Some multiplicative arithmetical functions

Transcription

1 Some multiplicative arithmetical functions an invitation to number theory K. N. Raghavan knr/ IMSc, Chennai August 203

2 Standard form of prime factorization of a number; GCD and LCM STANDARD FORM AS A PRODUCT OF PRIMES: GCD AND LCM As we all know well, every positive integer is uniquely a product of primes. Given a positive integer n, we can write: n = p r q s or, alternatively, n = p r p r k k Here we assume tacitly that p, q,... are distinct primes and that r, s,... are positive (sometimes only non-negative) integers; in the latter case, that p, p 2,... are distinct primes and that r, r 2,... are positive (sometimes only non-negative) integers. Such an expression for n is said to be in standard form. If m = p r q s and n = p r q s are in standard form where the exponents are assumed to be non-negative then the GCD or HCF of m and n, denoted (m, n), and the LCM of m and n are given by (m, n) = HCF of m and n = p min(r,r ) q min(s,s ) LCM of m and n = p max(r,r ) q max(s,s ) Since {r, r } = {min(r, r ), max(r, r )}, {s, s } = {min(s, s ), max(s, s )},..., it follows that m n = their HCF their LCM.

3 The functions τ and σ THE FUNCTIONS τ(n) AND σ(n) Let n = p r p r 2 2 be in standard form. Then the number of divisors τ(n) of n is (r + )(r 2 + ), for the standard form of any divisor is p r p r 2 2 with 0 r r, 0 r 2 r 2,..., and there is a one-to-one correspondence between divisors and the choices (r, r 2,...). the sum of the divisors σ(n) is ( + p + + p r )( + p p r 2 2 ) = p r + p pr2+ 2 p 2 To justify the above formula, recall how to sum a geometric series. Writing S = + p + p p r, we have: p S = p + p p r + p r + p r+ S = + p + p p r + p r so that ps S = + p r+ and S = (p r+ )/(p ).

4 Definition of multiplicative arithmetical functions MULTIPLICATIVE ARITHMETICAL FUNCTIONS We ve just seen that if n = p r p r 2 2 is in standard form, then τ(n) = (r + )(r 2 + ) σ(n) = pr + p pr2+ 2 p 2 A function (say, taking real values) on positive integers is called an arithmetical function; an arithmetical function f is called multiplicative if f (mn) = f (m)f (n) whenever m and n are coprime (i.e., relatively prime). The functions τ and σ just defined are multiplicative arithmetical functions. Indeed, if m = p r p r 2 2 and n = q s q s 2 2 are the standard forms for coprime integers m and n, then p r p r 2 2 q s q s 2 2 is the standard form of mn.

5 perfect numbers defined; even perfect numbers PERFECT NUMBERS A positive integer n is called perfect if the sum σ(n) of its divisors equals 2n. For example, 6 ( = 2) and 28 ( = 56). Suppose n is an even perfect number. Write n = 2 k m with m odd and k. We have σ(n) = σ(2 k )σ(m) = (2 k+ )σ(m) Since σ(n) = 2n, we get: (2 k+ )σ(m) = 2 k+ m. Thus 2 k+ divides m. Put l := m/(2 k+ ), so that m = (2 k+ )l and σ(m) = 2 k+ l. If l >, then m has at least 3 distinct divisors,, l, and m; σ(m) + m + l = + (2 k+ )l + l = + 2 k+ l > 2 k+ l = σ(m) But this is a contradiction. So l =, which means that m = 2 k+ and σ(m) = 2 k+ ; since σ(m) = m +, we conclude that m = 2 k+ is a prime. Thus the even perfect numbers are precisely those of the form n = 2 k (2 k+ ) with 2 k+ a prime so

6 Even perfect numbers EVEN PERFECT NUMBERS We ve just seen that the even perfect numbers are precisely those of type n = 2 p (2 p ) with p an integer such that 2 p a prime. The observation is due to the great mathematician Leonhard Euler ( ). For 2 p to be a prime, it is necessary that p be prime: for, if p = rs, then 2 p = 2 rs = (2 r )(2 r(s ) + 2 r(s 2) r + ) So to generate even perfect numbers, we must take p to be prime and check if 2 p is prime. For the values 2, 3, 5, 7,, 3, 7, 9, 23, 29 of p, the respective values of 2 p are: 3, 7, 3, 27, 2047 = 23 89, 89, 307, , , Thus the first few even perfect numbers are: 2 3 = 6, 4 7 = 28, = 496, = 828, =

7 Mersenne primes and perfect numbers MERSENNE PRIMES AND PERFECT NUMBERS A prime number of the form 2 p is called a Mersenne prime after a certain Frenchman Marin Mersenne who lived in the 7th Century. As we just saw, in order that 2 p be prime, it is necessary that p be prime. But the primality of p is by no means sufficient: 2 = 2047 = 23 89, for example. We know precious little about Mersenne primes: It is not known if there are infinitely many Mersenne primes. The total number of known Mersenne primes is 48 (the latest being discovered in Janurary 203! It has 7,425,70 digits!). It is not even known whether 2 p is composite for infinitely many primes p. As to perfect numbers, our knowledge of them is equally thin: Since the known number of Mersenne primes is 48, it follows that the known number of even perfect numbers is also 48. We do not know even a single odd perfect number. Nor do we know that they do not exist.

8 Primality testing PRIMALITY TESTING: THE AKS TEST As you can see, it would help, in the search for Mersenne primes, to be able to tell quickly whether a given large number (with thousands of digits or even larger) is prime. This question of whether there is a quick test for primality is a fundamental one and was open for a long time (although efficient algorithms were known to check primality of numbers of the form 2 p with p prime). Screenshot of Wikipedia page:

9 Some open problems SOME OPEN PROBLEMS A similar test for whether or not a number is square-free (that is, whether the square of any prime divides it) is not known. For this and many other simply stated open problems in mathematics, one could see the video of a lecture by Joseph Oesterlé on matsciencechannel YouTube: Some simple open problems in Mathematics. Whether there are infinitely many prime pairs that are 2 apart, like: 3, 5 5, 7, 3 7, 9 29, 3 is a very famous problem called the twin prime problem which has been open for several hundred years and towards which there has been spectacular recent progress (by Yitang Zhang, April 203). from the NY times of May 20, 203

10 The functions µ and φ THE MÖBIUS FUNCTION µ AND THE EULER TOTIENT FUNCTION φ The Möbius function µ(n) of a positive integer n = p r p r k k in standard form with all exponents r,..., r k assumed to be positive is defined to be 0 if at least one of the exponents r,... r k is 2 ( ) k otherwise (that is, if either r =... = r k =, or if k = 0, i.e., n = ) It is easily seen to be multiplicative. Its usefulness will soon become clear. The Euler totient function φ(n) of a positive integer n is the number of integers among,..., n that are coprime to n. We take φ() = (by definition, if you wish). We will presently show that for n = p r p r k k in standard form with r,..., r k all > 0, In other words, φ(n) = n ( p ) ( p k ) φ(n) = n p pk = (p )p r (p k )p r k k p p k It follows immediately from the formula above for φ that it is multiplicative.

11 Proof of the formula for φ PROOF BY INCLUSION-EXCLUSION OF THE FORMULA FOR φ If d n, then there are n/d integers from among,..., n that divisible by d: namely, d, 2d, 3d,..., (n/d)d = n. Suppose that n = p k is a prime power. To compute φ(n) we need only delete all multiples of p from,..., n, and count how many remain. Since the number of these multiples is n/p = p k, we get φ(p k ) = p k p k = p k (p ) = p k ( p ) Now suppose n = p r p r 2 2 has two prime divisors p and p 2. To compute φ(n), we must now delete all multiples of p and also all multiples of p 2 from,..., n, and count how many remain. The number of multiples of p here is n/p ; the number of multiples of p 2 is n/p 2. Are there numbers that are multiples of both p and p 2? Yes, of course. They are precisely those that are multiples of p p 2, and their number is n/p p 2. We therefore get φ(n) = φ(p r p r 2 2 ) = n n p n p 2 + n p p 2 = n( p p 2 + p p 2 ) = n( p )( p 2 )

12 The inclusion-exclusion principle formulated PROOF BY INCLUSION-EXCLUSION (CONTINUED) Now suppose n = 30 = The red, blue, and green circles in the picture consist respectively of the multiples of 2, 3, and 5. Those integers from,..., 30 that are relatively prime to 30 lie outside all three circles. Thus φ(30) = 8. The priniciple of inclusion exclusion : let S = S 0 be a set, and S,..., S k be subsets. Then the cardinality of the complement of the union of the subsets is: k (S \ (S S k )) = ( ) j S i... S ij j=0 In case k = 3, for example, we get: i <...<i j k S\(S S 2 S 3) = S ( S + S 2 + S 3 ) + ( S S 2 + S S 3 + S 2 S 3 ) S S 2 S 3

13 Proof of the inclusion-exclusion principle PROOF OF THE INCLUSION-EXCLUSION PRINCIPLE J. J. Sylvester (84 897) The justification for the principle of inclusion exclusion is this. Consider the contribution of a given element, say s, of the set S to the expression on the right hand side. If s does not belong to any of the subsets S,..., S k, then it contributes only to S and to nothing else, so its total contribution is. Now suppose that s belongs to some of the subsets, say m of them, namely S l,..., S lm. Then s contributes to 2 m terms on the right: those with j m and i,..., i j chosen from l,..., l m. Since there are ( ) m j of such choices, the total contribution is k j=0 ( )j( ) m j = ( ) m = 0. This finishes the justification.

14 Proof of the formula of φ by inclusion-exclusion PROOF BY INCLUSION-EXCLUSION (CONTINUED) We now give the formal proof of the following formula for φ(n), when n = p r p r k k is in standard form with r,..., r k all positive: φ(n) = φ(p r p r k k ) = n ( ) ( ) = (p )p r (p k )p r k k p P k Apply the inclusion-exclusion priniciple with S = {,..., n} and subsets S,..., S k, where S j consists of all those elements of S that are multiples of p j. If an element of S has a common factor with n, then at least one of the primes p,..., p k divide it. Thus φ(n) counts the cardinality of the complement in S of the union S S k. The intersection S i S ij consists of simultaneous multiples of p i,..., p ij ; since the p i are mutually relatively prime, this intersection consists of precisely of the multiples of p i p ij. The cardinality of S i S ij is thus n/p i p ij. Plugging this into the formula inclusion-exclusion formula, we get φ(n) = k ( ) j n k = n ( ( ) j ) = n ( ) ( ) p i p ij p i p ij p p j j=0 j=0

15 New multiplicative functions from old OTHER MULTIPLICATIVE FUNCTIONS We have so far defined four multiplicative functions: τ, σ, µ, and φ. These are by no means all. In fact, there are infinitely many of them: for example, n n α for any real number α is mutliplicative; moreover, the value of a multiplicative function on prime powers determines it, but these values can be fixed arbitrarily without any restriction. Given an arithmetical function f, we may define another one f as follows: f (n) = d n f (d). If f is multiplicative, then so is f. Indeed, if m and n are coprime, and D, E, and F the sets of divisors respectively of m, n, and mn, then D E F given by (d, e) de is a bijection. Thus f (mn) = f (d) = f (d d 2) = ( f (d )) ( f (d 2)) = f (m) f (n) d mn d m d 2 n d m d2 n If c denotes the constant function with value (that maps every positive integer to ), then c (n) = d n c (d) = d n = τ(n). If ι is the identity function (that maps every positive integer to itself), then ι(n) = d n ι(d) = d n d = σ(n).

16 µ and φ TWO MORE COMPUTATIONS We claim that µ =, where (n) is zero for all n except () =. Proof: Since µ is multiplicative, so is µ. We clearly have µ() = µ() = (from the definition of µ). Thus we need only show that µ(p k ) = 0 for every prime p with k positive. Again from the definition of µ, we have µ(p k ) = 0 j k µ(pj ). But µ(p j ) 0 only when j = 0 and j =. In the former case it is and in the latter it is. Thus µ(p k ) = 0 when k is positive. Let us compute φ. Since φ is multiplicative (by our formula), so is φ. It is thus enough to compute φ on a prime power p k. As is easily seen, φ(p j ) = p j p j for j > 0. So φ(p k ) = φ() + φ(p) + φ(p 2 ) + + φ(p k ) + φ(p k ) = + (p ) + (p 2 p) + + (p k p k ) = p k. This proves that φ = ι, the identity function. Summary of our computations: c = τ, ι = σ, µ =, φ = ι.

17 Möbius inversion formula THE MÖBIUS INVERSION FORMULA The Möbius inversion formula gives an answer to the following natural question: suppose that f is an arithmetical function; given f, can you determine f? If so, how to do it? Answer: f (n) = d n f (d)µ(n/d) = µ(d) f (n/d) = µ(d ) f (d 2) d n d d 2 =n For the proof, expand the right hand side by substituting for f : µ(d) f (n/d) = µ(d)( f (e)) = f (e)( µ(d)) = d n d n e (n/d) e n d (n/e) e n f (e) µ(n/e) But, as we just saw, µ =, which means that µ(n/e) is non-zero only for e = n in which case it is. Thus the last summation in the above display reduces to f (n), and the Möbius inversion is proved.

18 Convolution product MÖBIUS INVERSION REORGANIZED: CONVOLUTION PRODUCT To better organize the idea of Möbius inversion, we introduce the following convolution product on arithmetical functions: f g (n) = f (d)g(n/d) = f (d )g(d 2) d n d d 2 =n f = f c (where c is the function that takes value everywhere). Indeed, f c (n) = d n f (d)c (n/d) = d n f (d) = f (n). The convolution product is commutative and associative. It admits an identity, namely the function that takes value at and 0 elsewhere: f = f = f. (Routine exercise) If f and g are multiplicative, so is f g. On the one hand, µ c = c µ = µ from the first item above; on the other, µ =, as seen on an earlier slide. Thus c and µ are inverses of each other with respect to convolution. The Möbius inversion formula can now be written as: f = f µ, and its proof as: f µ = (f c ) µ = f (c µ) = f µ = f = f