HEEMPORIA STATE 0 I*' ..RESEARCH C STUDIES b 0 0 THE: GRADUATE PUBLICATION OF THE KANSAS STATE TEACHERS COLLEGE, EMPORIA A General Method for Determining Simultaneously Polygonal Numbers L. B. WADE ANDERSON, JR. KANSAS STATE TEACHERS COLLEGE EMPORIA, KANSAS 66801 J A General Method for Determining Simultaneously Polygonal Numbers L. B. WADE ANDERSON, JR. Volume XXII Fall, 1973 Number 2 THE EhIPOHIA STATE HESEARCH STUDIES is publisl~ed quarterly by Tlle School of Gr'ld~~ateand ~rofeGsion,~lStudies of thc Kans'ls Statc Teachers Collt~ge, 1200 Commercial St., Emporia, hamas, 66801. Entered as second-class matter Scptc.inl)er 16, 1932, at the post office at Emporia, Kansas, under the act of Augu\t 24, 1912. Postage paid at Empori,i, Kansas. KANSAS STATE TEACHERS COLLEGE EMPORIA, KANSAS JOHN E. VISSER President of the College SCHOOL OF GRADUATE AND PROFESSIONAL STUDIES JOHN E. PETERSON, Intejrim Dean EDITORIAL BOARD WILLIAMH. 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Numbere Simultaneously Polygonal L. B. Wade Anderson, Jr. " For the purpose of this study, the term, polygonal number, will refer only to positive integers and is defined as follows: let a) be an arithmetic sequence whose first term is 1 and whose common dif- ference is m - 2, where m is a positive integer greater than 2. The sequence of partial sums, {s.) , associated with (a,) is called a sequence of m-gonal numbers or the sequence of polygonal numbers with m sides. For example, when m = 3, the arithmetic sequence to be considered is {a. ) = { 1, 2, 3, . ., k, . } , and the as- sociated sequence in this case is {s.) = (1, 3, 6, . , ., r (i + 1)/2, . ) This is the sequence of 3-gonal (triangular) numbers. For simplicity, the rth term of the sequence of m-gonal numbers will be denoted by pi. Table I is a general listing of pr, . Table I was obtained using the following well-known formulas for arithmetic sequences and series: a, = 1 + (k - 1) (m - 2) and s = (r/2) (2 + (r - 1) (m - 2)). Historically, the numbers were named polygonal' because they can describe, for a given m, a nest of regular polygons of m sides having a common vertex and with r = 1, ;, 3, . points for each side. The diagrams shown below in Figure 1 illustrate polygons which are repre- sentative of the first four triangular, square, and pentagonal numbers. FIGURE I POLYGONS ILLUSTRATING THE FIRST FOUR TRIANGULAR, SQUARE, AND PENTAGONAL NUMBERS * The author is a member of the faculty of the Department of Mathematics at Junction City, Kansas, High School, and recently completed his Ph.D. degree at Michigan State University. This study originated in a thesis, under the direction of Professor Lester E. Laird, presented to the Department of Mathematics at Kansas State Teachers College, Emporia, for the Master of Arts degree. TABLE I POLYGONAL NUMBERS m-gonal 1 m 3m-3 6m-8 10m-15 15m-24 . rz (m - 2) -r (m - 4) 2 If P denotes the set of all polygonal numbers, it is apparent that P is the set of all positive integers except 2. An integer, w, will be called si&ultaneously polygonal if, and only if, there exist integers r and q such that for distinct integers m and n it is true that w = p; = p:. Let the set of all' simultaneously polygonal numbers be denoted by P2. The following facts immediately present themselves: (1) P2 is a proper subset of P. (2) 1 P2, since for all m > 2, p 1 = 1. (3) If w E P2 then exactly one of the following hold: ( i) r = q = 1 or ( ii) r # q, where w = pi, = p . An investigation of the possible ways a given number may be polygonal helps to determine the nature of the set Pz. Let w be allv integer. If w is the rth m-gonal number, then w=p;= (r/2) [2 + (r - 1) (m - 2)] and hence 2w - r [2 + (r - 1 (m - 2) Now, if any.given w is to be polygonal, then 2 w will have to be expressed as a product of two factors. One of these factors is r, and the other is 3 + (r - 1) (m - 2). The following theorem shows that r must be the smaller factor. Theo~em1: If w=p:,, then r<2+ (r-1) (m-2). Proof : By definition, m - 2 -2 1. Hence, by multiplying each member of the inequality by r - 1, (1.- 1) (m - 2) => r - 1 is obtained. Adding 2 to both members yields (r - 1) (m - 2) -t 22.r- + 1. And it follows that r< 2 + (r-1) (m-2). QED It is now clear that the smaller factor of 2 w is r, and that by subtract- ing 2 from the larger factor, the product (r - 1) (m - 2) is obtained. Using the preceding fact, m is easily determined. As an exampke of this method; the problem of deciding the number of ways 36 may be polygonal is examined. The first step is to express 2 x 36 as a prod- uct of two factors in all possible ways: 2x36 = 3x24 = 4x18 =6xJ2 = 8x9 The first factorization, 2 x 36, is then considered. Since r must cor- 1.espond to the smaller fartor, r - 2. Bv subtracting 2 from the larger factor, 34 is obtained. Hence, (r - 1j (rn - 2) must be 34 in this case, and m is, therefore, 36. Thus, this factorization indicates that 36 = p ?j6. Similarly, the factorization 3 x 24 indicates that r = 3 and (r - 1) (m - 2) = 22, which implies that m = 13. From this factorization, it is concluded that 36 - p f3 . Not all factorizations are indicative of a manner in which 36 is polygonal. It will be noticed that the factorization, 4 x 18, shows that 36 is never the 4th element of an m-gonal sequence; since in this case r = 4, (r - 1) (m - 2) - 16, and since 3 = r - 1 does not divide 16, there can be no integral value for m - 2 and, hence, no value for m. Table I1 indicates all of the ways in which 36 is polygonal. Hence, 36 is polygonal in exactly four ways. It is possible that some factorizations can be elminated from consideration. The following theorem indicates that consideration need only be given those factorization of 2 x w, where the smallest factor is less than or equal to 34 ( 4- - 1). Theorem 2: If w - p', then rz1/2 ( dm-1). Proof: If w = (r/2) [2 + (r-1) (m--2)], then solving for m - 2, m - 2 = eb'i=d . But, also, by definition m - 2 -> 1. r (r- 1) - 2(w-r) > > > r (r-1) So, ------ = 1 or 2 (w - r) Ir (r - I). Hence w = ------ + r (r-1) 2 r or 2wZ r' + r and completing the square 8w + 1 -24rz + 4r + 1 = (2r + 1)2. Thus, 8w + 1.2- Zr + 1; and there- fore, l/s ( vm-1) 2 r. 'Q~~ A number that is not an element of P2 is 26. This is apparent, since 2 x 26 = 4 x 13 are the only factorizations of 2 x 26. The first factorization shows that 26 = p g6 , but since 3 does not divide 11, this is the only way 26 is polygonal. It also follows that, if w is any prime, then the only factorization of 2 x w is 2 x w and, hence, w = p :. This is the only way w is polygonal. Thus, P2 contains no primes. Furthermore, the above method reveals that there are twenty-seven composites less than 150 that are not elements of P2: 4, 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68, 74, 77, 80, 86, 98, 110, 116, 119, 122, 125, 128, 134, 140, 143, and 146. The following con- jecture seems appropriate at this point: -- - Conjecture: With the exception of 4, there does not exist a composite integer that is not an element of Pz that is not congruent to 2 modulo 3. In partial support of his conjecure is the following theorm: Theorem 3: If w is a composite and is congruent to 0 modulo 3, then w is an element of P2. Proof: If w = 0 (mod 3), then there exists a positive integer k such that w = 3k and then Zw = 2 (3k). Factorizations of 2w include 2 (3k) and 3 (2k). The first factorization implies r = 2 and (r - 1) (m - 2) = 3k - 2 and hence m - 2 = 3k - 2 which implies m = 3k = w or w = p:. The second factorization implies r = 3 and (r-1) (m-2) = 2k-2 and hence m-2 r k-1 so that m = k + 1 and w = p 3, + .