Vol. 81, No. 4, October 2008 MATHEMATICS MAGAZINE 141=3.47 39693=3.101·131 14273478 =2.32.232.1499 = 11 . 67.107.181 142=2.71 39694=2.89 . 223 14273479 143=11 . 13 39695=5.17 . 467 14273480 =23.5.13. 27449 = 3.29.359.457 144=24.32 39696 =24.3.827 14273481 145=5·29 39697=7.53.107 14273482=2.113.137.461 146=2.73 39698=2.23.863 14273483=7.53.79.487 147=3.72 39699 =32.11.401 14273484 =22.3·19. 62603 148=22·37 39700 =22 .52.397 14273485 =5.31.71. 1297 Consecutive Integers with Equally Many Principle Divisors • The Shot Made Round the Table • Probabilistic Reasoning Is Not Logical • Wazir Circuits on an Obstructed Chessboard An Official Publication of The MATHEMATICAL ASSOCIATION OF AMERICA EDITORIAL POLICY M<�thematics Magazine aims to provide lively and appealing mathematical exposition. The Magazine is not a research journal, so the terse style appropriate for such a journal (lemma-theorem-proof-corollary) is not appropriate for the Magazine. 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He did his undergraduate work at Pfeiffer College and his graduate work at the University of North Carolina at Chapel Hill. His thesis advisor was Patrick Eberlein. lrl's mathematical interests include differential geometry, recreational mathe­ matics, and the history of mathematics. Since the inception of the North Carolina State Mathematics Contest thirty years ago, he and his colleague Ben Klein have written nearly half of the contests. Ben Klein is Dolan Professor of Mathematics at Davidson College. He began teaching at Davidson in 1971 and will retire at the end of the 2007-08 academic year. He did his undergraduate work at the University of Rochester and his graduate work at Yale University. His thesis advisor was G. A. Hed­ consecutive integers with two principle divisors, with three principle divisors, and with four principle divisors. How long a run can we have with equally many principle divisors? Our lead article examines this and related questions. lund. Ben's graduate work and his early interests were in analysis, but, over the years, his interests AUTHORS have expanded to include probability and combinatorics. He is active in the Advanced Placement Roger Eggleton is Calculus Program and attended his 18th consecutive Calculus Reading in June 2008. Professor of Mathematics at Illinois State University, where he convenes the Discre!P Mathematics group, is advisor for the Math Club, and coaches the Putnam team. He received his PhD at University of Calgary (1973), with Richard Guy as advisor. He has published over sixty research papers, mainly in number theory, graph theory, and combinatorics. Among them are fourteen joint papers with Jim MacDougall. His wide-ranging other interests include active involvements with languages and philately. 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Postmaster: Send address changes to Membership/ Subscriptions Department, Mathematical Association of America, 1529 Eighteenth Street, N.W., Washington, D.C. 20036-1385. Printed in the United States of America ARTICLES Consecutive Integers with Equally Many Principal Divisors R O G E R B. E GGL E T O N I l l i no i s State U n iversity No rmal, IL 6 1 790, U . S. A . J AM E S A. M A C D O U GA LL The U n iversity of N ewcast l e N ewcast l e, N . S. W. 2 3 08, Austra l i a 1. I ntrod u ction Classifying the positive integers as primes, composites, and the unit, is so familiar that it seems inevitable. However, other classifications can bring interesting relationships to our attention. In that spirit, let us classify positive integers by the number of principal divisors they possess, where we define a principal divisor of a positive integer n to be any prime-power divisor pa 1 n which is maximal (so p is prime, a is a positive integer, and pa+ I is not a divisor of n). The standard notation pa II n can be read as "pa is a principal divisor of n." The Fundamental Theorem of Arithmetic is usually stated in a form emphasizing how primes enter the structure of the positive integers, such as: Every positive integer is the product of a unique finite multiset ofprimes. (Recall that a multiset is a collection of elements in which multiple occurrences are permitted.) Alternatively, the Fundamental Theorem of Arithmetic can be stated in a form that focuses on how maximal primepowers enter the structure of the positive integers, such as: Every positive integer is the product of a unique finite set of powers of distinct primes. Consequently every positive integer is the product of its principal divisors, and every finite set of powers of distinct primes is the set of principal divisors of a unique positive integer. Of course, the number of principal divisors of n is equal to the number of distinct prime factors of n, but here the principal divisors are the simple structural components of n, whereas the distinct prime factors are but a shadow of that structure. Readers who find the present paper of interest might find similar interest in [6], where upper bounds on the sum of principal divisors of n are established by elementary means. For each integer n :::: 0, let Pn be the set of all positive integers with exactly n principal divisors, so Po = {1}, and PI = {2, 3, 4, 5, 7, 8, 9, 1 1 , 13, 16, 17, 19, 23, 25, 27, 29, 3 1 , 32, 37, . . . }, p2 = {6, 10, 12, 14, 15, 18, 20, 2 1 , 22, 24, 26, 28, 33, 34, 35, 36, 38, . . . }, p3 = {30, 42, 60, 66, 70, 78, 84, 90, 102, 105, 1 10, . . . }, P4 = {210, 330, 390, 420, 462, 5 10, . . . }, etc. 235 236 MATH EMATICS MAGAZI N E In particular, P1 comprises the prime-powers, or principal integers; P2 comprises the products of two coprime principal integers, or rank 2 integers; and so on. Collectively, we call {Pn : n 2: 0} the rank sets of positive integers. Clearly the rank sets are a partition of the positive integers, by the Fundamental Theorem of Arithmetic. Thus it is interesting to look at the occurrence of runs of consecutive integers within each rank set: this is one of the relationships immediately brought into focus by the classification. For compactness, let us write afr 1 to denote the run of r consecutive integers beginning with a, where a and r are positive integers, so a [r 1 = {a + i : 0 :S i < r}. We call r the size of the run. In particular, the run afr 1 is nontrivial if r 2: 2, and a lrl C P11 is a maximal run in Pn if it is nontrivial and Pn contains neither a - 1 nor It is easy to see that the size of runs in Pn is bounded. For if M is the product of the first n + 1 primes, then M E Pn l and any run of M consecutive integers contains a multiple of M, so any run in Pn has+ size less than M. Thus, for each integer n 2: 0 there is a positive integer r (n) which is the maximum size attained by runs in Pn. Trivially, r(O) = 1. We have already seen that r(1) 2: 4 and r(2) 2: 5 , and we shall soon see that in fact r(l) = 4. Our main objective is to study r(2), which we shall determine "within 1". Later we shall also discuss r(n) for n 2: 3. 2. Maximal runs in P1 and P2 Returning to the principal integers, it is clear that any nontrivial run in P1 contains an even integer so, being principal, any such integer must be a power of 2. Since 2 and 4 are the only powers of 2 that differ by 2 , any maximal run of principal integers greater than 5 must contain exactly one even integer, so 2f41 is the unique longest run in P1 , and r(l) = 4. The long-standing conjecture credited to Catalan, that 8 and 9 are the only two consecutive integers which are nontrivial prime powers, was recently proved by Mihailescu [11]. From this it follows that any maximal run of principal integers greater than 9 must contain a power of 2 and any adjacent number in the run must be a prime. It is well known that 217 - 1 can only be prime when n itself is prime, and 2" + 1 can only be prime when n is a power of 2: when n 2: 3 these two conditions are mutually exclusive, so any maximal run of principal integers greater than 9 has just two members. Primes of the form 211 + 1 are Fermat primes. The only known Fermat primes are 3 , 5 , 17, 257 and 65537 , but no proof is known that there are no others. Primes of the form 211 - 1 are Mersenne primes. Currently 44 Mersenne primes are known [7]. A distributed computing project known as GIMPS (the Great International Mersenne Prime Search) has made numerous additions to this list in recent years, but no proof is known that infinitely many such primes exist. Consequently, although we know 2 141 and 7 l31 are the only maximal runs of more than two principal integers, it is not known whether P1 contains infinitely many runs of size 2. What happens with maximal runs of rank 2 numbers? This is less familiar territory, so one does not know quite what to expect. We shall prove: THEOREM 1. There is no run of 10 consecutive integers in P2 . V O L . 8 1 , N O . 4, OCTO B E R 2 008 237 Hence r (2) :S 9. Is this a "sharp" result? It turns out to be "within 1" of the exact value of r(2) . Our methods appear unable to decide on the existence of runs of size 9 in P2 , but our results strongly suggest the following: CONJECTURE 1 . In P2 there is no run of size 9 and the only maximal runs of size greater than 6 are 141[8l, 212[8l, 323[7l and 23Q2[7l. It may also be true that 91 [6l is the only maximal run of size 6 in P2 , but we have less information about runs of size 6 than about longer runs. 3. St0rm er's theorem s For any set of primes P let S(P) be the set of positive integers with all their prime factors in P, that is, S (P) is the multiplicative semigroup of positive integers generated by P. Some results about the integers in S ( P) tum out to be among the main tools we require to prove Theorem 1. If D is a non-square integer, the Pell equation x 2 - Dy 2 = c always has positive integer solutions when c = 1 (see [12], for example). However, for other values of c it need not have positive integer solutions; for instance, this is the case when c = -1 if D = 3 (mod 4). But if c E {I , -1} and x 2 - Dy 2 = c does have positive integer solutions, then all such solutions are generated by the smallest positive solution (x , y0 ), usually called the fundamental solution or minimal solution [12, 16]. In 1 8970 Carl St0rmer published the following theorem [13] which identifies a remarkable property of the fundamental solution. ST0RMER' S PELL EQUATION THEOREM. Let D be a non-square positive inte­ ger, let P be the set of prime divisors of D, and let c E {1, -1}. If the Pell equation x 2 - Dy 2 = c has positive integer solutions and (x0 , y0 ) is the fundamental solution, then y0 is the only solution for y that is a possible member of S(P). For example, if D = 2 then S(P) comprises all the powers of 2. Thus the fundamental solution (x, y) = (3, 2) is the only solution to x 2 - 2y 2 = 1 in positive integers with y a power of 2; similarly (1, 1) is the only solution to x 2 - 2y 2 = -1 in positive integers with y a power of 2. St0rmer used his theorem to prove the following result [14], first published in 1 898. ST0RMER'S NEIGHBORING PAIRS THEOREM. For given positive integers A, B, m, n, a 1 , • • • , am, b 1 , • • • , bn , there are at most finitely many sequences of positive integers X1, • • • , Xm, Yi, ..., Yn such that All solutions follow from the fundamental solutions to a finite set of Pell equations determined by A , B, m, n, a,, . . . , am, b 1 , . . . , bn . In principle it is routine to set up the Pell equations referred to in the Neighboring Pairs Theorem, and to determine the fundamental solutions of those that do have solutions, so St0rmer effectively provided an algorithm for solving the Diophantine inequality in the Neighboring Pairs Theorem. Subsequently he gave simpler proofs of both theorems in a paper [15] that we commend to the reader. We shall soon solve two special cases of the neighboring pairs problem by methods illustrating St0rmer's arguments: the results are stated below as a Corollary to the Neighboring Pairs Theorem. If P is finite, then by St0rmer's Neighboring Pairs Theorem there are only finitely many pairs of consecutive integers in S(P). D. H. Lehmer [9] gave a new proof of 238 MATH EMATICS MAGAZ I N E St!11rmer's theorem for the case of consecutive integers, and explicitly computed the last pairs of consecutive integers in S(P) when P is any initial subset of the primes with largest member 41 or less [10]. Some related tabulations are given in [4], and some elementary arguments establishing special instances of St!11rmer's result are given in [5] and [8]. The latter paper, by Halsey and Hewitt, discusses the fascinating connection between fundamental frequency ratios in Western music and consecutive pairs of integers in S(2, 3, 5). TAB L E 1: Last max i m a l r u n s i n S(P) S(P) S(2,3) S(2,3,5) S(2,3,5, 7) Last maximal runs 8[21 , 2[ 3 1 80[21' 8[3 1' 3[41 4374[21 , 48[ 31 , 7[41 For our present purposes, we note in TABLE 1 the last maximal runs of various sizes in S(P) when P is an initial subset of the primes with largest member 7 or less. We shall also need the following consequence of St!11rmer's theorem: COROLLARY 1 . The only pairs of integers satisfying 13° - 5b i = 2 are {3, 5} and {25, 27}, and the only pair satisfying W - 7b l = 2 is {7, 9}. Proof St!11rmer's method applied to pairs of integers {3°, 5b} satisfying 13a - 5bi = 2 puts them equal to x - 1 and x + 1 , so their product is of the form x 2 - 1 = Dy 2 , where y E S(3, 5) and D E {3a5b : 1 :::; a, b :::; 2} is non-square, so D E {15, 45, 75}. For these three values of D, the corresponding Pell equations have fundamental solutions (xo, y0 ) = (4, 1), (161 , 24) and (26, 3) respectively. Now St!11rmer's Pell Equation Theorem shows that y0 = 1 and 3 are the only y solutions of these Pell equations that lie in S(3, 5). The corresponding x solutions yield the pairs {x - 1, x + 1 } = {3, 5} and {25, 27}, so these are the only pairs {3a , 5b} that differ by0 2. 0 Similarly, pairs of integers {3a , 7b} satisfying 13° - 7b l = 2 must correspond to fundamental solutions of x 2 - 1 = Dy 2 , where y E S(3, 7) and D is non-square, so D E {21 , 63, 147}. The corresponding Pell equations have fundamental solutions (xo, Yo) = (55, 12), (8, 1) and (97, 8) respectively. Now y0 = 1 is the only y solution of these Pell equations that lies in S(3, 7). The corresponding x solution yields the pair • {x - 1 , x + 1 } = {7, 9}, so this is the only pair {3°, 7b} differing by 2. 4. Constraints on runs in P2 This section builds a proof of Theorem 1 . Our method is to establish several properties of runs in P2 that allow us to close in on the possible structure of long runs. Finally we accumulate enough constraints to show that no run of size 10 or more could possibly satisfy all the constraints, so we can conclude that every run in P has size less than 10. Consequences of St!11rmer's Neighboring Pairs Theorem are used2 in several key steps, including some of the results in TABLE 1 and our Corollary to the Neighboring Pairs Theorem. PROPERTY 1 . Any run of consecutive integers in P2 contains at most one multiple of6. V O L . 8 1, NO. 4, OCTO B E R 2008 239 Proof Since 89 and 97 are the first two consecutive primes with difference greater than 6, it follows that between any two consecutive multiples of 6 less than 96 there is at least one prime. Since primes are in P1 , no run in P2 can contain two consecutive multiples of 6 less than 96. On the other hand, 8 and 9 are the last two consecutive integers in S(2, 3), so 48 and 54 are the last two consecutive multiples of 6 in P2 . The • property follows. The next three properties concern multiples of 5 in P2 that occur within runs which contain a multiple of 6. PROPERTY 2. Only one maximal run of consecutive integers in P2 contains a multiple of6 and a multiple of5 which differ by 2 or 3; that run is 158[51. Proof If 6a and 5b are members of P2 that differ by 2, then 6a E S(2, 3) and 5b = I Oc E S (2, 5) for some integer c. Then 3a and 5c are consecutive integers in S(2, 3, 5), and 80[21 is the last such pair. Hence the corresponding pairs {6a, 5b} in P are {10, 12}, {18, 20}, {48, 50} and {160, 162}. For the first three pairs, the intervening2 number is in P1 ; hence the only maximal run in P2 that contain integers 6a and 5b with 16a - 5bl = 2 is 158[5 1 . Similarly, if 6a and 5b are members of P that differ by 3, then 6a E S(2, 3) and 5b = 15c E S(3, 5) for some integer c, so2 2a and 5c are consecutive integers in S(2, 3, 5). The corresponding pairs {6a, 5b} in P2 are {12, 15}, {15, 18}, {45, 48} and {72, 75}, but in each case there is an intervening prime, so no maximal run in P2 • contains integers 6a and 5b with 16a - 5bl = 3. PROPERTY 3 . No run of consecutive integers in P2 contains a multiple of 6 and a multiple of5 which differ by 4. Proof First suppose 6a and 20b are members of P2 that differ by 4, so 6a = 12c E S(2, 3) for some integer c, and 20b E S(2, 5). Then 5b and 3c are consecutive integers in S(2, 3, 5). Since 80[21 is the last such pair, the corresponding pairs {6a, 20b} in P2 are {20, 24}, {36, 40}, {96, 100} and {320, 324}. The first three pairs have an intervening prime, while the last pair is not in a run in P because 322 E P • Thus, no run of consecutive integers in P2 contains a multiple of 62 and a multiple of320 differing by 4. Now suppose 6a and 10d are members of P2 that differ by 4, and dis odd. Then 13a - 5dl = 2, so a is also odd. But 6a E S(2, 3) and 10d E S(2, 5), so a is a power of 3 and d is a power of 5. By Corollary 1 , {3, 5} and {25, 27} are the only pairs of proper powers of 3 and 5 that differ by 2. The corresponding pairs {6a, 10d} in P2 are {6, 10} and {50, 54}, but each has an intervening prime, so neither pair is contained in • a run in P2 . The property follows. It is noteworthy that the pair {320, 324}, appearing in the proof of Property 3, actually corresponds to a near miss: P2 contains the two maximal runs 319[3 1 and 323[7], and the only intervening integer is 322 E P3 • The two bordering integers are 3 1 8 E P3 and 330 E P (both multiples of 6), and their neighbors 3 17 and 33 1 are consecutive primes. 4 PROPERTY 4. Any run of consecutive integers in P2 contains at most one multiple of5. Proof On the contrary, suppose there is a run of consecutive integers in P that contains two multiples of 5. Let R be that portion of the run which begins and2 ends with two consecutive multiples of 5. Since R has size 6, we have 6a E R for some integer a. But 6a E P2 so it is distinct from the multiples of 5. If 6a differs by 2 from the nearer multiple of 5, these two members of R belong to the maximal run 158[5 1 , by Property 2. But this does not contain two multiples of 5, so is disjoint from R. Hence MATH EMATICS MAGAZ I N E 240 6a must be adjacent to the nearer multiple of 5. But then it must differ by 4 from the other multiple of 5 in R, and Property 3 shows that no run in P2 contains two such • numbers. Hence, by contradiction, R does not exist. Since any run of 10 consecutive integers contains two multiples of 5, Property 4 immediately implies our target result: THEOREM 1 . There is no run of 10 consecutive integers in P2 • COROLLARY 2. Runs in P2 have maximum size r(2) = 8 or 9. Although our methods do not seem to be strong enough to decide whether runs of size 9 exist in P2 , consideration of multiples of 7 yields further properties, revealing more of the structure of P2 • In particular, we are led to discover the examples of runs in Conjecture 1 . We shall pursue this in the next of size 8 in P2 which are incorporated section. 5. Fu rth er constraints on runs in g Now that we know P2 has no runs of size 10 or more, the study of long runs of rank 2 integers can proceed by asking: What is the structure of any run of size 7 or more in P2 ? Since any run of 7 consecutive integers must contain a multiple of 7, at least one multiple of 6, and at least one multiple of 5, the relative placement of these multiples will now be considered. PROPERTY 5 . Only two maximal runs of consecutive integers in P2 contain a multiple of6 and a multiple of7 which differ by 2 or 3 : those runs are 54[51 and 141 [81 . Proof If 6a and 7b are members of P2 that differ by 2, then 6a E S(2, 3) and 7b = 14c E S (2, 7) for some integer c. Then 3a and 7c are consecutive integers in S(2, 3, 7). From the fact that 4374[21 is the last pair in S(2, 3, 5, 7), a straightforward calculation verifies that 63 [21 is the last pair in S(2, 3, 7). The corresponding pairs {6a, 7b} in P2 are {12, 14}, {54, 56} and {96, 98}. For the first and third pair, the intervening number is prime; hence the only maximal run in P2 that contains integers 6a and 7 b with 16a - 7bl = 2 is 54[51. Similarly, if 6a and 7b are members of P that differ by 3, then 6a E S(2, 3) and 7b = 21c E S(3, 7) for some integer c, so2 2a and 7c are consecutive integers in S(2, 3, 7). The corresponding pairs {6a, 7b} in P2 are {18, 21}, {21 , 24}, {8 1 , 84}, {144, 147} and {189, 192}. For all but one of these pairs, there is an intervening prime; hence the only maximal run in P2 that contains integers 6a and 7b with 16a - 7bl = 3 • is 141 [81. PROPERTY 6. Any run of consecutive integers in P2 contains at most one multiple of7. Proof On the contrary, suppose there is a run R of consecutive integers in P2 that of contains two consecutive multiples of 7. Since they are in P2 , neither is a multiple 6, so there is a multiple of 6 between them but not adjacent to either of them. Thus the multiple of 6 and the nearer multiple of 7 differ by 2 or 3, so R must be contained in 54[51 or 141 [81 , by Property 5. But each of these runs contains only one multiple of 7, • so by contradiction it follows that R does not exist. PROPERTY 7. There are no runs of consecutive integers in P2 that contain a multiple of6 and a multiple of7 which differ by 4.