Vol. 5 No 2 - Pi Mu Epsilon

PSILONVOLUME 5 SPRING 1970 NUMBER 2Two Algorithms for ExpressingCONTENTSn1 ikas a Polynomial in ni=lLouis D. Rodabaugh, Ph. D. ...................... 49Elliptic Curves Over Local FieldsBruce L. Rienzo..................................A Decimal Approximation to T Utilizing a Power SeriesTim Golian and John Hanneken.....................Some Comments on Terminologies Related to DensenessR. 2. Yeh........................................ 79A Necessary and Sufficient Condition for CertainTauberian TheoremsA. M. Fischer.................................... 80A Characterization of Homeomorphic Tl SpacesW. M. Priestley... ............................... 86Problem Department ........................................... 87Initiates.................................................... 100Copyright 1970 by Pi Mu Epsilon Fraternity Inc.e c

Two ALGORITME FOR ExwzssI~G yi k AS A poL~Non1AL in nPI MU EPSILON JOURNALTHE OFFICIAL PUBLICATIONOF THE HONORARY MATHEMATICAL FRATERNITYKenneth Loewen, EditorLouis D. Rodabaugh, Ph. D.University of AkronIn this paper we shall illustrate and verify two algorithms fordetermining the coefficients in the polynomial representation offor each non-negative integer, k. and each positive integer, n.These coefficients will be placed in an infinite triangle as follows:ASSOCIATE EDITORSRoy B. DealLeon BankoffOFFICERS OF THE FRATERNITYPresident: J. C. Eaves, West Virginia UniversityVice-president: H. T. Kames, Louisiana State UniversitySecretary-Treasurer: R. V. Andree, University of OklahomaPast-President: J. S. Frame, Michigan State UniversityCOUNCILORS:E. Maurice Beesley, University of NevadaL. Earle Bush, Kent State UniversityWilliam L. Horkness, Pennsylvania State UniversityIrving Reiner, University of IllinoisChapter reports, books for review, problems for solution and solutions to problems,and news items should be mailed directly to the special editors found in this issueunder the various sections. Editorial correspondence, including manuscripts, shouldbe mailed to THE EDITOR OF THE PI M U EPSILON JOURNAL, 1000 Asp Avenue,RobM'215, The University of Oklahoma, Norman, Oklahoma 73069.PI MU EPSILON JOURNAL is published semi-annually at The University ofOklahoma.SUBSCRIPTION PRICE: To individual members, $1.50 for 2 years; to nonmembersand libraries, $2.00 for 2 years. Subscriptions, orders for back numbers andcorrespondence concerning subscriptions and advertising should be addressed tothe PI MU EPSILON JOURNAL, 1000 Asp Avenue, Room 21 5, The University ofOklahoma, Norman, Oklahoma 73069.so that if n is any positive integer, and k is any non-negativeinteger, then:1Since 1 i k = lk = 1, we know thati=lr(2) 1 R i = 1 for all positive integers r.i=lFrom the well-known identity3 n k = @ - (i - l l j for all positive integers k.i=lcan be derived the equationsandfor all positive integers r.

5 ) RrVs = Fs-1i=lRr-i,s-iPr+lMfor all integers r,ssuch that 1 < s < r.The P appearing in equation (5) are the elements of Pascal's"'*nTriangle, placed and numbered as follows:^lIf we now divide each column sum by 5 (the number of the row whichwe are determining). we shall have the fifth row of our coefficienttriangle:11 1 15' 2"' 3" 0' -30'-The appearance of the coefficient triangle, down to and includingthe fifth row, is therefore as follows:p21 p22'31 p32 p33so that'Ql '42'Q3 'QQ'51 '52 '53 '5~'55......................Equations (2). (41, and (5) provide the basis for our FIRSTALGORITHM. We here illustrate this algorithm by using it in orderto determine the fifth row of our coefficient triangle on the assumptionthat we have already determined the first four rows. We place ablank row beneath the fourth row, and place a "1" in the sixth rowin the fifth column numbered from the right. We place as a columnof multipliers the elements of the sixth row of Pascal's Triangle,beginning with a "1" in our sixth row and proceeding upward withalternating signs:Our SECOND ALGORITHM is very largely based onTHEOREM 1: If r,s are integers and 1 < s < r, then(7-r - 1Rr,s = r - s + 1 -Rr-l,s.Proof: Let m be any integer greater than 1. Then (7) holds forr = m and s = 1, because(8)=-=-.-= 1 m-1 1 m- 1Rm,l m m m-1 m - l + l Rm-l,l 'Next, we assume that (7) holds not only for R but also form.1the top q elements of the column headed by RmSl; that is, that (7)(=R, ).h01dsf0rRm,~sRm+~,~~Rm+~,~~~~~yRm+q-~,q-~~andRm+q-~,q,qThen:MULTIPLIERS= - 1 k- i) + k - q + 1 Rk+1 i=lk-i,q+l-i pk+2,2+i- 1 k- i (k+l)!Rk+1 i=1 ) + k - q + 1 k-i,q+l-i rl+i)! (k-i)!

k 1k - q + l ' kThus(lo) %+l,q+l =That is, (7) holds also for the (q+l)th element (counting from thetop down) of the column headed by R 1This concludes the Second-Principle Induction, and thereforealso the proof, that (7) holds for every element of the column headedSince m was an arbitrary integer greater than 1, we seeby RmSl .that (7) holds for every element of the coefficient triangle not inthe rightmost column. THEOREM 1 is therefore proved.The validity of our SECOND ALGORITHM for the determination ofthe coefficient triangle is now established. We describe this algorithm:IUse (4) to construct the first upper-right-to lower-left diagonal:.I1I11Determine R j,jl for each positive integer j. in either of thefollowing two ways:a) Apply the FIRST ALGORITHM to only the right-most column;b) When the jth row is known except for the element R j,j'use equation ( 2) ;Use equation (7) to determine R r.S for every r and s in I suchthat 2 < s

k(12) q = aon t a n k-1 +1. + ;Ln t a .Therefore h divides nq.q. In other words, 3 is an integer. SincehSince (n,h) = 1. this implies that h divideswe see that f ik is divisible by n,i=lLemma 1 and the fact thatestablish immediatelyCorollary 1: If k,n are positive integers and (n, (ktl)!) = 1, thenLemma 2:n-1ik is divisible by n.i=lIf h,n are positive integers. (2ntl. (ktl)!) = 1, and kis even, then f ik is divisible by (2ntl)!.i=lProof: By Corollary 1,2n k(15) 7 i 50 (mod(2n+l)).i=lSince k is even, then for every i c (1. 2, ... , n} we have(16) ik S [(2ntl) - ilk (mod (2ntl)) .Hence2nk(17) 1 ik E 2 -1 i (mod (2ntl)) .i=l ~=lFrom (15) and (17) we have-(18) 2! i k S O (mod (2ntl)) .iil- --k- From this, since ((2ntll.2) = 1, we see that (2ntl) divides 7 i ,i=1THEOREM 2:If k,n are positive integers, k 1 3, and k is odd, then,in the right-hand member of (ll),Proof: If, in (11). we select n as an odd integer, 2mtl. such that(2in+l, (ktl)!) = 1, then we can writeFrom this we see thatif the expression(21) k(2mtl)(lk-1t 2k-1t ... tmk-I)2m+l21 ik is divisible by (2mtl) if and onlyi=l2is divisible by (2m+l) . This is seen to be the case whenever theexpression(22) lk-I t 2k-1 + ... t mk-Iis divisible by (2mtl). The latter is the case, however, as we seefrom Lemma 2. We have, therefore, that if n = 2m t 1 and(2m+l(k+D!) = 1, then the right-hand member of (11) is equal to2sn for some integer s. From this it follows that2Thus n divides akn and hence n divides a,. From the way in whichn was selected we see that a, is divisible by every positive integerwhich is relatively prime to (k t I)!.\ = 0.MEETING ANNOUNCEMENTIt follows, of course, thatQ.E.D.Pi Mu Epsilon will meet in late August, 1970, at the Universityof Wyoming, Laramie, Wyoming, in conjunction with the MathematicalAssociation of America. Chapters should start planning NOW to senddelegates or speakers to this meeting, and to attend as many of thelectures by other mathematical groups as possible.The National Office of Pi Mu Epsilon will help with expensesof a speaker OR delegate (one per chapter) who is a member of Pi MuEpsilon and who has not received a Master's Degree by April 15, 1970,as follows: SPEAKERS will receive 54 per mile or lowest cost, confirmedair travel fare; DELEGATES will receive 2 1/24 per mile or lowest cost,confirmed air travel fare.Select the best talk of the year given at one of your meetingsby a member of Pi Mu Epsilon who meets the above requirement and havehim or her apply to the National Office. Nominations should be in ouroffice by April 15, 1970. The following information should be included:Your Name; Chapter of Pi Mu Epsilon; school; topic of talk; what degreeyou are working on; of you are a delegate or a speaker; when you expectto receive your degree; current mailing address; summer mailing address;who recommended by; and a 50-75 word summary of talk, if you are aspeaker. MAIL TO: Pi Mu Epsilon, 1000 Asp Ave., Room 215. Norman,Oklahoma 73069.

f(x) = 0 hasthree distinctreal roots.y2=x3-3x2) f(x) = 0 hasonly onereal root.y2=x3-3x+3 ?ELLIPTIC CURVES OVER LOCAL FIELDSBruce L. RienzoRutgers UniversityElliptic curves may be put into the standard form y2 = x + Ax + B,called the Weierstrass form. In this form, the points on the curvedefined over a field k form an abelian group under an appropriatecomposition law. This group law also works for singular curves,provided we avoid the singular point.Considering the curves over finite fields of P elements, we see2that there can be only p possible curves. We then may program thegroup law on a computer and run off all possible cases. Looking atthese results, we can then make same conjectures as to the number ofpoints on the curve mod p.Having found the solutions mod p, we procede to develop a methodfor lifting these solutions to solutions mod pN, for arbitrary N.This gives solutions in the p-adic fields.Finally, we develop the Nagell-Lutz Theorem, for p-adic fields.By this theorem, points of finite order in the group must have integercoordinates.1. Elliptic Curves and the Group Law.51.1 Weierstrass FormRather than having to work directly with elliptic curves, wemay first put them into a standard form. An elliptic curve, definedover a field k of characteristic not 2 or 3, is birationally equivalent3to a plane cubic curve of the form y 2 = x + Ax + B, provided the curvehas a point defined over k.Thus, we will not need to consider general elliptic curves, onlythose of the form y2 = x + Ax + B.be in the Weierstrass form.by y2 = f(x), where f(x) is a cubic.Curves of this form are said toWe will often denote the Weiestrass formWhat do these curves look like? This question can be asked forvarious different fields. We will restrict our attention to pointswhich are rational over the field. First, consider the field of realnumbers.- -*- 51.2 The Real Ground FieldFor the field of real numbers there are several cases dependingon the roots of f(x) = 0.3f(x) = 0 has adouble root atx=a and a distinctsingle root. Thepoint (a.0) isthen a singularpoint of the curve.y2=x3-3x+251.3 Projective Space4) f(x) = 0 hasa triple root atx=a. The curve thenhas a cusp at (a.0);this is also asingular point.2 3y =xWe will be considering these curves from the point of view ofprojective geometry. That is, we will be including points at infinityon the curve. Putting the equation into homogeneous form gives2 3 2 3Y Z = X + AXZ + BZ . The points at infinity are the points withZ = 0. But this means that x3 = 0. Thus there is only one infinitepoint on the curve (the point (0, 1, 0) in projective coordinates);and the line at infinity intersects the curve at this point withmultiplicity three.51.4 The Group ~ a w ~Consider a fixed elliptic curve defined over a field k. If wecan devise a way of making the points of the curve into a group, wemay then study the points by studying the structure of the group.We will see that we can in fact define such a group operation. Itwill turn out to be commutative, so we will call the operation additionand denote it "+",Geometrically, the group operation for non-singular curves isbased on the fact that, counting multiplicities, any line defined overk intersects the curve in exactly three points (over the algebraicclosure of k). What this means is that if Pl and P2 are two pointson the curve, we may draw the line through them. and this will giveus a third point associated with P and P2.Unfortunately, this easily defined composition is not a groupoperation. For one thing, it has no identity. However, we may remedythis situation by first fixing some point 0 on the curve to serve asthe identity element of the group. Then when we get the third pointof the curve on the line through P and P2, we simply draw the linethrough this point and the point 0. The third point on this line willbe the desired point Pl + P2....^

- - -It is clear immediately that this addition law is commutative.(The line through P and P is certainly the same as the line through2P2 and P .) To show that the point 0 is indeed the identity, we let1P be a point on the curve and find P + 0. The line through P and 0intersects the curve in some third point Q. We then consider the linethrought 0 and Q. But this must be the same line. Thus the thirdpoint must be P. That is, P + 0 = P as desired.To get inverses, we draw the line through 0 twice (i.e. the linetangent to the curve at 0) and let S be the third point. Then if Pis any point, the third point of the line through P and S is thepoint -P. (The third point of the line through P and -P is S.) Thenthe third point of the line through 0 and S is 0. So P + (-PI = 0.The hard part is to show associativity. We omit this proof here,referring to ate^ for a proof. This difficulty may be avoided completelyby using the definition of elliptic curves in cassels4.We may choose any point on the curve to be the fixed point 0.If we choose the point at infinity, then the lines through 0 are justthe vertical lines (and the line at infinity). That is, the linethrough 0 and P = (x,y,z) has the point (x,-y,z) as its third pointof intersection with the curve.Inverses are now simple to compute. The point S described aboveis now the point 0. (The line tangent to the curve at the point atinfinity is the line at infinity, Z = 0. But we have seen that Z = 0intersects the curve 3 times at the point 0. Thus, S = 0.) So, ifP = (x,y,z) then -P is the third point of the line through P and 0which is just (x.-y,z).We may now restate the addition law. If P and P are two points1 2on the curve, let the line through them have P3 = (x,y,z) as its thirdpoint. Then P + P2 = (x, -y, z). That is, Pl + P2 = -P3; orPl + P2 + P = 0, where P -, P2, and P3 are collinear.It will be useful to have an actual formulafor the addition of two finite ~oints. Letpl = (x~,Y~.~) = (xl,yl) and p2= (x2,y2,1) =(x2,y2). If xl = x2 and yl = -y2, the points areinverses and P + P2 = 0. Otherwise, consider the- -line through Pl and P , Say its equation is Y -Yy = Ax + v. If P # P , then the slope A = -. If P = P , thex -X2 1line through P and P is the tangencat that point.1 2gives A = ft(x)- 2~ .Then y 2 = f(x).. In either case, v = yl - Ax (=y2 - Ax2). To get the third pointP3 = (x3,y3), we plug y = AX + v into y2 = x 3 +AX + B:(Ax + v ) = x 3 + Ax + BThis is a cubic in x whose roots are just the x-coordinates ofthe three points of intersection of the curve with the line. Theroots must equal the negative of the coefficient of the second orderterm. i.e. x + x , + x3 = A2. Thus the group law becomes:"1-"2 f '(XIwhere A = -when xl # x2, and A = -when Pl = P2, and wherex1-x2 2~v = y, - Ax,.J. J.N z z If the curve is given in the form y2 = x 3 + ax 2 + bx + c,then x 1 + x2 + x3 = A2 - a. So the group law is x3 = A2 - a - (x 1+x 2 ).These formulas could now be used to prove associativity.51.5 Singular curvesWe have described the group operation for non-singular curves.What can be said about the singular cases? We needed the fact thata line intersects the curve in exactly three points. This is stilltrue provided the line does not pass through the singular point.If P. and P. are two points on the curve, then the line throughJ. £them does not pass through the singular point. (The singular pointis in effect a double point, so any line through it can intersect inonly one other point of the curve.)Thus our group operation holds for points other than the singularpoint. That is, the complement of the singular point forms a group.2. Local and Finite Fields52.1 P-adic fieldsMany of the most interesting results on elliptic curves come fromlooking at the curves over p-adic fields. We will not discuss thetheory of p-adic numbers here. (For an explanation of p-adic numberssee a number theory text such as Borevich and Shafarevichl.)by:We will be using the exponential p-adic valuation, which is givenv (pnu/v) = n where p 4 u and p 4 V.P~f a, 6 are non-zero p-adic numbers, thenv (aB) = v (a) + V (6)P P Pv (a+6) 2 min[v (a),v(B)]PPwith equality if v (a) # v (0).P PIf v,,(a) 2 0, then a is a p-adic integer.rIf v_(a) = 0, then a is a unit of the ring of p-adic integers. (SinceI'we will in general be working with a fixed p, we will often write just\)(a) to denote the valuation. )

Solution of y 2 = x 3 + Ax + B in D-adic numbers is eauivalentNto solution of y2 E x 3 + Ax + B (mod p 1 for arbitrarily high N.We consider first N = 1.52.2 Solutions mod p -- Finite FieldsIn this section we will be considering the group of points onthe curve over the finite field of p elements (i.e. the field of numbersmod p where p is a prime. In general, we will avoid p = 2 and 3,since these cases present special problems with regard to singularities.(Note that fields of characteristic 2 and 3 were excluded from thediscussion on page 1 of this paper.)For a given p, there are only p2 curves of the form y 2 = x3+Ax+~.(A and B can each take only p values.) Thus it is possible to programthe group law on a computer and run off all the possible cases.Before seeing the actual results, how many points might we expectthe curve to have for a given p? The "Riemann hypothesis"2 givesthe number of points as N = p + 1 - a where la1 < 26. In other words,p + i - 2 6 ~ ~ ~ ~ + 1 + 2 6 .Let's look first at p = 5. Then N should fall in the range2 & N 5 10. The following chart gives the number of points in thegroup for each possible A and B.0 1 2 3 4A single digit indicates that theB\group for that particular curve is cyclic5' 2.2 2 10 2.4of that order. For examnle. . .3 I 6 9 7 4 % 8y 2 = x + 2x + 1 has 7 solutions(including the ~oint -.6 4 6'5 3at infinitv). andits groupis cyclic of order 7. 3 6 4 6 ' s 3When the entry is expressed as a 4 6 9 7 4' Bproduct, the group is the direct productof cyclic groups. For example,y2 = x 3 + 4x has 8 solutions, and its group is the direct product ofa cyclic group of order 2 and one of order 4.m "An "s" indicates that the curve has a singular point.For example,y' = x + 2x + 3 has 7 solutions, one of which is singular. Its groupis cyclic of order 6. (The group does not include the singular point.)Inspection of the chart shows many interesting features. First,all values are within the predicted 2 d N

52.3 Solutions mod p NNow that we have the solutions mod p, we need a way of liftingNthem to solutions mod p . We do this one step at a time, i.e. fromp to p2, then from p 2 to p 3 , etc.nt1mod pn to solutions mod p .In general, we want to lift solutionsAny solution mod p n must also be a solution mod p n .solutions mod pntlThus allare in the form (xotspn,yo+upn) where (xo.yo) isa solution mod p n , and s and u are between 0 and p-1.(y+u~")~ 2 f (x0tspn) (mod pn+l)2yo +2youpntu2p2n E f(xotspn)yo2+2youpn 5 fdtotspn)2youpn 5 f(xo+spn)-yo 2(mod pn+l)(md on+')(mod pn^)Then,The right side is divisible by p n since f(xo)-yo2 Z 0 (mod pn).2y u 50f(xots~n)-~o 2pn(mod PProvided yo # 0 (mod p), this is a simple linear congruence and soeach value of s gives exactly one value of u.so there are p solutionsSuppose yo s 0 mod p).n 22y0upn 2 f ( x+~P )-yoBut now, y0pn E 0 (mod pn), soUsing the expansion:f(xotspn)-yo2 ; 0Then, as beforeThere are p such values,(mod pn+l)(mod pntl)2 2nf(x0tspn) = f(x) t spnf *(x0) + S+ f1'(x0) t a . ,,gives f(x) t spnff(xo) - 0 (mod pntl)If both y and ft(x0) 5 0 (mod p), then the point (xo,yo) is0a singular point.Otherwise, each point mod pn lifts to P pointsmod p n , and we need only solve a linear congruence to find then.3. Nagell-Lutz TheoremIn this chapter, we give the major theorem on the structure ofthe group for curves over local fields. The proof given here generallyfollows the proof given by ~utzl.Let r be the group of points on the curve y2 = x 3 + Ax t B overa p-adic field, where A and B are p-adic integers.Lemma 3.1Each rational point P = (x,y) in l' has coordinates in theform (~p-2n,6p"3n), where n 2 0 is an integer and E,6 are p-adic integers.6 and 6 are units if n > 0.-8and n=0.If x is a p-adic integer, then so is y; and then 6=x, 6=y,3Otherwise, v(x) < 0, and we have v(x ) = 3v(x) < 0 ^. v(Ax).2 3 3Also, v(x3) < 0 < v(B). Therefore, v(y ) = v(x +Ax+B) = v(x 1.2v(y) = 3v(x) and so we must have v(x) = -2n and v(v) = -3n withn > 0. Thus, x and y are in the form x = (p-2n and y = 6p-3n where6 and 6 are units.For any rational point P on the curve, let n(P) be the integer nof the above lemma. Let l' denote the set of points P with n(P) 2 m(That is, with v(x) 5 -2m and v(y) & -3m).Theorem 3.2-1Pis a subgroup of l'.Let Pl = (xl,y2) and P2 = (x2,y2) be two points in rm. Let= P t P and say P3 = (x3.y3).1Suppose n(Pl) # n(P2). We may assume n(P2) > n(Pl). The addition2 2(y2-yl)2 - (x tx )(x2-x1l2 = x x tx x tA(xltx2)t2B-2y~formula x3 = 1 2 2 1 2 1(x2-xl) 2 X22-2X X +X 22 112 2gives v(x3) = v(x2 xltx2x12t~~xl+x2~t2~-2y~l~ - v (x~~-~x 2x 1 tx 1 ).Thus for n(P) # n(P-1, we havef (x0)-yo 2 t sft(x0) 5 0 (mod PpnProvided f'(xo) 3 0 (mod p), there is exactly one s which solves thislinear congruence. This value of s and any value of u gives a solution.there are p values of u, so again there are p solutions.Suppose n(Pl) = n(P2). Then Pl = P3 - P2, so if n(P3) # n(P2) we haven(P) = min[n(P) ,n(P)I. Thus, for n(P) = n(P2), we have n(P3) 2 n(Pl).In either case, we have n(P3) 2 min[n(Pl),n(P2)l. Therefore, since

-. n(P 2 m and n(P ) > m, we have n(P3) 2 m, so P3 is in rm.1 2 -rm istherefore a subgroup of T.Theorem 3.3 I' has finite index in r, for integers m > 0.mLet Pl = (xl,yl) and P = (x2,y2) be two points in r.need to consider conditions under which (P2-Pl) E rm,We may assume the P,P2 f! r , say n(Pl) = n(P2) = n < m. Put-2n -3n) -2n -3n)P + ( CP ,6^p and P2 = (C2p ,i2p with612 = C13 t A5p4" t Bp6" and 6; = 6; + AC2p4n 6nt BP , where El,C2, 6. and i2 are units.From the addition formula, n(P-P) 2 m if and only ifWeIn particular, when n = m - 1 this shows that rm-l/l'm is finite form > 1. [The lifting procedure described in 2. shows that the indexof rm-l in rm is exactly 0. Also, it should be noted that these argumentsrequire that singularities be avoided.] This shows that I' is of finiteindex in rl.We still must show that r/rl is finite. Let n(P) = n(P) = 0.We claim that n(P 2- P 1 )> m when C2 2 El and 6- E 61 modulo a sufficientlyhigh power of p. Say C2 2 El, 6 2 5 6 1 (mod for sufficientlylarge r.2If not, n(P 2 -P 1 ) < m implies ~ (6~t6~) i. r and ~ ( c ~ ~ +A)> ~ C r ~ C ~by the above argument. Then, ~(26~) = ~((6,+6~)-(6,-6~)).> r andBut62+61 2^1so n P ) m if and only if v(-) = n+v&) < n-m.2 1 62-61 2-"1Thus, v(C -6 ) .> m-n and ~ (6~-6~) 2 m-n is clearly a necessary2 - 1 -condition for n(P2-P) i m.If P # 2, S t 6 is a unit, so v(6,+6)is a sufficient condition.If p = 2, write(62+61)(62-61) = 6^-6 1 = 6; -£4n+ A(E2-E1)p .= 0. Then, (E2-C1) i m-n2 2 2 4nSince P # 3, E2 +62~ltC12 is a unit; and then so is 5 tC2CltCl tAp .From the above, we knowHowever, we have the identity2-@A3 - 27B2 = (x3tAx+B)~(x) + (3x tA)Q(x) where P(x) = 1BAx - 27B.2and Q(x) = -6Ax + 9Bx - @A2.&'

If yl  y2  0 (mod p), and the curve is non-singular mod p, wedt)The curve may be parametrized as follows: x = à y = -may write its equation in the form y2 = (x-xo)g(x) where xlEx2Exot t 3(mod p) and g(xo) is a unit. Consider 4where ~(t) = (1 t At t ~ t ~ ) " ~ ~(t) . may be represented as a power2wY2- Yl-= xl-xoseries: ~(t) = 1 t 1 [The series may be derived from theg(xl) Here g(x ) and g(x2) are units.1 i=2 - -"2 - xl1 1 2series (ltu)"'We may assume that v(xl-x) 2 v(x2-x). Suppose firstp-adicly for t  0 (mod p). From the formula for t, we see that tv(x -x ) > v(x-x). Then v(x -x ) = v((x2-xo) + (x-x))1 0 2 1 is the parameter of a point in rm if and only if v(t) 2 m.= min[v(x2-xo) ,v(xo-xl)l= 1 t 5 u - ,. u t & u 3 - . . . I This series convergesThis means that -x2-x12x -X2 1is a unit.= v(x-x)x2-x0 is a unit and"l^o -=0 (mod 0). Therefore,x2-x1On the other hand, if v(x -X ) = v(x2-x), we may write1 0Let P 1' P 2 have parameters tl,t2 resp. Put 5 = e(tl),E = ~ ( t ) nl , = n(P), and n2 = n(P2). Let P3 = P 1 + P2 and haveparameter t3, We need to express the addition law in terms of theparameters; that is, we need t3 in terms of tl and t2.We may assume nl 5 n2.Since g(x) is a polynomial, x2-x2 divides g(x2)-g(xl).g(x)-g(x)is an integer. x 2 xo (mod P), so x2-xo  0 (mod P 1.2x2-X2Thus the product 50 (mod 0). However g(xl) is a unit, so the sum isa unit.£ ÂY -Y1In either case, ÑÑà is a unit. We knowx2-x12 2Y2+Y1 Y2 -Y1 1- = --where y  y (mod p). Therefore,x2-x1 X2-X1 Y2-YlSOIf we let 8 = 1By the addition formula,-- - -y2-y1, we get - =2 3 3t tt1t2tt2^-tl t2 81i=:1 x 2 -X 1 t1t2(t1tt2)The right side of the above may be expressed as a power series usingt ...2 2We know v(tl) j n, v(t2) 2 n, and u(tl tt1t2tt2^) = v(tl ) 2 2n1.2 2Thwçfore v(tlt2(tl ttlt2+t22)) = v(tl)tv(t2)tv(tl ttlt2tt22) 2 3nltn2.(1-u)-1/2 = 1 t F U t ^'

and. If p # 2, 9 is an integer (since the denominators of the y are powers12 2of 2.) Then v(tlt2(tl +tlt2+t2 )9) 2 3n1+n2For nl> 1 the series converges, and theni-n1(21-4) 5 i-(21-4) = 4-i 5 2, since the series starts at i = 2.2Again we have v2(tlt2(tl +t1t2+t2)) 2 3n +n2. But now,2 2v (t t (t +t t +t )9)2 3nl+n2-2. Mow if Pl,P2 e rm with m2 1,2 1 2 1 1 2 2v2(t3-(t 1+t 2 )) 2 Sm-2. Proceding as before, v(t(1P)-tt(P)) i Sm-2,(t(2P)2 21^7 - 1) 2 4"-3'v2(# - 1) ; 4.-3.SoagainforP e rand 1 =rps p If r, we haven(1P) = n(P) + s.Consider now multiples of a point P. By induction on 1 in theabove equation. we get (writing t for the parameter of P):v(t(1P) - tt(P)) 2 Sm. If 1 = p, v(t(pP) - P ~(P)) 2 Sn, and soThen, by induction on s, v t(pSp) ( ~P t(P)1, (w- 1) 2 4n - 1. Thus for all integers- 1) 2 4n - 1. Therefore, for P c rl (i.e. n 2 11, if1 = rps where pj r then we have v(t(1P) - 1t(P)) = v(t(tP) - rpSt(p))= v(# - 1) + virpStip)>-> (bn-1) + s > n + s.That is, n(JlP) = n(P) t s.For P = 2, 6 is not in general an integer. Therefore, we mustmake sure that it is not too bad---that is, that v,(B) is not toomuch below zero.1 1 2 - iGoing back to (l~u)~" = 1 + Ã - p + &.I' -... = 161. ,is0iwe see that v2(6i)i -21. In the expansion for e(t), the u tern expandsto terms in t4i (and higher order terns). The coefficient of thet i term is ; so v (y ).> -21, whereby v2(yi)> -i.2 2121-3 2i-3- t2 -tlHe have 9 = 7- -- i=2^ 5-5*2i-3- 2i-3" ('2 t1 ) = v2(t-"-4+tlt'~-5+...+t 1 '1-5- > v2(t121-4) (2i-4)n.Thus, v2(ith term of 8) 2 (2i-4)nl-i.What this means is that n acts on the points of rm in exactlythe same way v acts on the p-adic integers.following theorem:In fact, we have theTheorem 3.5 For m? 1, rm is isomorphic to the additive group ofp-adic integers.&IWe need to show that there exists a Po E rm such that anyP e rm can be uniquely expressed as P = CPo where ( is a p-adic integer.m m+1From the values of t 5 0 (mod p ) choose a to 5 0 (mod p ).iLet Po be the point with to as its parameter. Let P. = p Po and tibe the corresponding parameter.n(Pi) = n(P) + 1, i.e. P E Fmi.t . s 0 (mod pm+i).The preceding result givesSo,Let P be any point in r , and let t s 0 (mod pm) be its parameter.Let no be the unique integer nod p such that t noto (mod pm+l).~ e P^ t= P - n oo Then t() = t(pl)) 5 t - n ot (mod psm)(1) (mod pm^ 1.S0'P E^1.m+ 2Let n be the unique integer mod p such that t") z nltl (mod p ).Let P(2) = P(l) -(2)n E + let P(~) have parameter t .Continue by induction:5 0

Here I n ip i is a p-adic integer, unique since the ni are unique mod p.Corollary 3.6 A point P E I' of finite order is not in I'. That is,It must have integer coordinates.These results over p-adic fields have interesting consequencesfor the group of points on the curve over the field of rational numbers.- -Theorem 3.7 (Nagell-Lutz) Let yz = x + Ax + B be non-singular andhave integer coefficients. Then all rational points P = (x,y) offinite order have integer coordinates such that y = 0 ory21 - 4 - ~ 2~7 ~ ~ .If P is of finite order in the group of rational solutions,%of finite order in the group of p-adic solutions for each p.Thus by the above corollary, x and y are integers in every p-adic field.But then they must be integers in the field of rationals.If P is of order 2, then y = 0.Otherwise, consider the point 2P. It is non-zero and of finiteorder. Thus it too has integer coordinates. The addition law givesf'(x) 2the x-coordinate of 2P as (-1 - 2x. For this to be an integer2vwe must have 2y1f '(XI and then y If '(XI.- 4 - ~ 2~7 ~ ~But we have the identity= f(x)P(x) + fV(x)Q(x) given in the proof of Theorem 3.3.y2 = f(x) so certainly y[f(x). Now, y f(x) and y[f'(x).ylf(x)~(x)+ fv(x)Q(x). That is yl-4f14 - 2 7~~.Footnoted ReferencesChapter 1 Chapter 2Therefore,'~assels, p. 211. '~orevich and Shafarevich, Chapter 1.ate, Chapterl.2~assels, p. 242."~assels, p. 210.3~assela, p. 211.Chapter 3'tassels, p. 213'~utz, pp. 239-244.BibliographyBorevich, 2.1.. and I.R. Shafarevich, Number Theory, Academic Press,New York, 1966, 435 pp.Cassels, J.W.S., "Diophantine Equations with Special Reference toElliptic Curves", Journal of the London Mathematical Society,Vol. 41, London, 1966.. 193-291.Lutz. Elizabeth. "Sur lvequation y2 = x3-Ax4 dans les corns D-adioues".~ournal fur die -reine uid angewandte ~athematik,~oi. 177,Walter de Gruyter E Co., Berlin, 1937, pp. 238-247.Tate, John, Rational Points on Elliptic Curves, Haverford College,1961, 107 pp.Student paper presented at the meeting of Pi Mu Epsilon in Eugene,Oregon, August, 1969.A DECIMAL APPROXIMATION TO n UTILIZING A POWER SERIESTim Golian and John HannekenOhio UniversityCAN A CIRCLE BE SQUARED? This question has puzzled mankind sinceantiquity. Even before the 17th century mathematicians believed thatthe key to answering this question lie in a very special number - pi,the ratio of the circumference of a circle to its diameter. Sincethat time, mathematicians have tried to find a unique value for pi.Their attemps can be divided into three distinct periods.In the first period, which was from the earliest times to themiddle of the 17th century, the principle aims of mathematicians'studies were directed toward the approximation of pi by calculationsof perimeters or areas of regular inscribed and circumscribed polygons.From the middle of the 17th to the middle of the 18th century,the calculus of infinite series was utilized in the development ofexpressions for pi.The last period, extending more then 150 years, pertained primarilyto investigating and characterizing pi. In 1761. J.H. Lambert provedthe irrationality of pi and in 1882 transcendence was established byF. Lindeman.In the following development of IT the specific objective relatesto the second period, and thus the basic relations introduced in thatera will be examined. Early expressions such as:do not converge rapidly enough for practical use. For example, thelatter expression, according to Newton, would require 5 BILLION termsto accurately calculate the value of pi to 20 decimal places. Theserelations were replaced by relations based upon the power seriesx5 x 7arctan (x) = x - *Â + - - - + . . . (-1 x < 11, which was discovered3 5 7by James Gregory in 1671.There are nine Important relations based on Gregory's series.These are:1 14 2 3#1. = arctan - + arctan -Charles Hutton - 17761 1#2. = = 4 arctan - - arctan - John Machin - 17064 5 239

1 1 1#3. -^ = 8 arctan -- 4 arctan -- arctan -4 10 5151 3#4, 5 = 5 amtan 7 + 2 arctan -1 1#5. v- = 4 arctan à - arctan - + arctan -79 Euler - 1755S. Klingenstierna - 1730Euler - 17644 70 991 1 1#6. ^- = arctan -+ arctan -+ arctan - L.K. Schulz von Strassnitzky -4 2 5 818441 1#7, = 2 arctan 7 + arctan - 7 Button - 17761 1#8. v- = 3 arctan - + arctan t arctan S.L. Loney - 18934 4 201 123g1#9. = = 12 arctan - + 8 arctan ¥ge - 5 arctan - Gauss4 18The or-oofs of these nine relations follows easily from the nextexample. The relations are found in "The Evolution of Extended DecimalApproximations to n," Wrench, Jr., J.W., The Mathematics Teacher,December, 1960, pp. 644 - 650, which did not contain the proofs.1 1 1$1. SHOW: 2 arctan - = arctan - + arctan -10 5 5151Let: ~zarctan- o < A < ;10C = arctan - 515tan B + tan Ctan (B- 1 - (tan B)(tan C)1 1- +-tan (B + C) = 5 51520tan (B + C) = Etan (2A) = 2 tan A 21 - tan A20tan (2A) = Etan (2A) = tan (B + C)1 1 12A=BtC Therefore: 2 arctan10- = arctan -5+ arctan -5151 1#2. SHOW: = 4 arctan ,- - arctan - 2391Let: A = arctan - 5 O L A e L ~B = arctan - 239 l O < , B < i2 tan (2A)tan (4A) =1 - tan 2 (2A)4 tan A21 1 1or: arctan = 2 arctan - - arctan g^g-1022 2(1 - tan A )- ~ 4 tan Atan (4A) = 1 - - (4 tan A)(l - tan A)tan A21 - 4 tan A - -4 24tan (4A) =(57(d -120 --(d 25tan (4A) - tan Btan - = 1 + tan (4A) tan B-120 -- 124 - 4tan (4A - B) = 119 239 =tan = 14tan ; = tan (4A - B)= 4~ - B4Therefore: = - 41 1arctan 5 - arctan -4 23915151arctan -14 101 1Since: = 4 arctan g - arctan -4 2391 1 1And; arctan F = 2 arctan10-- arctan -5151 1 1Therefore substituting: ^- = 8 arctan -- 4 arctan -- arctan -4 10 515 239$3. SHOW: ^- = 8 arctan - - 4 arctan - -239This, therefore, completes the proof of relation number three.

In this development relation number three will be used. Therelation was originally discovered in 1730 by S. Klingenstierna andrediscovered by Schellbach in about 1830. In 1926 it was used byC.C. Camp to evaluate IT/^ to 56 places. Pi was calculated to 10021places on a Pegasus computer by G.E. Felton on March 31, 1957, at theFerranti Computer Center in London. Thirty-three hours of computertime were required to accomplish this. A later check revealed thatthe computer erred and the result was only accurate to 7480 decimalplaces. This relation was later replaced by more efficient relations,such as relation number two.-where Ris the error in approximating the sum of theseries after its first n terms. R may also be writtenas:1 (-I)~" ai, which would also be a convergenti=ntlalternating series. As shown in Theorem #1, the magnitudeof each of the partial sums of R is less than the magnitudeof a , and consequently 1 R I < lantl 1 .After choosing which relation to use, the next logical step isto determine bounds on the error. Theorem #3: If IR~(X~)I < 5 then a decimal approximationTheorem #1:The magnitude of the partial sums of a convergent continuallydecreasing alternating series is less than the magnitudeof the first term.Proof:Consider the convergent alternating series,for arctan (xo) correct to p-1 places can be obtained byusing the first n terms of the power series (this followsdirectly from rules of round off).FIND THE NUMBER OF TERMS NECESSARY TO OBTAIN IT CORRECT TO 16 DECIMAL PLACES.1 1 1Since IT = 32 arctan -- 16 arctan - 4 arctan -10 239 -with the two partial sums,S = ( a - a,) t (a3 - a,,)t (a5 - a6) t ... t (an-l - an)Sntl = al - (a2 - a3) - (a4 - as) - ... - (an - anti).The quantities in parenthesis are positive because0 < a 1 < a for all positive integers (definition ofconvergence). Since all quantities are positive S ) 0and S 1 < al for all positive integers. Furthermoresince Sntl = Sn t anti then S < Sntl and 0 < Sn < Sntl < al.Theorem #2: -nt11f 1 (-l)it1ai=al-a2ta3- .,. t (-1) ant ...i=lis a convergent alternating series, then the error (R)in approximation the sum of the series after its firstn terms is less than the absolute value of the firstneglected term (antl). --Proof:- --+ Consider the convergent alternating series as:thwhere S , T, and Q,, are the respective remainders after the m ,wh, and v h term.than (5)(10-) for 16 place accuracy.that 132.~~By theorem #3, the total remainder must be less- 16*Tw - 4 .~~132% - 16.T~ - 4 .~~1when: m = 7 I ~ z - s ~ I< 5 *lo-17. NOW,32I~nl< 213.344 x 10-Thus choose a, w, and v such+ 161 ~ 14 ~ IQV 1 1. According to theorem #2,Since each remainder taken separately is greater than (5)(10-17)then at least one more term must be taken from each series.when: m = 8 32-Sm < 1.888 xv = 3 4-Q < 1.284 x 10-17.Therefore, 321~~1 t 161~~1t 41~~1< 3.188 x 10-17< (5)(10-17).

Thus, n will be correct to 16 decimal places when 8 terms of arctan1rn1 1and 3 terms of arctan -and 3 terms of arctan -yy are calculated515according to the idenity for n, which is: -BOUNDS ON ERROR DUB TO ROUNDING OFF.1 1 1n = 32 arctan - - 16 arctan g^g- - 4 arctan101Maximum error in arctan z: 8 terms, each with $0.5 error in the lastdigit used, or (8)(0.5) = 4; therefore, maximum error in 32 arctan1-10is (32)(4) = 128, andMaximum error in arctan & 3 terms, each with $0.5 emor in thelast digit used, or (3)(0.5) = 1.5; therefore, maximum error in116 arctan Is (16)(1.5) = 24, and1Maximum error in arctan .gg-; 3 terms, each with $0.5 error in thelast digit used, or (3)(0.5) = 1.5;14 arctan -yy is (4)(1.5) = 6,therfore, maximum error inTotal maximum error due to rounding off is 128 + 24 + 6 = 158Thermfore,calculations oust be carried out to 20 decimal places toassure 16 place accuracy.term # positive terms negative terms1 0.100000000000000000002 0.000333333333333333333 0.000002000000000000004 0.000000014285714285715 0.000000000111111111116 0.000000000000909090917 0.000000000000007692388 0.00000000000000006667sum 0.10000200011111880349 0.00033334761995677662term # positive terms negative terms1 0.001941747572815533982 0.000000002440377758283 0.00000000000000552070sum 0.00194174757282105468 0.000000002440377758280.00194174757282105468-0.000000002440377758280.00194174S13244329640116 arctan TTc-~0.03106792211909274240term id positive terms negative terms1 0.004184100418410041842 0.000000024416591787083 0.00000000000025647231sum 0.00418410041866651415 0.00000002441659178708

1 1 1n = 32 arctan -- 16 arctan -- 4 arctan -10 515 2393.18939687971718485984SOME COMMENTS ON TERMINOLOGIESRELATED TO DENSENESSR. Z. Yeh, University of HawaiiAccording to theory the answer is only correct to 16 decimal places,therefore: n as 3.1415926535897932Established value accurate to 16 decimal ~laces: rift; 3.1415926535897932(from page A9 of Handbook of Chemistry and Physics, The ChemicalRubber Co., Cleveland, 47th Ed., 1966).Student paper presented at the meeting of Pi Mu Epsilon in Eugene,Oregon, August, 1966.NEED MONEY? AND MATCHING PRIZE FUNDThe Governing Council of Pi Mu Epsilon announces a contest forthe best expository paper by a student (who has not yet received amasters degree) suitable for publication in the Pi Mu Epsilon Journal.The following prizes will be given$200. first prize$100. second prize$ 50. third prizeproviding at least ten papers are received for the contest.In addition there will be a $20. prize for the best paper fromany one chapter, providing that chapter submits at least five papers.The Governing Council of Pi Mu Epsilon has approved an increase inthe maximum amount per chapter allowed as a matching prize from $25.00to $50.00. If your chapter presents awards for outstanding mathematicalpapers and students, you may apply to the National Office to match theamount spent by your chapter--i.e., $30.00 of awards, the NationalOffice will reimburse the chapter for $15.00, etc. ,--up to a maximumof $50.00. Chapters are urged to submit their best student papersto the Editor of the Pi Mu Epsilon Journal for possible publication.These funds may also be used for the rental of mathematical films.Please indicate title, source and cost, as well as a very brief commentas to whether you would recommend this particular film for other PiMu Epsilon groups.The definitions of denseness, nowhere-denseness, and densenessin-itselfcan be very confusing to the students learning about themfor the first time. Perhaps more than anything the terminologies areat fault.The familiar topological descriptions of sets, such as compactness,connectedness, openness and closedness, are either invariant ornon-invariant with respect to subspace topologies. We recall thatgiven a topological space X a subset A is said to be dense in a subsetB if the closure of A contains B; in particular A is said to be densein X (or dense everywhere) if the closure of A is X. A subset A issaid to be nowhere-dense in B if the complement in B of the closureof A is dense in B; in particular A is said to be nowhere-dense in Xif the complement of the closure of A is dense in X. Obviously,denseness and nowhere-denseness are non-invariant concepts. Forexample, the set of all rationals is dense in the real x-axis, nowheredensein the entire xy-plane, and neither in the union of the x-axisand the first quadrant. The often used phrase "dense everywhere",though convenient, is not really appropriate. It is almost as bad asif one were to say that A is "open everywhere" when one really meansthat A is open with respect to the topology of X. The word "nowhere"can be confusing also, especially when one has to consider "nowheredense=? in the whole space X? or in some set B?" We also recallthat a subset A of a topological space is said to be dense-in-itself ifevery point of A is a limit point of A. It is not difficult to showthat if A is dense-in-itself with respect to the subspace topology ofsome set containing A, it is dense-in-itself with respect to the subspacetopology of any set containing A. Denseness-in-itself is thus aninvariant concept, and the term is suggestive of this. Only the word"dense" used here has nothing to do with the same word used earlier. Oneshould keep this in mind or else substitute a new term for "dense-initself".Hocking and Young [l] points out that the terminology "dense initself" (meaning of course dense-in-itself, since every set is triviallydense in itself) is misleading. ~u~undji [2] paranthetically calls anowhere-dense set a -. The choice of a noun instead of an adjective,however, might obscure the fact that nowhere-denseness is only a noninvariantconcept.References1. J. G. Hocking and G. S. Young, Topology, Addison-Wesley, Reading,Mass., (1961) p. 88.2. J. Dugundji, Topology, Allyn and Bacon, Boston (1966) p. 250.

LEMMA 2. (1) 3 (2).Proof: (l)z$V e > 0, 3xWx > x': Ig(x)/~(x)-Sl < e. Hencem< g(x)-l / eA(xt)e^dt + x-lg(x)- / ~g(t)-S~(t)1x'/x 0x'< E t [ / g(t)+€A(t)dtl/[xg(x)0le-^dtAs a consequence of Lemma 1, and since e is arbitrary, it followsthat (1) +p (2). which was to be shown.Lemmas 3 and 4 are devoted to showing that (2) =) (1).In Lemmas 3 and 4 we shall take u=+l if ATI;is constant, arbitrarily take v=+l).LEMMA 3. Define B(q) = A^jg sup g(xq)/g(x).exists for every q and 14 BtqU) = 1.q 5.u=-1 if Alf (if AIf (2). then B(q)CASE i. If A* then gs, and v=-1. B(q) exists for q 21 simply becauseg is non-increasing. If q > 1, then by Lemma 1 we have(x/q)g(x) < (x/q)g(x/q) < xg(x)from which we infer that ~(q-) exists and that 1 $, BC~") A q.CASE ii. If A/*, from (2) we see: VE > 0, 3x'aVq > 0, Vx > x':2by (4). where c=(C+l) exp(l/log2).Divide this by g(x), take thelim sup as x-, the limit as E+O, and note that g?I. This results in-2 -1/q1 S, B(q) 1 + 2coq(l-qlog2) e log q.The lemma follows immediately.LEMMA 4. Define CL and CH respectively as $^jg inf andA(x)& sup of ^y ; then (2) CL ^, 1/â 2 CH. In the interest of brevityonly the firstinequality willbe shown in full detail, the proof ofthe second is conceptually the same.Proof: Consider an arbitrary q (1 < q < l/log2). Define ~ ( t = ) 4(2-t-2-2t)and select an N > 2, then set H(t ) = pN(t and h(t) = H(t ) -~~-'~tt)where 0=0(q) = maxfp(q-l),p(q)}.obviously 0 < 1. FurthermoreSince p(t)+ (O&ts) and p(t)+ (ts),R(5) H(t) and h(t) are both of the form 1 vre-tvr (or > 0).r=l(6) H(t) 0 (t20) and h(t) ^, { H(t) (tk0)0 tl'cq-l,q)'For an additional ease of notation also defineIt follows from (6) that(7) 0 < J(x) > J (x) > j (x) 2 j(x).q qIt will be clear that (8) through (13) hold for any particular e > oif x is sufficiently large. Now observe thatconsequently B(q) exists.= 2-^cxq),Put q = l/log2 in (3) and obtain(4) g(xq) < (~+e)~exp(l/log2)2qg(x).Now consider 1 < q 0 and for x sufficiently largewhere c= 4[~(~~~/lo~2)+l]/lo~2 > 0. In view of (7), this leads to

of (5) [since x is sufficiently large]where the last step is a result of Lemma 4.divide by g(xqV)qJ to obtainq1(11)Combine (9) and (10) andNow let us momentarily consider J Since p"(t) 5 0 (Oit^),q'p(0) = 0 and p(1) = 1 it is evident that p(t) t (0s). ThusCombine (11) and (12) and then take limits as x-x and c+0;obtain(13) tL i, ~t~(q~">ql"~ - c (~t1)we readilyBecause N was chosen as any integer .> 2, it can be taken large enoughso that the last term in (13) is arbitrarily close to zero (recallthat 8 < 1). Finally take the limit as q+l and apply Lemma 3. Thisproves the first inequality of Lemma 4.To prove the second inequality, alter the definitions of J(x),j(x), J(x) and j(x) by replacing A with g. Then in parallel to2.for3.4.A.E. Ingham, "On Tauberian Theorems", Proc. London Math. Soc.(3) 14A (1965) 157-73.The author is indebted to editors of the Jour. Am.these references:Math. Soc.A. Edrei and W.H.J. Fuchs, Tauberian Theorems for a class ofmeromorphic functions, Proc. Int. Conf. in Function Theory, Erevan,Armenian S. S. R. (1966). 339-58.D. Drasin, Tauberian Theorems and slowly varying functions, Trans.Amer. Math. Soc. 138(1968), 333-56.5. D. Shea, On a complement to Valiron's tauberian theorem for theStieltjes transform, Proc. Amer. Math. Soc. (1969). 1-9.Student paper presented at the meeting of Pi Mu Epsilon in Eugene,Oregon, August, 1969.UNDERGRADUATE RESEARCH PROPOSALSBernard McDonaldUniversity of OklahomaDevelop a theory for the n x n matrices over a field having theproperty such that every submatrix has non-zero determinant. Isthe Vandermond matrix of this form?Two matrices over a finite field GF(~) are to be considered equivalentif they differ by a row or column permutation. Count the numberof equivalence classes and number of matrices in each class.Determine a cannonical matrix for the ring M(R) of n x n matricesover a principal ideal domain R under operation on the left byunimodular matrices and on the right by permutation matrices.Let m > 0 be square free. Take a = a + b 6 , where a and b arefrom Q, the set of rationals. Let G be the multiplicative groupof and H be Q - (0). Is the quotient G/H finitely generated?from which tH A 1/C is a simple deduction.REFERENCES1. G.H. Hardy, "Tauberian Theorems Concerning Power Series and Dirichlet'sSeries Whose Coefficients Are Positive", Proc. London Math. Soc.(2) 13 (1913) 174-91.

A CHARACTERIZATION OF HOMEOMORPHIC Tl SPACESW. M. PriestleyBeginning students of topology appreciate the following theorem,whereas writers of elementary textbooks apparently do not.THEOREM: Let (X,b), (Y,J) be tl spaces. (X,d) and (Y,J) arehomeomorphic (hordered sets., and (3 , G) are isomorphic as partiallyPROOF: (d) If f:X + Y is a homeomorphism, then I: * d definedby I(G) = (f(x) 1 x &GI for G e. ç is an order isomorphism.(6 ) If I: a *J is an order isomorphism, consider the complementarylattices and J' of closed subsets and the induced orderisomorphism 1': % * a)' defined for F e. J' by I1(F) = I(F1)', where S'denotes the complement of the set S. In Tl spaces singleton sets areclosed. They are also minimal in the sense that each is preceded byexactly one other set (the empty set 1(1) in the ordering C. An orderisomorphism sends minimal elements into minimal elements, and it thereforemakes sense to define a function f:X + Y by (f(x)} = I1((x}) forx e X. f is one-one and onto since I' is, by an elementary argumentsimilar to that given in [l]. It is a simple exercise to show that foreach F â j', f(F) = It(F), from which it follows that both f and f-1are continuous.PROBLEM DEPARTMENTEdited byLeon Bankoff, Los Angeles, CaliforniaThis department welcomes problems believed to be new and, asa rule, demanding no greater ability in problem solving than that ofthe average member of the Fraternity, but occasionally we shall publishproblems that should challenge the ability of the advanced undergraduateor candidate for the Master's Degree. Solutions should be submittedon separate, signed sheets and mailed before August 1, 1970Address all communications concerning problems to Leon Bankoff,6360 Wilshire Boulevard, Los Angeles, California 90048.PROBLEMS FOR SOLUTION232. Proposed by Solomon W. Golomb, University of So. Calif., LosAngeles .Find a direct combinatorial interpretation of this identity:The example of X = (l}, Y = (1,2), = a! = (1(1,1(1'} shows the Tlhypothesis to be essential.Compare Kelley's final remark on p. 130 of [2].ReferencesChan Kai-Meng, An alternative formulation of an unsolved problem ofset theory, Amer. Math. Monthly, 76(1969) 53.John L. Kelley, General topology, van Nostrand, Princeton, 1955.MOV I NG?-BE SURE TO LET THE JOURNAL KNOW!Send your name, old address with zip code and new addresswith zip code to:Pi Mu Epsilon Journal1000 Asp Ave., Room 215The University of OklahomaNorman, Oklahoma 73069233. Proposed by Charles W. Trigg, San Diego, California.The director of a variety show wanted to give the female impersonatora job, but questioned his ability to dance with thehigh-kicking Folies Bergere chorus. In reply to the director'sguery, the impersonator's Spanish agent said:"SI/HE = .CAN CANCAN...,but .CAN be less than one-fourth effective in his demonstrationtoday ."If each letter of the cryptarithm uniquely represents a digitin the scale of eleven, what is the sole solution?234. Proposed by Charles W. Trigg, San Diego, California.Show that when the nine positive digits are 2 8 7distributed in a square array so that no column, 6 1 4row, or unbroken diagonal has its digits in order 5 3 9of magnitude, the central digit must always be odd.235. Proposed by James E. Desmond, Florida State University.Prove that an+l divides (ab + c) (ad)n - ctad)" for integersa > 0, b, c, d > 0 and n > 0.

Proposed by Erwin Just, Bronx Community College.If k is a positive integer, prove that (616k+2/2) - 1 is nota prime.Proposed by Leonard Barr, Beverly Hills, California.The diameter of a semi-circle is divided into two segments,a and b, by its point of contact with an inscribed circle.Show that the diameter of the inscribed circle is equal to theharmonic mean of a and b.Proposed by David L. Silverman, Beverly Hills, California.A necessary and sufficient condition that a triangle exist isthat its sides, a, b, and c satisfy the inequalities (1) a < b + c,(2) b < a + c, (3) c < a + b. Express (1). (2), and (3) ina single inequality.Proposed by David L. Silverman, Beverly Hills, California.A pair of toruses having hole-radius = tube-radius = 1 arelinked. a) What is the smallest cube into which the torusescan be packed? b) What convex surface enclosing the linked toruseshas the smallest volume? c) What convex surface enclosing thelinked toruses has the smallest area? d) What is the locus ofpoints in space equidistant from the two links?SOLUTIONS(Spring 1969) Proposed by Gregory Wulczyn, Bucknell University.Prove that a triangle is isosceles if and only if it has a pairof equal ex-symmedians. (Editorial note: See Mathematics Magazine,Problem 637, November 1966, May 1967 and January 1968, for thecorresponding problem involving symmedians.)Solution by the Proposer.Let a, b, c denote the sides opposite vertices A, B, C of thetriangle and let x and x denote the lengths of the ex-symmediansissuing from A and B and terminated by the opposite sides.I. If a = b, we haveb sin C"a = sinoc sin A"b = sinop. 1711Then, since b sin C = c sin B,b sin C c sin B c sin Axa=-=sin(B-C)=sin)=xb11. If x = xb, then[Davis, "Modern College Geometry",214. (Spring 1969) Proposed by Charles W. Trigg, San Diego, California.Find the unique nine-digit triangular number Adigits and for which n has the form abbbb.Solution by the Proposer.In Awhich has distinct= n(n + 1)/2, the last three digits of n(n + 1) determinethe last two digits of A . Thus we find that for b = 0, 3,6, 9, duplicate digits terminate A . NowTherefore, since A has nine digits, n < [ /2(987654321) 1 = 44444,and a < 4. Furthermore, n > [ /2(102345678) I = 14307. Consequently,there are only seventeen possible values of n, all of which yielda A having duplicate digits except A25555 = 326541790.Answers (without solutions) were also supplied by Carl A. Argila,TRW Inc., Houston, and by Kenneth A. Leone, Michigan StateUniversity.215. (Spring 1969) Proposed by Leon Bankoff, Los Angeles, California.In an acute triangle ABC whose circumcenter is 0, let D, E, Fdenote the midpoints of sides BC, CA, AB, and let P, Q R denotethe midpoints of the minor arcs BC, CA, AB of the circumcircle.Show thatSolution by Alfred E. Neumann, New York City.2It is known that OD+OE+OF = R + r. Since 1 cos (A/2) = 2 + r/2R2and 1 sin (A/2) = 1 - r/2R, we haveAlso solved by Guy Gardner, USAFAcademy, Colorado; Gregory Wulczyn,Bucknell University; and theproposer.bsinC csinA csinB~-=sin(A-C)=slnoIt follows that sin A sin (B-C) = sin B sin (A-C), which simplifiesto sin (A-B) = 0. Hence A = B, and the triangle is isosceles.The proposer also supplied a geometric solution.

216. (Spring 1969) Proposed by Erwin Just, Bronx Community College.Prove that the Diophantine equationx9 + 2y9 + 3z9 + 4w9 = khas no solution if k E (11, 12, 13, 14, 15, 161.Solution by the Proposer.Since 0(27) = 18, x18 5 1 (mod 27) when (x, 27) = 1. This9implies (x9 - l)(x + 1) 2 0 (mod 27), from which it may readilybe concluded that x 9 5 *1 (mod 27). On the other hand, if(s,27) # 1, then it must follow that x 9 5 0 (mod 27). Thus,in all cases either x 9 s 0 (mod 27), x 9 Z 1 (mod 271, or x 9 2 -1(mod 27).As a result, when the given Diophantine equation is viewed asa relation among the integers (modulo 271, it is apparent thatnone of the permitted values of k will enable the equation tobe true. Since there can be no solutions (modulo 27), it followsthat the given equation has no solutions in integers.217. (Spring 1969) Proposed by C.S. Venkataram, Sree Kerala VannaCollege, Trichur, South India.A transverse common tangent of two circles meets the two directcommon tangents in B and C. Prove that the feet of the perpendicularsfrom B and C on the line of centers are a pair of common inversepoints of both the circles.Solution by the Proposer.Let the direct common tangents meet in A. Then the two circlesare plainly the incircle and excircle opposite to A of triangleABC. Therefore let us denote their centers by I, I , respectively..Let M, N be the feet of the perpendiculars from B, C on 11,the line of centers, and let Y, Z be the points of contact ofthe incircle with AB, AC respectively. Join BI, CI.Adopting the usual notation for a triangle ABC, we obtain readilythat: IN = CI cos NIC = CI cos (A/2 C/2) = CI sin (B/2)* IM = BI cos BIM = BI cos (A/2 + B/2) = BI sin (C/2).- --BTherefore IN-IM = (BI sinF)(CI sin:) = IY-IZ = r 2So N,M are inverse points with respect to the circle (I).Similarly, they are inverse points with respect to the circle(I,).218. (Spring 1969) Proposed by Charles W. Trigg, San Diego, California.Find the three 3-digit numbers each of which is equal to theproduct of the sum of its digits by the sum of the squares ofits digits.Solution by the Proposer.- - -of a^ + b^ + c^ is attained when a = b = c.If three digits, a, b, c, have a fixed sum, the minimum value3(5)[3(~~)] > 1000, then a + b + c < 15.2 2N = (a + b + c)(a + b + c2) = (a + b + c)2 2(a + b + c)(a + b + c 2 - 1) E 0 (mod 9).Since(mod 9). soWe need consider only those digit sets whose sum E 0, and thosethe sum of whose squares E 1 (mod 9). In the latter case, onesquare must be= 1 and each of the other two squares 5 0 (mod 9).It is necessary to examine only the twenty-four sets, 009, 018,027, 036, 045, 117, 126, 135, 144, 225, 234, 333, 001, 008,031, 038, 061, 068, 091, 331, 338, 361, 391, 661, to see if theproduct of the sum of the digits by the sum of the squares ofthe digits in any of these sets is equal to one of the sixpermutations of the set.The three solutions are:803 = ll(73).133 = 7(19); 315 = 9(35); andAlso solved by Carl A. Argila, TRW Systems, Houston and by KennethLeone, Michigan State University.219. (Spring 1969) Proposed by Stanley Rabinowitz, Polytechnic Instituteof Brooklyn.2Consider the following method of solving x - llx + 36x - 36 = 0.Since (x3 - llx 2 + 36x)/36 = I, we may substitute this valuefor 1 back in the original equation to obtain2x3 - llx + 36x(x3 - llx 2 + 36x)/36 - 36 = 0,3 -or x4 - lox + 25x 2 - 36 = 0, with roots -1, 2, 3, and 6.find that x = -1 is an extraneous root.Generalize the method and determine what extraneous roots aregenerated.Solution by Charles W.Trigg, San-Diego, California.The polynomial equation'f(x) = 0 has a constant term a . When"the method" is applied to this equation by multiplying the terman-kxk by 1, that is, by [f(x) - an]/(-an),we havekf(x) - an-.x + a^xk[f(x) - an]/(-an) = 0.This simplifies to(an-kxk - an) f(x) = 0.WeAlso solved by Alfred E. Neumann, New York City, who found theproblem stated but not solved in Forder's "Higher Course Geometry",page 182, problem 48.

Consequently, the extraneous roots introduced by "the method"are the roots of an-kxk = a n"Also solved by the Proposer.220. (Spring 1969) Proposed by Daniel Pedoe, University of Minnesota.a) Show that there is no solution of the Apollonius problem ofdrawing circles to touch three given circles which has only sevensolutions. b) What specializations of the three circles willproduce 0, 1, 2, 3, 4, 5, and 6 distinct solutions?The solution to problem 220 will appear in the next issue.221. (Spring 1969) Proposed by Murray S. Klamkin, Ford ScientificLaboratory.Determine 8 vertices of an inscribed rectangular parallelepipedin the sphere(x - xl)(x - x2) + (y - yl)(y - y2) + (z - zl)(z - z2) = 0.Solution by Charles W.Trigg, San Diego, California.Obviously, the following eight points fall on the surface ofthe sphere:A(xls Y s zl)s B(xls y1, z2)s CCx2, y1, z2)s D(x2, y1, zl)sA'(xls Y , zl)$ B'(xls Y , 2). C'(xs Y , 2)s Dt(x2, Y , z).Clearly, MI = lyl- y21 = BBI = cct = DD',- z 1 = CD = A'B' = C'D',AB = [ z2AD = [x - x21 = BC = AID1 = B'C',so ABCD-A'B'C'D' is a parallelepiped. Also,2 2(A'B*)~= (x1 - xl) + (yl - y2) + (z1 - 9 = (AA-)~+ (AB)~,2 2(A'D)~ = ( X - x ) + (yl - y2) + (zl - z1l2 = AD)^ + (AA')~,and the tree face angles at A are right angles. Therefore,ABCD-A'B'C'D1 is an inscribed rectangular parallelepiped.Also solved by the Proposer.-d e fa b cvwxyzNo one of a, c, d, f can be zero.Keep M < N.To avoid duplication of pairs,If M = 101, then e may be any one of the ten digits, and d + f < 9.Thus there are 10(8 + 7 +... + 1) or 360 accompanying values ofN.If M = 102, then 2d + f < 9 and d, e, f < 5.5(4 + 4 + 3 + 1) - 1 or 59 accompanying values of N >M.Hence, there areFor other possible values of M

Solution by Murray S. Klamkin, Ford Scientific Laboratory.It follows by symmetry thatTotal for all 36 values of M 801For each M, N the corresponding MI, N' necessarily also appearsin the tabulation.B) If (abc)(def) = uvwxyz and (cba)(fed) = zyxwvu, then (c) (f) = pn,where n = p t 1. The only possible terminal duos are 2(6) = 12,3(4) = 12, 5(9) = 45, and 7(8) = 56.Now in PI, 399(499) = 199101, so 3. 4 may not be a terminal duo.Also, 299(699) = 209001, but (b2)(e6) = lOObe + 20(e + 3b) + 12,so in P the penultimate digit is not zero, which rules out 2,6 as a terminal duo.If (5b5)(9e9) or (7b7)(8e8) provide a solution, the P = PI, sothe product must be palindromic and therefore divisible by 11Hence, any solutions must come from (5b5)(979), (7b7)(858),or (737)(8e8). There are only four such solutions:where c = fi.We first transform the rectangular coordinates (x,y) into ellipticcoordinates (E,n) (see Stratton, Electromagnetic Theory, McGraw-Hill, N.Y., 1941, pp. 52-54.)Herer tr1 2£EÑ 2c. n=- 1-22cand the region of integration is c > 1, -1 < n < 1.AlsoThus,2 EZ-,,~ 2 z 112dxdy = c { - a -1 dcdn.€2 1-nzNo other solutions appear when the products (5b9)(9e5), (7b8)(8e7),(5b7)(9e8), and (5b8)(9e7) are exhausted.11. Solution by Carl A. Argila, TRW Systems, Houston, Texas.Given any three digit integer, I, we define the function B asfollows :Integrating with respect to n;Now let 6 = cosh0 giving-where [A] is the greatest integer in A. Note that B(1) is justI written backwards. We wish to find all pairs of three digitintegers, M and N, for with B(M) and B(N) are also three digitintegers and for whichB(HxN) = B(tt)xB(N).By means of a simple computer program we determine that thereare 805 distinct pairs of three digit numbers which satisfy thiscondition.83. (Spring 1956) Proposed by G.K. Horton, University of Alberta.EvaluateDifferentiating the known integralmKo(a) = f e-a cOshOdO (KO - modified Bessel function)0twice with respect to a, we obtain1K = ~ ( a )0a1 0Whence,m- -K (a) = f cosh20 e* c0sh8de.

and2I = { 2K2(2c) - $C1(2c) - K(2c) }.Now just replace c by fi .i T X = &)n=l nsTherefore, since bothandTitchmarsh (1.2.1) page 491. (Fall 1956) Proposed by Nathaniel Grossman, California Instituteof Technology.Prove that1 a 2- Hd) = n*T(n)d/nwhere T(n) denotes the number of divisors of n. a(n) is thesum of the divisors of n, and $(n) is the Euler Totient function.I. Solution by James E. Desmond, Florida State University.It is well known that a, $ andTare multiplicative number-theoreticfunctions. As shown in (Calvin T. Long, Number Theory, D.C.Heath and Co., Boston, 1965, p. 103),is multiplicative.F(n) = 1 o(n/d)$(d)d/nWe note thata(pr-s)~ps) = pr - ps-lfor any prime p and integers r .> s > 0.a2 "kWrite n in standard form, n = p, p, ...pk . ThenTherefore FCP~) = pr*~(pr),We note that the result appears without proof in History of theTheory of Numbers by Leonard E. Dickson, P. 285, and is generalizedthe corresponding coefficients of nmust be equal:Reference: E.C. Titchmarsh, The Theory of the Riemann ZetaFunction, Oxford, Clarendon Press, 1951.Also solved by Marco A. Ettrick, Brooklyn, N.Y.; Murray S. Klamkin,Ford Scientific Laboratory; Bob Prielipp, Wisconsin State University;Cary C. Todd, Buies Creek, North Carolina, and Alfred E. Neumann,New York City.111. (Spring 1960) Proposed by M.S. Klamkin, AVCO RAD, and D.J. Newman,Brown, University.It is conjectured by at most N - 2 super-queens can be placedon an NxN (N > 2) chessboard so that none can take each other.A super-queen can move like an ordinary queen or a knight.(It should have been stipulated that N is even. For N = 5,Michael J. Pascual has shown that one can place 4 super-queens.)Comment by Martin Gardner, Hasting-on-Hudson, N.Y."In 1965 a reader of Scientific American Column, Hilario FernandezLong, (of Fernandez Long y Reggini, Esmeralda 356, Buenos Aires)sent me the following counter-example to the conjecture---losuper-queens on the 10 x 10.11. Solution by Solomon W. Golomb, University of Southern California.-iFor R (s) > 2, the following identities hold:Titchmarsh (1.2.12) page 6Titchmarsh (1.3.1) page 8He said a computer program had shown this to be a unique solutionfor 10 super-queens on the 10 x 10."

Comment by the Editor.Solomon W. Golomb notes that if n 210 is either a prime or oneless than a prime, there is a construction which places n mutuallynon-attacking super-queens on the n x n board. Furthermore,for n prime, the board may even be regarded as a torus! Inthe example shown above, if a row is added above the board anda column to the left, a super-queen can be placed in the upperleft corner thus rendering the solution applicable for a torus.Also solved by George S. Cunningham, University of Maine; RichardE. Sot, University of Toledo; and Stanley Rabinowitz, Far Rockaway,N.Y.(Spring 1961) Proposed by Robert P. Rudis and Christopher Sherman,AVCO RAD.Given 2n unit resistors, show how they may be connected usingn single throw (SPST) and n single pole double throw (SPDT)(the latter with off position) switches to obtain, between asingle fixed pair of terminals, the values of resistance ofi and i-l where i = 1, 2, 3 ,..., 2n.Editorial Note:Two more difficult related problems would beto obtain i and i using the least number of only one of theabove type of switches.166. (Fall 1964) Proposed by Leo Moser, University of Alberta.Show that 5 points in the interior of a 2-by-1 rectangle alwaysdetermine at least one distance less than sec lSO.Solution by Charles W.Trigg, San Diego, California.In the 2-by-1 rectangle ABCD connect the midpoint E of a longside DC to the extremities and midpoint F of the opposite side.From A draw lines making angles of 30Â with AE, meeting DC inG and EF in H. Also, from B draw lines making 30Â angles withBE, meeting DC in K and (by symmetry) EF in H. Thus the trianglesAGH and BKH are isosceles, and consequently are equilateraltriangles inscribed in.unit squares. As may be seen from righttriangle AHF, each side of the triangles is sec lSO. The fivepoints A, G, H, K, B are as widely separated as possible in oron the boundary of the 2-by-1 rectangle. Clearly, any movementof one of these points will reduce the distance between it andat least one of the other points. Since the boundary is excludedin this problem, it follows that at least one distance betweentwo of the points is less than sec ISO.This method follows that of Dewey Duncon in dealing withsubstantially the same problem in Mathematics Magazine, 23(March, 19501, page 206.Solution by C.W.Dodge, University of Maine, Orono.The accompanying circuit is minimal since, for the series resistance2n connection, switch 2n* must be closed with all others open,and for the parallel resistance l/2n connection, all other switchesmust be closed. Thus the number of permanent connections isa maximum. We see that 2n - 1, SPST switches and 1 SPDT switch. .The series resistances are obtained by closing switch 2n* andalso switches 2n - 2 and 2n - 1, 2n - 2, 2n - 4 and 2n - 1,2n - 4, ... , 2, 2n - 1, none,'for 1, 2, 3, 4, ..., 2n - 2, 2n - 1,2n ohms resistance, respectively.The parallel resistances requireclosing switches 1 and 2n* 1 and 2 and 2n, 1 and 2 and 3 and2n*, ..., 1 through 2n, for 1, 1/2, l/3, ..., l/2n ohms resistance,respectively.Finally, observe that the lone SPDT switch does not need tohave an off position.Solution I1 by C.W. Dodge, University of Maine, Orono, Maine.First we show that 3 points on a unit square determine at leastone distance not exceeding sec lSO. The maximum distance betweenthe points occurs when one point A is at a vertex of the squareand the other two points X and Y lie on sides BC and CD of thesquare to form an equilateral triangle. By symmetry it followsthat angle BAX = lSO = angle YAD. Then AX = XY = YA = sec lSO.Since now angle YXC = 4S0, then CY > (sec 1S0)/2. Reflectingthis figure in side BC produces a 1 by 2 rectangle with 5 pointsthereon and inside determining distances of at least sec lSO.By symmetry, sec lSO is the largest value this minimum distancecan have. It follows that 5 points all strictly interior tothe rectangle cannot obtain this minimum value.

LOUISIANA BETA, Southern UniversityClaude Eubank. Jr. Robert Johnson Luicien HirabeauEverett Gibson Stephen McGuire Phyllis NorrisRosie Hoskins Jean-Robert Mirabeau Gwendolyn VealShirley WashingtonAlona WinbushLOUISIANA EPSILON, McNeese State CollegeMEW INITIATESLeo P. Boutte Anand Datiyar Kuang-Nan LinBarbara Godwin hnathan Lalitha Bill OliverEdward Guimbellot Joseph Lee Mary Lou POllardApril Larry Till RyanALABAHA ALPHA, University of AlabamaLinda AtchleyLuther Bailey, 111Alice BarkerJames Ban*Bruce BemanRichard Boswell, Jr.Richard BradleyEdward Champion, Jr.Johnny CookStephen CurryTerry DawsonCharles Dyas, Jr.Catherine EnglemanGeorge Feigley, Jr.Betty FitchHerbert Forsythe, Jr.Eddie FridayBen J. 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