NUMBERS COMMON TO TWO POLYGONAL SEQUENCES


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1 NUMBERS COMMON TO TWO POLYGONAL SEQUENCES DIANNE SMITH LUCAS Chia Lake, Califoria a iteger, The polygoal sequece (or sequeces of polygoal umbers) of order r (where r is r > 3) may be defied recursively by (1) (r,i) = 2(r, i  1)  (r, i  2) + r  2 with (r,0) = 0, (r,l) = 1. It is possible to obtai a direct formula for (r,i) from (1). A particularly simple way of doig this is via the Gregory iterpolatio formula. (For a iterestig discussio of this formula ad its derivatio, see [ 3].) The result is (2) (r,i) = i + (r  2)i(i  l ) / 2 = [(r  2)i 2  (r  4)i]/2. It is comfortig to ote that the ft square umbers the polygoal umbers of order 4 actually are the squares of the itegers. Usig either (1) or (2), we cam take a look at the first few, say, triagular umbers (r = 3) 0, 1, 3, 6, 10, 15, 21, 28, 36, 45,. Oe observatio we ca make is that three of these umbers are also squares amely 0, 1, ad 36. We ca pose the followig questio: Are there ay more of these "triagularsquare" umbers? Are there ideed ifiitely may of them? What ca be said about the umbers commo to ay pair of polygoal sequeces? We shall begi by aswerig the last of these questios, ad the retur to the other two. Suppose that s is a iteger commo to the polygoal sequeces of orders r A ad r 2 (say r A < r 2 ). The there exist itegers p ad q such that s = [(rj  2)p 2  ( r i  4)p]/2 = [ (r 22)q 2  (r 24)q]/2, so that (3) ( r i  2)p 2  (r t  4)p = (r 22)q 2  (r 24)q, This paper is based o work doe whe the author was a udergraduate research participat at Washigto State Uiversity uder NSF Grat GE
2 Feb NUMBERS COMMON TO TWO POLYGONAL SEQUENCES 79 ad i fact, sice both sides of the equatio (3) are always eve, every pair of oegative itegers p, q which satisfy (3) determie such a iteger s. As a quadratic i p, this has itegral solutios, so sice all coefficiets are itegers the discrimiat ( r i _ 4) 2 + 4(r 42)(r 22)q 24(rj  2)(r 24)q must be a perfect square, say x 2, so that x 2 = 4(r A  2)(r 22)q 24(r t  2)(r 24)q + (r t  4) 2. As a quadratic i q, this also has itegral solutios, ad the discrimiat ad hece 1/16 of the discrimiat must agai be a perfect squre, say y 2, so that (4) y 2  ( r i  2)(r 22)x 2 = (r t  2) 2 (r 24) 2  (  2)(r 22)(rj  4) 2, where p ad q are give by ( 5 ) P = (ri  4) + x 2(r 42) q = (r x  2)(r 24) + y 2(r t  2)(r 22) Although it ca be show, by solvig (5) for x ad y ad substitutig ito (4), that every s o lutio of (4) gives a solutio of (3), it should be oted that some of the iteger solutios of (4) may ot give iteger values for p ad q. Nevertheless, (4) ad (5) give us all possible cadidates for iteger solutios of (3). Now (4) is i the form of PelPs equatio, y 2  dx 2 = N, which has a fiite umber of itegral solutios i x ad y if d is a perfect square while N does ot vaish. For the the left side ca be factored ito (y  ax)(y + ax), where a is a iteger; ad N has oly fiitely may itegral divisors. So we already have a partial aswer to our questio. If (r A  2)(r 22) is a perfect square ad the quatity o the right side of (4) is o zero, we have oly fiitely may cadidates for itegers commo to the two sequeces of orders r 1 ad r 2. O the other had, if (r t  2)(r 22) is a perfect square ad the right side of (4) is zero, the (4) reduces to a liear equatio i x ad y: y = ±N/Tr"i  2)(r 22)"x Sice the coefficiet of x is a iteger, this has ifiitely may itegral solutios. A aalysis of the right side of (4) reveals that, with T± / r 2, this quatity vaishes oly whe oe of r A ad r 2 is 3 ad the other is 6. I that case, (4) becomes y 24x 2 = 0, or y = ±2x; ad equatios (5), with y replaced by ±2x, become p = (x  l)/2; q =? (1 ± x)/4.
3 80 NUMBERS COMMON TO TWO POLYGONAL SEQUENCES [Feb. At this poit it is ot too hard to see that for ifiitely may itegers x, the above equatios yield oegative itegral values for both p ad q. Therefore, there a r e i  fiitely may hexagoaltriagular umbers. I this case, however, we have take the log way aroud; for it ca be show directly, usig (3), that ideed every hexagoal umber is also a triagular umber. It remais for us to ivestigate what happes whe (r 12)(r 22) is ot a perfect square (ad here the right side of (4) is ecessarily ozero). If this is the case, the there are ifiitely may positive itegral solutios to (4) if there is oe such solutio [ 2, p. 146 ], But i fact we ca always exhibit at least oe solutio amely x t = r l 9 y t = r ^ r j  2) correspodig to p = q = 1. We still have the job, however, of showig that ifiitely may of these solutios of (4) give us iteger solutios of (3). Cosider the related equatio (6) u 2  (r i  2)(r 22)v 2 = 1. With (r t  2)(r 22) ot a perfect square, this has ifiitely may itegral solutios, geerated by u + v N/(r x  2)(r 22) = (uj + v t sli^  2)(r 22)), where u l9 v A is the smallest positive solutio [2, p. 142]. We obtai u l9 v 4 by ispectio. I particular, u 2, v 2, give by u 2 + v 2 N/(r x  2)(r 22) = (u x + v t *J(r t  2)(r 22) ) 2, is a solutio of (6), ad by expadig the right side ad comparig coefficiets, we get (7) u 2 = u + (ri  2)(r 22)vJ v 2 = 2u 1 v 1 Now ifiitely may (but ot ecessarily all) of the positive solutios of (4) are give by (8) y x + l N / ( r 12 ) ( r 22 ) = (u, + v. ^(r t  2)(r 22) )(y + x^ {r t  2)(r 22)) where u., v. is ay positive solutio of (6) [2, p. 146], say u 2, v 2. Agai comparig coefficiets, we get (9) V l = U 2\ + ( r l " 2 ) ( r 2 " 2 ) v 2 X ' x i = v 0 y + u 0 x, +1 2 J 2 ' with the side coditios x 4 = r 4, y A = r 2 (r A  2).
4 1973] NUMBERS COMMON TO TWO POLYGONAL SEQUENCES 81 Cosider the first of equatios (9). This ca, by addig a suitable quatity to each side, be chaged to y +l + ( r i " 2 ) ( r 2 " 4 ) + ( r i " 2 ) ( r 24 H%  1) = u 2 (y + (r 12)(r 24)) + (r x  2)(r 22)v 2 x, ad usig (6) ad (7) 9 we get y + < r i +l " 2)(3?2 " 4 ) = U 2 ( y + ( r i " 2 ) ( r 2 " 4 ) ) + 2(l>1 " 2 ) ( r 2 " 2 ) u l v i x (10)  20 (/_ r 120>2/ ) 2 ( r 22 ow ) ( r 24,w.2 ) v 2 Recallig that y 1 = r 2 (r x  2), clearly y t = (rj  2)(r 24) (mod 2(rj  2)(r 22)) ; ad lettig = k i (10), we see that if y k = (r A  2)(r 24) (mod 2(r t  2)(r 22)) for some iteger k, the each term o the right of (10) is divisible by 20^  2)(r 22). Hece the left side of (10) is divisible by this same quatity, ad y k+l s ~ ( r i " 2 ) ( r 2 " 4 ) ( m o d 2 ( r i " 2 ) ( r 2 ~ 2 ) ) e By mathematical iductio, ad with referece to the secod of equatios (5), all of the y f s give by (9) produce positive itegral values for q. Similarly, the secod of equatios (9) ca be trasformed ito X 1 + ( r i " 4 ) + (U2 " 1 ) ( r i " 4 ) + V 2 ( r i " 2 ) ( r 2 " 4 ) = V 2 ( y + ( r i " 2 ) ( r 2 " 4 ) ) ad agai usig (6) ad (7), we get + u 2 (x + (r t  4)), ( 1 1 ) x +l + (l>1 " 4 ) = v 2 ^ + ( r i " 2 > (r 2 " 4» + u 2< x + r i  4 ))  2v 2 (ri  2)(r 22)(r t  4)  2u 1 v 1 (r 12)(r 24) Sice y = (r*  2)(r 24) (mod 2{v t  2)(r 22)) for all, certaily
5 82 NUMBERS COMMON TO TWO POLYGONAL SEQUENCES [Feb. y = (rj  2)(r 24) (mod 2(v t  2)). We have that x t =  ( r i  4) (mod 2(ri  2)), sice xj = v 1 ; ad it ca be see from (11) that if x k s fri  4) (mod 2(r A  2)) for some iteger k, the x k+l 5 " ( r i " 4 ) ( m o d 2 ( r * " 2 ) ) * That i s, 2(r A  2) divides x + (r 14) for every positive iteger. To summarize, for (r t  2)(r 22) ot a perfect square, we have exhibited (i (9)) ifiitely may but ot ecessarily all of the solutios to the Pelltype equatio (4); ad all of these give positive itegral solutios p,q of (3). These, i tur, give itegers s which are commo to the two polygoal sequeces of orders r A ad r 2. I view of the above, we ca ow state the followig theorem: Theorem. Give two distict itegers r t ad r 2, with 3 < rj < r 2, each defiig the order of a polygoal sequece, there are ifiitely may itegers commo to both s e  queces if ad oly if oe of the followig is true: i. ri = 3 ad r 2 = 6 S or ii. (r A  2)(r 22) is ot a perfect squre. I practice, give particular itegers r t ad r 2, we ca get all of the solutios of (4) by usig at most fiitely may equatios of the form (8), with a differet x 1} y t for each oe. Some of these equatios ca be elimiated or modified to leave out those solutios which give oiteger values for either p or q. We may the obtai equatios geeratig all pairs p,q for which (r^p) = (r 2,q); ad, if desired, fiitely may equatios geeratig the umbers s commo to the two sequeces. The procedure for fidig all solutios of (4) is arduous ad depeds erratically o the actual values of rj ad r 2. For the geeral machiery, see G. Chrystal [ l, pp ]. Now we ca easily aswer our questios about triagular squares. Lettig r x = 3 ad r 2 = 4, (r t  2)(r 22) becomes 2, which is ot a perfect square. There are, the, ifiitely may triagular squares. A s a matter of fact, this result has bee kow for some time. To exhibit these umbers, we ote that sice the coefficiet of q i (3) becomes 0, we ca get a formula like (4) by applyig the quadratic formula oly oce. The result is or x 28q 2 = 1
6 1973] NUMBERS COMMON TO TWO POLYGONAL SEQUENCES 83 (12) x 22y 2 = 1, where p = (x  l ) / 2 ad q = y/2. Coveietly eough, (12) is already i the form of (6); ad sice x t = 3, y t = 2 is the smallest positive solutio, all oegative solutios of (12) are give by (13) x + y N/2 = (3 + 2<s/2) ( = 0, 1,. 2, ). Certaily the "ext" solutio is give by x ^ + y ^ ^ 2 " = (x + y N/"2)(3 + 2 N / I ), +1 ^+l v J / v J ' ad by comparig coefficiets we get x f = 3x + 4y, +1 xi v = 2x + 3v J +1 ^ J with (from (13)) x 0 = 1, y 0 = 0 9 It follows by iductio from (14) that all values of y are eve oegative itegers, ad all x f s are odd positive itegers. Therefore, for ay solutio p,q of (3) i oegative itegers ad with r t = 3, r 2 = 4 there exists a ( = 0, 1, 2,... ) such that (15) P = P = (x  l ) / 2 q = q = y /2 where x, y are give by (14)* Furthermore, p, q give by (15) forms a oegative itegral solutio for ay, sice the x f s are always odd ad all of the y f s are eve, All triagular square umbers, the, are give by (16) S = fc + p ) / 2 = <. Solvig (14) with x 0 = 1, y 0 = 0, we get x = [(3 + 2<\/2) + (32 \ / 2 ) ] / 2 y = [(3 + 2N/2)  (32N/2) ]/2Nj2, ad combiig these with (15) ad (16), we obtai ( N/2) + (1712\/2)  2 S " ~ 32"
7 84 NUMBERS COMMON TO TWO POLYGONAL SEQUENCES Feb where s is the triagularsquare umber. th Likewise, we ca compute a formula for the triagularpetagoal umber. The result is  (2  N / 3 ) ( N / 3 ) H + (2 + N / 3 ) ( A/3) 114 S ~. This agrees with a result recetly published by W. Sierpiski [ 4 ]. I am thakful to Dr. D. W. Bushaw, whose suggestios ad ecouragemet made the writig of this paper possible. REFERENCES 1. G. Chrystal, Algebra: A Elemetary TextBook, Vol. 2, Adam ad Charles Black, Lodo, William J. LeVeque, Topics i Number Theory, Vol. 1, AddisoWesley, Readig, Mass., C. T. Log, "O the Gregory Iterpolatio Formula," Amer. Math. Mothly, 66 (1959), pp W. Sierpiski, "Sur les Nombres Petagoaux," Bull. Soc. Royale Scieces Liege, (1964), pp [Cotiued from page 71. ] is give by m M" 1 = E ^ D k k=0 m* (VII) 2 TT ^ = l/tae 01 * + B e 2 t > k=0 We ow ote that for Case 2, where A + B = 0, Eq. (VIE) does ot exist for t = 0, ad hece there is o iverse operator M". Thus, a sufficiet coditio for M~ (see CD) to exist is that A + B ^ 0, i. e., Case 1. For A + B ^ 0, oe readily fids that (VIII) (A + B)m* = (c 2  CiTH, \ ( ^ 2 \  A ^ ) where H, (x A) is the Euleria polyomial cited i (*). May more idetities ca be quoted. Ideed, for m, = 0, 1,, oe has [Cotiued o page 11&* ]
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