An Exploration of the Arithmetic Derivative

Transcription

1 An Eploration of the Arithmetic Derivative Alaina Sandhu Final Research Report: Summer 006 Under Supervision of: Dr. McCallum, Ben Levitt, Cameron McLeman 1 Introduction The arithmetic derivative is a recently defined operator on the integers whose properties directly relate to some of the most well known conjectures in number theory. While the definition of this function may have originated well into the past, perhaps the first serious analysis of the arithmetic derivative was included in the Putnam Prize Competition in 1950, and further refined by EJ Barbeau s Remark on an Arithmetic Derivative [Ba61] in Upon first glance, the arithmetic derivative is a simple function defined using the unique prime factorization of integers and the product rule from calculus. This is quite deceiving, however, as the properties and behavior of the derivative are directly related to some of the oldest and most studied conjectures in elementary number theory. The arithmetic derivative operator is defined to be the unique map which sends every prime integer to 1 and that satisfies the Leibnitz rule: For all a, b Z, (ab) = a b+ab, which maintains some familiar properties from calculus such as (n k ) = kn k 1 n. Already from this definition, we can see the link to number theory. For eample we can ask whether or not that for any a N, there eists a solution to the differential equation n = a. A proof of Goldbach s conjecture would imply this statement, as the derivative of the product of two primes (p 1 p ) is their sum (so if a = p 1 + p, then (p 1 p ) = n). As another eample, we can ask whether or not there are infinitely many solutions to the differential equation n = 1. A proof of the Twin Prime Conjecture would imply this, since if p is a lower twin prime, then (p) = p + is the upper twin prime, whose derivative is thus 1 (so (p) = 1). These are but a few of the related conjectures that can be eplored in terms of the arithmetic derivative, and our research has uncovered yet another (see Theorem 1). There are potentially many more relationships to be eplored and redefined in terms of this function, a topic requiring further research. The goal of this paper is to familiarize the reader with the properties of the arithmetic derivative and propose further conjectures with regards to the nature of the function, as well as its implications on previously established 1

2 conjectures. Specifically, we demonstrate conjectures which are contingent upon the eistence, and characterization, of solutions to differential equations, thereby centering much of the research on the behavior and solutions of differential equations. We conclude with partial results and proposals for future work in the hope of encouraging future research of the arithmetic derivative. Definitions and Background The arithmetic derivative of a non-negative integer is defined as follows: 0 = 0. p = 1 for any prime p. (ab) = a b + ab for any a, b N (Leibniz rule). We now provide an eplicit formula, ensuring that the function is well-defined: Theorem 1. ([AU03], Theorem 1) For any natural number n, if n = k is the prime factorization of n, then n = n k i=1 i=1 pei i e i p i. (1) Proof. We can write any such n by n = m i=1 p i, where the p i are now no longer necessarily distinct. We proveed by induction on m. When m = 1, n is prime, and hence n = 1. The induction hypothesis states for any k = m N, if n = m i=1 p i, then n = n m i=1 1 p i. We now consider what happens when we add another prime p m+1 : (np m+1 ) = n p m+1 + n(p m+1 ) = n + np m+1 m m+1 = np m+1 i=1 1 p i=1 i Going back to the original epression, n = m i=1 pei i, our formula becomes n = n m e i i=1 p i, since our sum has a summand of 1 p i for each power of p i dividing n, giving a total of ei Lemma 1. 1 = 0. p i. Proof. 1 = (1 1) = = 1, so 1 = 0. 1 p i

3 .1 Bounds for the Arithmetic Derivative Theorem. [[AU03], Theorem 9] For any positive integer n If n is composite, n n log n. () n n. (3) Furthermore, if n is a product of k factors larger than 1, then n kn k 1 k. (4) We shall not prove this as the proof does not contain information relevant to our paper. It is important, however, to recognize that the first derivative of every number is bounded by an eplicit function of n. This will be used below in describing solutions to differential equations. 3 Differential Equations As mentioned in the introduction, there are many conjectures within number theory that may be epressed in terms of the arithmetic derivative. As the relationships of such conjectures translates into more comple differential equations, we shall begin by solving simple equations. 3.1 Solutions to n = a Theorem 3. The only positive integer n which satisfies n = 0 is n = 1. Proof. This is a direct result of the definition of our function, since every other positive integer has at least one prime factor. Theorem 4. The only solutions to n = 1 in natural numbers are prime numbers. Proof. A composite number can be epressed as the product of prime numbers, of which the derivative of (by the product rule) is the sum of at least two positive integers, which is greater than 1. Now we turn our attention to the eistence of solutions to the equation n = a, where a > 1. First, we observe from Theorem that there can only be a finite number of solutions to n = a because all potential solutions are bounded above by a 4. One direct relation of the differential equation n = a is the Goldbach Conjecture, which states that every even number larger than 3 is the sum of two distinct prime numbers. In terms of our function, we can restate it as follows: Conjecture 1. For any a N, there eists a solution to the equation n = a. 3

4 If the Goldbach Conjecture were proven, it would allows us to represent a = p 1 +p, so if we take the derivative of the product (p 1 p ) = p 1p +p 1 p = p + p 1 = a, and thus the differential equation n = a has a solution. Additionally, there eists solutions for n = a for certain odd numbers a: Theorem 5. If a is prime, then n = a has a solution, namely (a ). Proof. ((a )) = (a ) + (a ) = a + = a. Note, that this is not an if and only if statement; that is, there eists some numbers such that a is not prime but there are still solutions to the differential equation n = a. If we restrict this theorem by requiring that a is an upper twin prime, the twin prime conjecture implies the following conjecture: Conjecture. There are infinitely many solutions to the differential equation n = 1. The reasoning follows directly from the above Theorem 5, since if a is an upper twin prime, its derivative will be Solutions to n = n There also eist unique solutions to the differential equation n = a, where we restrict a to be n, epressing it as n = n. We find solutions to this equation are of the form, p p (where p is prime). To illustrate that these are indeed solutions: (p p ) = p p p 1 p = p p. (5) This unique property of p p is seen in the following theorems: Theorem 6. ([AU03], Theorem 4) If n = p p m for some prime p and integer m > 1, then n = p p (m + m ) and lim k n (k) =. Proof. Assume n = p p m, then n = (p p ) m + p p m = p p (m + m ) > n. Further, proof by induction shows that n (k) n + k. Theorem 7. ([AU03], Theorem 5) Let p k be the highest power of prime p that divides the natural number n. If 0 < k < p, then p k 1 is the highest power of p that divides n. Furthermore, each derivative n, n, n,..., n (k) is distinct. Proof. Let n = p k m. Then n = kp k 1 m+p k m = p k 1 (km+pm ), and because k < p, the inside term is not divisible by p, therefore the entire term is only divisible by p k 1. From this argument, we can see that n can only be divisible by p k, and etending this pattern ensures that each derivative is distinct. Theorem 8. If n = p pk m for some prime p and integers k, m > 1, then n = p pk (km + m ). Proof. (p pk m) = pkp (pk 1) m + p pk m = p pk (km + m ) 4

5 Theorem 9. ([AU03], Theorem 6) For n N, n = n if and only if n = p p, where p is a prime. As an immediate consequence, there is an infinite number of solutions to the equation. Proof. We ve already seen in (5), that if n = p p, n = p p = n. Conversely, assume n = n. Then by Theorem 7, if p n at least p p n or else it would contradict n = n. By Theorem 6, we conclude that this occurs when n = p p. Now that we are familiar with some differential equations and properties of our function, we shall introduce the main topic of our research and eplore the associated differential equations. 4 Germain primes and Cunningham chains A prime number p is called a Sophie Germain prime if p + 1 is also prime. It is conjectured that there are infinitely many primes of this form, the largest to date being A sequence of Sophie Germain primes, {p, p+1, (p+1)+1,...} is called a Cunningham chain of the first kind, where all but the last prime in the sequence are Sophie Germain primes. Similarly, there eists Cunningham chains of the second kind, which consist of primes of the form {p, p 1, (p 1) 1,...}. The longest known Cunningham chains of both first and second kind were found in 005 and are of length 16. Conjecture 3. There eists infinitely many Cunningham chains of length k, for any k N. We now eplore the properties of Sophie Germain primes and Cunningham chains within the derivative. For every prime number, the following is true: ( 4 p) = 4 (p + 1). (6) This is because ( 4 p) = 4 3 p + 4 p = 4 (p + 1). Now, assuming that p is a Sophie Germain prime, we see that the derivative will yield 4 (p + 1), where now p + 1 is prime as well. This makes the Sophie Germain property detectable by differential equations: Theorem 10. For any positive integer m, we have ( 4 m) 4 (4m + 3), with equality if and only if m is a Sophie Germain prime. Proof. Consider m N s.t. m is not prime. Then ( 4 m) = 4 3 m + 4 m = 4 (m + m ) ( 4 (m + m )) = 4 3 (m + m ) + 4 (m + m) = 4 (4m + m ) + 4 (m + m ) = 4 (4m + m + (m + m) ) > 4 (4m + 3). 5

6 Now consider m N where m is prime. Then ( 4 m) = 4 3 m + 4 m = 4 (m + m ) = 4 (m + 1). ( 4 (m + 1)) = 5 (m + 1) + 4 (m + 1) = 4 (4m + + (m + 1) ) 4 (4m + 3), with the last inequality being an equality if and only if m + 1 is also prime, i.e. if and only if m is a Sophie Germain prime. So if we epress n = 4 p, the differential equation n = 4n + 48 is satisfied if and only if p is a Sophie Germain prime. Thus, a reasonable conjecture is the following: Conjecture 4. There are infinitely many solutions to the differential equation n = 4n + 48 where n = 4 p. This conjecture is equivalent to the conjecture that there eists infinitely many Sophie Germain primes. We can now etend this pattern to Cunningham chains. Theorem 11. For any Cunningham chain of length k, ( 4 m) (k) 4 ( k m + k 1) with equality if and only if {m, m + 1, 4m } are prime. Proof. (By induction) For a Cunningham chain of length 1, ( 4 m) = 4 3 m + 4 m = 4 (m + m ) 4 (m + 1) By Theorem 4, equality holds if and only if m is prime because m = 1. Assume that for a k-long Cunningham Chain, we have ( 4 m) (k) = 4 ( k m + k 1). Then ( 4 m) (k+1) = (( 4 ) k ) = ( 4 ( k m + k 1)) = 4 3 ( k m + k 1) + 4 ( k m + k 1) = 4 ( k+1 m + k+1 + 1) 4 ( k+1 m + k+1 1), with equality if and only if k+1 m + k+1 1 (the k-th term in the Cunningham chain) is prime. It is further clear from this proof that this will also hold for intermediate derivatives. Making the substitution n = 4 p, we conclude: 6

7 Theorem 1. If {p, p+1,..., p i 1 + i 1 1} is a Cunningham chain of length i and n = 4 p, then n (k) = k n + 4 ( k 1) for k = 1,,..., i. In particular, there is a bijection between Cunningham chains of length k and solutions n with n = 4 p to this differential equation and These differential equations are of special significance because they allow us to characterize the eistence of certain numbers and sequences in a different language. One particular re-wording is the following: Corollary 1. A Cunningham chain of length 17 eists if and only if there eists a prime number p such that n = 4 p satisfies the following differential equation: n (17) = 17 n Partial Results and Future Work 5.1 Bounds for the k th derivative of all n In this section, we bound the proportion of positive integers within the interval [1, ] whose repeated derivatives tend eventually to zero. Definition 1. We define the height, ht(n), of an integer n, to be the smallest non-negative integer k such that n (k) = 0. If no such k eists, we define ht(n) =. This definition allows us to phrase many interesting questions in terms of an integer s height. For eample, do there eist numbers with arbitrarily large height? We can also divide numbers into classes based on their particular height, and eamine whether they share other properties other than the same height. We turn to address some of those questions now. Define the functions: {0 n : ht(n) } f 0 () = f (k) {0 n : ht(n) = k} 0 () = They are linked as follows: f 0 () = f 0 0() + f 1 0() + + f k 0() +, (7) where we observe that for any, only finitely many of these terms are non-zero, so convergence is not an issue. We shall now eplore this function for some values of k : We know the only number whose height is zero is zero itself, thus: f (0) 0 () = 1 (8) 7

8 By Theorem 3, 1 is the only number whose first derivative is zero: f (1) 0 () = 1 (9) By Theorem 4, prime numbers, and only prime numbers, solve the differential equation n = 1, and thus solve n = 0, so f () π() 0 () =. (10) We have seen that p, where p is a lower twin prime, solves the differential equation n = 1, so they solve n = 0 as well. We have an asymptotic epression for the number of twin primes, however because we are counting the numbers p, we must only consider the number of twin primes : f (3) 0 () > π ( ) Π (π( )) = 4Π (π( )) 4Π ( ln ) = 4Π (ln ). As seen above, the epression p for lower twin primes will solve n = 0. We can integrate this epression to find that p, for lower twin primes, will solve n = 0. This is because (p ) = p, and then we follow the same pattern above. However, this time because we are counting the numbers of the form p, we can only count the number of twin primes, to ensure we are not counting numbers eceeding : f (4) 0 () > π ( ) Π (π( )) = Π (π( )) Π 3 (ln( )) 3 Π = (ln()) 8Π =. (ln()) To create a lower bound for f 0 () as, we can sum the preceding 8

9 epressions: f 0 () > π() + 4Π (π( )) + π ( ) + 1 ln + 4Π (ln ) + 8Π (ln ) = (ln ) + ln + 4Π + 8Π (ln ). We now repeat this process, bounding f 0 () below by constructing classes of integers with infinite height, and then estimate their density: Define f () = { n : ht(n) = } = 1 f 0 (). (11) The first such class of integers is the set of multiples of p p, since all of these 1 have infinite height by Theorems 6 and 9. Since in general, p of the positive p integers are divisible by p p, we can sum this quantity over all p (using inclusion eclusion to prevent overcounting) to conclude: f () > S P ( 1) S +1 p S (p p ) 1 = , for sufficiently large, where S ranges over all finite subsets of primes. We can further include all numbers of the form q 1 q less than or equal to where q 1, q are primes that satisfy q 1 + q 0 mod p p, again for any prime p. This is because in this case: (q 1 q ) = q 1 + q = p p m, so ht(q 1 q ) =, again by Theorems 6 and 9. Before we write the general formula, we will compute the term corresponding to p =. We are looking to find pairs q 1, q that satisfy q 1 + q 0 mod 4. One way to do this is by choosing a prime q 1 1 mod 4 and a prime q 3 mod 4, where q 1, q (so that q 1 q ). By Dirichlet s Theorem on arithmetic progressions, for sufficiently large, the primes will be evenly divided between those congruent to 1 mod 4 and 3 mod 4. Thus, asymptotically, the number of primes congruent to 1 mod 4 and less than (), is π( ), and the same follows for primes congruent to 3 mod 4. We conclude that the number of pairs q 1, q with the above properties, is asymptotically ( π( ) ). For the more general formula, Dirichlet s theorem says the primes are distributed evenly among the φ(p p ) different residue classes modulo p p. For our eample, we see that half the primes are 3 mod 4 and half 9

10 are 1 mod 4. This general formula then becomes φ(pp ) ( π( ) φ(p p ) ) = π( ) φ(p p ). Summing over all p, and again using inclusion-eclusion, we conclude (along with the above result) that f () > ( 1) S +1 1 p p + π( ) S P p S S P( 1) S +1 1 φ(p p ) p S π( ), for sufficiently large. Better estimates for both bounds can be obtained through further research into the patterns and behavior of the arithmetic derivative. 5. Derivative of a primorial Another interesting result is the implication for derivatives of primorial numbers. The n-th primorial is defined to be the product of all primes up through n. Definition. Let n = k i=1 p i for every prime number, so n represents the k th primorial. Then n = k i=1 p k i i=1 1 p i. One interesting relation to this definition is its relation to the infinitude of primes: Consider, for eample, the k-th primorial n. Because n contains each consecutive prime up through p n raised only to the first power, by Theorem 7, none of the preceding primes can divide n. We shall prove this as follows: Proof. Suppose {p 1, p,..., p k } are the only primes, and we represent n = k i=1 p i. When taking the derivative n, Theorem 7 restricts all the primes {p 1, p,..., p k } from dividing the derivative, because they are all raised to the first power. So n has at least one prime factor not included in {p 1, p,..., p k }, leading us to a contradiction. Therefore, we conclude that there are infinite primes. It would be interesting to eplore this property to find implications it may have on the behavior of very large primes. Throughout our paper we have show, and conjectured, that there eists many other primes, and sequences of primes, whose properties can be translated to the language of differential equations and arithmetic derivatives. We have already encountered one prime sequence, namely Cunningham chains, which could be defined and eplored within our function. Also, we believe there may be other properties in our derivative that lie in the pattern of prime arithmetic progressions. 10

11 6 Appendi The following are codes for various functions with the arithmetic derivative. The comment line above each code describes its function. This code will compute the arithmetic derivative of any integer or rational number: {f(n)=sign(n)*abs(n)*sum(i=1, matsize(factor(abs(n)))[1],} {factor(abs(n))[i,]/factor(abs(n))[i,1])} This code will compute the solutions, a for a = n: {I(n) = for(i=1, n^/4+1, if((f(i))==n, print(i)));} This code computes the number of solutions, a, to solve a = n: { i(n)= count = 0; k=n; for(i=1, k^/4+1, if((f(i))== k, count = count + 1)); print(count);} 11

12 7 Table of n (k) for n 100 and k 10 n n (1) n () n (3) n (4) n (5) n (6) n (7) n (8) n (9) n (10)

13 n n (1) n () n (3) n (4) n (5) n (6) n (7) n (8) n (9) n (10)

14 n n (1) n () n (3) n (4) n (5) n (6) n (7) n (8) n (9) n (10) References [AU03] Ahlander, Bo and Ufnarovski, Victor. How to differentiate a Number. Journal of Integer Sequences. Vol. 6 (003). [Ba61] Barbeau, E.J. Remark on an arithmetic derivative. Canadian Mathematical Bulletin. Vol. 4 (1961): [Bu97] Buium, A. Arithmetic analogues of derivations. Journal of Algebra. Vol. 198 (1997) : [HW80] Hardy, G.H. and Wright, E.M. An Introduction to the Theory of Numbers 5th ed., Oford University Press, [Ni91] Niven, Ivan et. al. An Introduction to the Theory of Numbers 5th ed., Wiley Tetbooks,