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An Exploration of the Arithmetic Derivative

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An Exploration of the Arithmetic Derivative An Exploration of the Arithmetic Derivative Alaina Sandhu Final Research Report: Summer 2006 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 1961. 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)0 = a0b+ab0, which maintains some familiar properties from calculus such as (nk)0 = knk−1n0. Already from this definition, we can see the link to number theory. For example we can ask whether or not that for any a ∈ N, there exists a solution to the “differential equation” n0 = 2a. A proof of Goldbach’s conjecture would imply this statement, as the derivative of the product of two 0 0 primes (p1p2) is their sum (so if 2a = p1 + p2, then (p1p2) = n). As another example, we can ask whether or not there are infinitely many solutions to the differential equation n00 = 1. A proof of the Twin Prime Conjecture would imply this, since if p is a lower twin prime, then (2p)0 = p + 2 is the upper twin prime, whose derivative is thus 1 (so (2p)00 = 1). These are but a few of the related conjectures that can be explored in terms of the arithmetic derivative, and our research has uncovered yet another (see Theorem 12). There are potentially many more relationships to be explored 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 conjectures. Specifically, we demonstrate conjectures which are contingent upon the existence, 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. 2 Definitions and Background The arithmetic derivative of a non-negative integer is defined as follows: • 00 = 0. • p0 = 1 for any prime p. • (ab)0 = a0b + ab0 for any a, b ∈ N (Leibniz rule). We now provide an explicit formula, ensuring that the function is well-defined: Qk ei Theorem 1. ([AU03], Theorem 1) For any natural number n, if n = i=1 pi is the prime factorization of n, then k X ei n0 = n . (1) p i=1 i Qm Proof. We can write any such n by n = i=1 pi, where the pi are now no longer necessarily distinct. We proveed by induction on m. When m = 1, n is prime, and hence n0 = 1. The induction hypothesis states for any k = m ∈ N, if Qm 0 Pm 1 n = pi, then n = n . We now consider what happens when we i=1 i=1 pi add another prime pm+1: 0 0 0 (npm+1) = n pm+1 + n(pm+1) m X 1 = n + np m+1 p i=1 i m+1 X 1 = np m+1 p i=1 i Qm ei Going back to the original expression, n = i=1 pi , our formula becomes 0 Pm ei 1 n = n , since our sum has a summand of for each power of pi dividing i=1 pi pi n, giving a total of ei . pi Lemma 1. 10 = 0. Proof. 10 = (1 · 1)0 = 10 · 1 + 1 · 10 = 2 · 10, so 10 = 0. 2 2.1 Bounds for the Arithmetic Derivative Theorem 2. [[AU03], Theorem 9] For any positive integer n n log n n0 ≤ 2 . (2) 2 If n is composite, √ n0 ≥ 2 n. (3) Furthermore, if n is a product of k factors larger than 1, then 0 k−1 n ≥ kn 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 explicit 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 expressed in terms of the arithmetic derivative. As the rela- tionships of such conjectures translates into more complex differential equations, we shall begin by solving simple equations. 3.1 Solutions to n0 = a Theorem 3. The only positive integer n which satisfies n0 = 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 n0 = 1 in natural numbers are prime num- bers. Proof. A composite number can be expressed 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 existence of solutions to the equation n0 = a, where a > 1. First, we observe from Theorem 2 that there can only be a finite number of solutions to n0 = a because all potential solutions are bounded a2 above by 4 . One direct relation of the differential equation n0 = a is the Goldbach Con- jecture, 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 exists a solution to the equation n0 = 2a. 3 If the Goldbach Conjecture were proven, it would allows us to represent 0 0 0 2a = p1 +p2, so if we take the derivative of the product (p1 ·p2) = p1p2 +p1p2 = 0 p2 + p1 = 2a, and thus the differential equation n = 2a has a solution. Additionally, there exists solutions for n0 = a for certain odd numbers a: Theorem 5. If a − 2 is prime, then n0 = a has a solution, namely 2(a − 2). Proof. (2(a − 2))0 = 20(a − 2) + 2(a − 2)0 = a − 2 + 2 = a. Note, that this is not an if and only if statement; that is, there exists some numbers such that a−2 is not prime but there are still solutions to the differential equation n0 = a. If we restrict this theorem by requiring that a is an upper twin prime, the twin prime conjecture implies the following conjecture: Conjecture 2. There are infinitely many solutions to the differential equation n00 = 1. The reasoning follows directly from the above Theorem 5, since if a is an upper twin prime, its derivative will be 1. 3.2 Solutions to n0 = n There also exist unique solutions to the differential equation n0 = a, where we restrict a to be n, expressing it as n0 = n. We find solutions to this equation are of the form, pp (where p is prime). To illustrate that these are indeed solutions: (pp)0 = p · pp−1 · p0 = pp. (5) This unique property of pp is seen in the following theorems: Theorem 6. ([AU03], Theorem 4) If n = pp · m for some prime p and integer 0 p 0 (k) m > 1, then n = p (m + m ) and limk→∞ n = ∞. Proof. Assume n = pp · m, then n0 = (pp)0 · m + pp · m0 = pp(m + m0) > n. Further, proof by induction shows that n(k) ≥ n + k. Theorem 7. ([AU03], Theorem 5) Let pk be the highest power of prime p that divides the natural number n. If 0 < k < p, then pk−1 is the highest power of p that divides n0. Furthermore, each derivative n, n0, n00, . , n(k) is distinct. Proof. Let n = pkm. Then n0 = kpk−1m+pkm0 = pk−1(km+pm0), and because k < p, the inside term is not divisible by p, therefore the entire term is only divisible by pk−1. From this argument, we can see that n00 can only be divisible by pk−2, and extending this pattern ensures that each derivative is distinct. Theorem 8. If n = ppk · m for some prime p and integers k, m > 1, then n0 = ppk(km + m0). Proof. (ppk · m)0 = pkp(pk−1) · m + ppk · m0 = ppk(km + m0) 4 Theorem 9. ([AU03], Theorem 6) For n ∈ N, n0 = n if and only if n = pp, 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 = pp, n0 = pp = n. Conversely, assume n0 = n. Then by Theorem 7, if p | n at least pp | n or else it would contradict n0 = n. By Theorem 6, we conclude that this occurs when n = pp. Now that we are familiar with some differential equations and properties of our function, we shall introduce the main topic of our research and explore the associated differential equations.
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