 2)2 n = 4M. One also gets the exact value of σ(σ(m)), whenever M is an even perfect number. Theorem 15. Let M be an even perfect number. Then, σ(σ(m)) = 4M M 2. Proof. Theorem 13 implies that M = 2 n 1 (2 n 1) for some positive integer n 2 such that 2 n 1 is a prime. Thus, we see that 2 2n 1 2 n 1 M = 0. Or equivalently, (2 n ) 2 2 n 2M = 0. Thus, 2 n is a positive root of the quadratic equation x 2 x 2M = 0 and hence 2 n = M. 2 We now prove the required equality using n 2 and the above value of 2 n. σ(σ(m)) = σ(2m) = (2 n+1 1)2 n = 4(2 2n 1 2 n 2 ) = 4[2 n 1 (2 n 1) + (2 n 1 2 n 2 )] = 4M + 2 n = 4M M. 2