250 problems in elementary number theory sierpinski

250 PROBLEMS IN ELEMENTARY NUMBER THEORY WACLAW SIERPINSKI ISBN ().444.()()()71·2 250 Problems, in Elementary Number Theory - WACLAW SIERPINSKI "250 Problems in Elementary Number Theory" presents problems and their solutions in five specific areas of this branch of mathematics: divisibility of numbers, relatively prime numbers, arithmetic progressions, prime and composite numbers, and Diophantic equations There is, in addition, a section of miscellaneous problems Included are problems on several levels of difficulty-some are relatively easy, others rather complex, and a number so abstruse that they originally were the subject of scientific research and their solutions are of comparatively recent date All of the solutions are given thoroughly and in detail; they contain information on possible generalizations of the given problem and further indicate unsolved problems associated with the given problem and solution This ancillary textbook is intended for everyone interested in number theory It will be of especial value to instructors and students both as a textbook and a source of reference in mathematics study groups 250 PROBLEMS IN ELEMENTARY NUMBER THEORY Modern Analytic and Computational Methods in Science and Mathematics A GROUP OF MONOGRAPHS AND ADVANCED TEXTBOOKS Richard Bellman, EDITOR University of Southern California Published R E Bellman, R E Kalaba, and Marcia C Prestrud, Invariant Imbedding and Radiative Transfer in Slabs of Finite Thickness, 1963 R E Bellman, Harriet H Kagiwada, R E Kalaba, and Marcia C Prestrud, Invariant Imbedding and Time-Dependent Transport Processes, 1964 R E Bellman and R E Kalaba, Quasilinearization and Nonlinear Boundary-Value Problems, 1965 R E Bellman, R E Kalaba, and Jo Ann Lockett, Numerical Inversion of the Laplace Transform: Applications to Biology, Economics, Engineering, and Physics, 1966 S G Mikhlin and K L Smolitskiy, Approximate Methods for Solution of Differential and Integral Equations, 1967 R N Adams and E D Denman, Wave Propagation and Turbulent Media, 1966 R L Stratonovich, Conditional Markov Processes and Their Application to the Theory of Optimal Control, 1968 A G Ivakhnenko and V G Lapa, Cybernetics and Forecasting Techniques, 1967 G A Chebotarev, Analytical and Numerical Methods of Celestial Mechanics, 1967 10 S F Feshchenko, N I Shkil', and L D Nikolenko, Asymptotic Methods in the Theory of Linear Differential Equations, 1967 11 A G Butkovskiy, Distributed Control Systems, 1969 12 R E Larson~ State Increment Dynamic Programming, 1968 13 J Kowalik and M R Osborne, Methods for Unconstrained Optimization Problems, 1968 14 S J Yakowitz, Mathematics of Adaptive Control Processes, 1969 15 S K Srinivasan, Stochastic Theory and Cascade Processes, 1969 16 D U von Rosenberg, Methods for the Numerical Solution of Partial Differential Equations, 1969 17 R B Banerji, Theory of Problems Solving: An Approach to Artificial Intelligence, 1969 18 R Lattes and J.-L Lions, The Method of Quasi-Reversibility: Applications to Partial Differential Equations Translated from the French edition and edited by Richard Bellman, 1969 19 D G B Edelen, Nonlocal Variations and Local Invariance of Fields, 1969 20 J R Radbill and G A McCue, Quasilinearization and Nonlinear Problems in Fluid and Orbital Mechanics, 1970 26 W Sierphlski 250 Problems in Elementary Number Theory, 1970 In Preparation 21 W Squire, Integration for Engineers and Scientists 22 T Pathasarathy and T E S Raghavan, Some Topics in Two-Person Games 23 T Hacker, Flight Stability and Control 24 D H Jacobson and D Q Mayne, Differential Dynamic Processes 25 H Mine and S Osaki, Markovian Decision Processes 27 E D Denman Coupled Modes in Plasms Elastic Media and Parametric Amplifiers 28 F A Northover, Applied Diffraction Theory 29 G A Phillipson Identification of Distributed Systems 30 D H Moore, Heaviside Operational Calculus: An Elementary Foundation 250 PROBLEMS IN ELEMENTARY NUMBER THEORY by W SIERPINSKI Polish Academy of Sciences AMERICAN ELSEVIER PUBLISHING COMPANY, INC NEW YORK PWN-POLISH SCIENTIFIC PUBLISHERS WARSZAWA 1970 AMERICAN ELSEVIER PUBLISHING COMPANY, INC 52 Vanderbilt Avenue, New York, N.Y 10017 ELSEVIER PUBLISHING COMPANY, LTD Barking, Essex, England ELSEVIER PUBLISHING COMPANY 335 Jan Van Galenstraat, P.O Box 211 Amsterdam, The Netherlands Standard Book Number 444-00071·2 Library of Congress Catalog Card Number 68·17472 COPYRIGHT 1970 BY PANSTWOWE WYDAWNIcrwO NAUKOWE WARSZAWA (pOLAND) MIODOWA 10 All rights reserved No part of this publication may be reproduced stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher, American Elsevier Publishing Company, Inc., 52 Vanderbil.t Avenue, New York, N Y 10017 PRINTED IN POLAND 113 SOLUTIONS We may assume that u ~ v, which implies U k+ v k + u v and consequently V k- < 2k; thus v k is odd, we have v = 1, and :!:k = ::: ~v k-l , < 2k/(k-l) < (because k > 3) Since ~ t/'-2(u-l) > (U-l)k-l, It follows that (u-l )k-l < 2\ which yields u-l < 3, hence, in view of the fact that u is odd, U = or u = The relation u = is impossible since we would then have Uk+V k = 2, contrary to (1) The relation U = is impossible, too, since it would yield tl'+v k u+v which is > 2k_1 (for k > 3) Suppose now that k is an even positive integer Since u and v are odd, the number ri'+v k gives the remainder upon division by 4, which is impossible since the left-hand side of (I) is divisible by This completes the proof 236* Proof due to A Rotkiewicz If 21n, then for positive integers k and I the number (2k+ 1)n+(21+ l)n is a sum of two nth powers of positive integers; as a number of the form 4t+2, it is not a difference of two squares; since 21n, it is not a difference of two nth powers of positive integers, either On the other hand, if 2,r n, then the numbers (211 + 1)211k = (2k+l)n+(2k)n, where k = 0, 1, 2, , are not the differences of two nth powers of positive integers In fact, if we had (2 n+1)2nk = xll_yn for positive integer x and y with x > y, then the numbers Xl = x/(x, y) and Yl = y/(x, y) would be positive integers, and being relatively prime, could not be both even It easily follows that 2,r (~-yi)/(Xl-Yl) and since we must have 114 250 PROBLEMS IN NUMBER THEORY which implies that We have, however, Xll-Yl > II ~-l~ 3n-1 XI-YI (since we cannot have Xl = 2, for then we would have Yl = 1, and 2"-11211 +1, which is impossible) We would therefore have 3n - < 2n +l, which is impossible for n ~ 237 We shall use the well-known formula rZ+22 + +n2 = n(n+l)6(2n+l) We have to find the least integer n > such that n(n+l)(2n+l) = 6m where m is a positive integer We shall distinguish six cases: n = 6k, where k is a positive integer Our equation takes on the form k(6k+l) (12k+l) = m2 • The factors on the left-hand side are pairwise relatively prime, hence they all must be squares If k = 1, then 6k + is not a square The next square after is If k = 4, we have 6k+ = 52, 12k+ = 72, and consequently, for n ~ 6k = 24 the sum 12+22+ +242 is a square of a positive integer, 70 n = 6k+ 1, where k is a positive integer In this case we have (6k+l) (3k+l) (2k+l) = m2 , and each of the numbers 2k+l, 3k+l, and 6k+l (which are pairwise relatively prime) must be a square The least k for which the number 2k+l is a square is k = 4; in this case, however, we have n = 6k+l > 24 n = 6k+2, where k is an integer ~ o We have in this case (3k+l) (2k+l) (12k+5) = m2 , and the numbers 3k+l, 2k+l, and 12k+5 (as pairwise relatively prime) must be squares If we had k = 0, the number 12k+5 would not be a 115 SOLUTIONS square On the other hand, for positive integer k, we have, as before, k ~ 4, hence n = 6k+2 > 24 n = 6k+3, where k is an integer ~ O In this case we have (2k+ I) (3k+2) (12k+ 7) = m2; we easily see that the numbers 2k+l, 3k+2, and 12k+7 are pairwise relatively prime, hence they must be squares We cannot have k = 0, 1,2 or since in this case the number 3k+2 would not be a square We have, therefore, k ~ 4, which implies n = 6k+3 > 24 n = 6k+4, where k is an integer ~ O We have in this case (3k+2) (6k+5) (4k+3) = m2, where the numbers 3k+2, 6k+5, and 4k+3 are pairwise relatively prime, hence they must be squares We cannot have k = 0, 1,2,3 since then the number 3k+2 would not be a square We have, therefore, k ~ 4, and consequently n = 6k+4 > 24 n = 6k+5, where k is an integer ~ O We have in this case (6k+5) (k+l) (12k+11) = m2, and the numbers 6k+5, k+l and 12k+11 are pairwise relatively prime, hence they all must be squares We cannot have k = 0, 1,2, since in this case the number 6k+5 would not be a square We have, therefore, k ~ 4, and n = 6k+5 > 24 We proved, therefore, that the least integer n > for which 12+ +22+ +n2 is a square is n = 24 REMARK It is rather difficult to show that n = 24 is the only positive integer for which 12+2z+ +n2 is a square On the other hand, the sum 13 +23 + +n3 is a square for every positive integer n, but one can prove that it is not a cube of a positive integer for any n 238 All positive integers except 1,2,3,5,6,7,10,11,13,14,15,19,23 It is easy show that none of the above thirteen numbers is a sum of a finite number of proper powers (these are successively equal to 22,23,3 2,24 = 42, 52, 33,25,62, ••• ) Now let n be a positive integer different from any of the above thirteen numbers 116 250 PROBLEMS IN NUMBER THEORY If n = 4k, where k is a positive integer, then the number n is a sum of k numbers 22 If n = 4k+ 1, then, in view of n ::F and n ::F 5, we can assume that k ~ 2; then n = 4k+l= 32 +4(k-2), where k-2 is an integer ~ o If k = 2, then n = 32, while if k > 2, then n = 32 +22+ +22, where the number of terms equal to 22 is k-2 If n = 4k+2, then since n is different from numbers 6, 10, and 14, we have k ~ and n = 4k+2 = 32 +32 +4(k-4) Again it follows that the number n has the desired property Finally, if n = 4k+3, then since n::F 3,7,11,15,19, and 23, we have k ~ and n = 32 +32 +32 +4(k-6), which again implies that n has the desired property 238a We have = 32_23, = 33_5 2, = 27_5 3, = 53_112 = 23_ _ 22, = 32-22, = 27_112, = 24-2\ = 52_42, 10 = 133-37 REMARK We not know whether the number is a difference of two proper powers It has been conjectured that every positive integer has a finite ~ number of representations as the difference of two proper powers 239 If a2 +b2 = c2 , where a, b, and c are positive integers, then multiplying both sides of this equality by the number a2(4I1Z-I)b411(2n+ 1)(11-1 )C4I1Z(2n-I) we obtain [(a2nb(2n+I)(I-I)c"(2n-I»2n]2+ [(a2n+lb2nZ-lc2nZ)2n-1]2 = [(a2n-lb2n(II-I)ClIlZ-2n+ 1)2n+ 1]2 240 There is only one such positive integer, namely n = We easily check that this number satisfies the equation (n-l)!+1 = n2 , and we also check that the numbers n = 2, 3, and not satisfy this equation For n = we obtain n2 > 6n-4 and we show by induction that the same inequality holds for every integer n ~ If n is an integer ~ 6, then (n-I) !+1 since n2 > 6n-4 > 2(n-l) (n-2) = 2(nZ-3n+2) > n2 Thus, we cannot have (n-I) '+1 = n2 for integer n ~ REMARK We know only two positive integers n > such that nZI(n-l) '+1, 117 SOLUTIONS namely numbers 13 and 563, and we not know whether there are more such numbers, or whether there are finitely many of them We know that every such number must be a prime Let us also note that for n = 5,6, and the numbers (n-I) '+1 are squares (of numbers 5, 11, and 71 respectively), and we not know whether there are any other such numbers 241 If for some integer n > I we had tn- tn = m where m is a positive integer, we would have (n -1)n2 = (2m)2, and since n2-1 and n2 are relatively prime, each of them would have to be a square, which is impossible since there are no two squares of integers, whose difference would be equal to one Let now n be a given positive integer The equation r-n(n+l)y2 = has infinitely many solutions in positive integers x and y In fact, one of these solutions is x = 2n+ and y = 2, while if for some positive integers x and y we have x 2-n(n+l)y2 = 1, then also [(2n+ l)x+2n(n+ l)y]2- n(n+ 1) [2x+ (2n+ l)y]2 If x and yare positive integers such that x 2-n(n+ l)y2 tnt2tny2 = t ntny2(2tny2+1) = t;y2,x2 = = = 1, then (tnyX)2 For instance, for n = we get t3t24 = 3OZ, t312400 = (3 '20·49)2, and so on 242 We have 210 = 1024> 103• It follows that 21945 = 25(21°)194 > 10 10J.194 = 10583 • Thus, 221945 > 210583 = (210)10582 > 103.10582, and the number of digits of the last number is greater than 10582 The number 1947 +1 has obviously the same number of digits as the number 1947 = 10 21946, and since the decimal logarithm of equals log102 = 0,30103 , we have 21946 = 101946loglOz = 10585,8 , and it follows that our number has 586 digits REMARK number The number F 1945 is the greatest known composite Fermat 118 250 PROBLEMS IN NUMBER THEORY 243 The number 211213 -1 has (in decimal system) the same number of digits as 211213 , as it differs only by one from the latter Thus, it suffices to compute the number of decimal digits of 211213 If a positive integer n is of the form n = lOX where x is real (of course, x ~ 0), then, denoting by [xl the greatest integer ~ x, we have lOX ~ n < lO[X] +1 , and it follows that the number n has [xl + decimal digits We have 211213 = 10112131011102, and since 10g102 = 0,30103 , we have 3375 < 11213log10 < 3376 Thus, the number 211213 (hence also 211213 _1) has 3376 decimal digits 244 We have 211212(211213_1) = 222425_211212 We compute first the number of digits of the number 222425 Since 22425 log102 = 22425 ·0,30103 = 6750,597 , we obtain (see the solution of Problem 243) the result that the number 222425 has 6751 digits, and we have 222425 = 106750 • 10°·597 Since 10°,597 > 101/ > 3, we get 106751 > 222425 > 10675°, which shows that the first digit of 222425 is ~ Thus, if we subtract from the number 222425 the number 211213, which has smaller number of digits, we not change the number of digits of the latter Consequently, the number 211212(211213_1) has 6751 digits 245 We have 3! = 6,3!! = 6! = 720,3!!! = 720! > 99! 100621 > 101242 Thus, the number 3!!! has more than thousand digits By the well-know theorem (see, for instance, Sierpinski [37, p 131, Theorem 6]), if m is a positive integer and p is a prime, then the largest power of p dividing m! is where [xl denotes the greatest integer of which divides 3!!! = 720! is ~ x It follows that the largest power 720 1+[72°] = 144+28+5+1 = 178 [ 72°]+[72°]+[ 25 125 625 ' while the largest power of dividing 720! is still greater (since already [ 7~0] = 360) It follows that the number 3!!! has 178 zeros at the end of its decimal expansion 246* The solution found by A Schinzel 119 SOLUTIONS For positive integers m which are powers of primes (with positive integer exponents), and only for such numbers In fact, if m = l', where p is a prime and k is a positive integer, then for f(x) = )(PI,!'>, in case p r x, by the Euler theorem, we have f(x) == (mod pk), while in case pix, in view of rp(pk) ~ pk-l ~ k (which can be easily shown by induction), we have rlX', and consequently, p"IXApk> Thus,j(x) == (modpk) If m is an integer> 1, and m is not a power of a prime, then m has at least two different prime divisors, p and q =F p Suppose that f(x) is a polynomial with integer coefficients, and that there exist integers Xl and Xz such that f(XI) == (mod m), while f(X2) == (mod m) We shall have, therefore, also (in view of plm and qlm) the relations f(xJ == (mod p) and f(Xl) == (mod q) Since p and q are different primes, by the Chinese remainder theorem there exists an integer Xo such that Xo == XI (mod p) and Xo == X2 (mod q) It follows that f(xo) == j(XI) == (mod p) and f(xo) == j(X2) == (mod q) The first of these congruences implies that we cannot have f(xo) == (mod m) Similarly, the second congruence implies that we cannot have f(xo) == (mod m) Consequently, f(xo) does not give the remainder upon dividing by m, nor does it give the remainder I Thus, if m is not a power of a prime, then there is no polynomial f(x) with integer coefficients which would satisfy the required conditions 247 We easily see that D < [(4m2+ I)n+m+ 1]2, hence the integral part of the number which implies that D-~ = 4mn+ I and Xl= YD-ao YD equals to ao = = (4m2+1)n+m, )lD+ao D-ao V v' Since ao is the integral part of the number D, we have ao < D < ao+ 1, which yields 2ao < VD+ao < 2ao+ I and since ao = (4mn+ I)m+n, we find 2m+ 2n 4mn+ < VD+llo D-cro 2n+1 < 2m+ 4mn+ I ; since (2n+ 1)/(4mn+ 1) ~ I, we see that the integral part of the number Xl = (V D+ao)/(D-~) is equal to al = 2m We have, therefore Xl = al+l/x2, and X2 = I/(xl-al)' 120 250 PROBLEMS IN NUMBER THEORY On the other hand, yD+ao Xl-at = 4mn+l 2m = YD- [(4mn+ l)m-n] 4mn+l ' and consequently, (4mn+l) [VD+(4mn+l)m-n] D- [(4mn+l)m-n]Z We easily check that which yields xl and since ao = VD+(4mn+l)m-n 4mn+l < yD < ao+l, or (4mn+l)m+n < VD < (4mn+l)m+n+l, we get 2m < Xz < 2m+ 4mn+l Consequently, the integral part of Xz equals az = 2m We have, therefore, X2 = az+ l/x3> which gives X3 = 1/(x2-a2) However, X2- aZ= yD+(4mn+l)m-n 4mn+l _ Y:D-(4mn+l)m-n m4mn+l Consequently, we have X3 - (4mn+l) [vD+(4mn+l)m+n] _ :ID (4 +1) + - 'D+ D-[(4mn+l)m+n]Z - J' + mn m n- V ao which implies that the integral part of X3 is 2ao, and that the number VI> has the expansion into the arithmetic continued fraction with the threeterm period, formed of numbers 2m, 2m and 2ao REMARK One can show that all positive integers D for which the expansion of yD into arithmetic continued fraction has a three:-term period are just the above considered numbers D See Sierpinski [32] 121 SOLUTIONS 24S Computing the values of functions cp(n) and den) for n ~ 30 from the well-known formulae for these functions i.e if n = qf'q~2 q~', then we easily see that the only values n ~ 30 for which cp(n) = den) are n = 1, 3, S, 10, IS, 24, and 30 We have here cp(I) = d(l) = 1, cp(3) = d(3) = 2, cp(8) = deS) = 4, cp(lO) = d(lO) = 4, cp(IS) = d(IS) = 6, cp(24) = d(24) = 8, cp(30) = d(30) = REMARK cp(n) cp(n) It was proved that there are no other solutions of the equation we have = den) in positive integers n It can be shown that for n > 30 > den); see P6lya and Szego [15, Section VIII, problem 45] 249 We easily check that for positive integer k and integer s;?: we have ( _1) ( _1) _ s+ (I.!) + k 1+ k+l 1+ k+s - 1+ k (1) A positive rational number w-l can be always represented in the form w-1 = where m and n are positive integers (not necessarily relatively prime) and n > g It suffices to take k = nand s = m-l; then the righthand side of (1) will be equal to w In this way we obtain the desired decomposition for the number w 250* We shall first prove that every integer k ;?: can be in at least one way represented in the form (1) where m is a positive integer, and the signs + and - are suitably chosen The assertion holds for the number since = 12+22_32+42_52_62+72 It is also true for the numbers 1, 2, and since = 12, = _12_22_3 2+42, = -12+22,4 = _12_22+3 Now, it suffices to prove that our theorem is true for every positive integer k, and since it is true for numbers 0, 1, 2, and 3, it suffices to prove that if the theorem is true for an integer k ;?: 0, it is also true for the number k+4 Suppose, then, that the theorem is true for the number k; thus, there exists 122 250 PROBLEMS IN NUMBER THEORY a positive integer m such that with the suitable choice of signs have relation (1) Since we have (m+l)2-(m+2)2-(m+3)2+(m+4)2 = + and - 4, we (2) it follows from (1) that k+4 = ±12±22± ±m2+(m+l)2-(m+2)2-(m+3)2+(m+4)2, that is, our theorem holds for the number k+4 Thus, it is true for every integer It follows from (2) that for every positive integer m we have (m+I)2-(m+2)2-(m+3)2+(m+4)2-(m+S)2+ +(m+6)2+(m+7)2_(m+8)2 = o Thus, in (I) we can replace the number m by m+8, hence also by m+16, and so on This shows that every integer k can be in infinitely many ways represented in the form (I), which was to be proved REFERENCES P Anning, Scripta Mathematica, 22 (1956),227 C L Baker and F J Gruenberger, The first six millions prime numbers, The RAND Corp., Santa Monica, pub! by the Microcard Foundation, Madison, Wise., 1959 J W S Cassels, On a diophantine equation, Acta Arithm., (1960/61), 47-52 - and G Sansone, Sur Je probleme de M Werner Mnich, Acta Arithm., (1960/61), 187-190 M Cipolla, Sui numeri compositi P che verificiano Ja congruenza di Fermat a P - == (mod P), Ann Mat Pura Appl (1904), 139-160 V A Demjanenko, On sums of four cubes, (Russian), Izv Vysiich Ucebnyh Zavedenif, Matematika (1966), 63-69 L E Dickson, History 0/ the Theory 0/ Numbers, vo! II, Carnegie Institution, 1920 P ErdOs On a problem of Sierpinski, Atti Accad Nazionale dei Lincei, 33 (1962), 122-124 - Quelques problemes de la Thiorie des Nombres, Monographies de l'Enseignement Math No 6, 126, Geneve, 1963 10 S Hyyro, Ober das Catalansche Problem, Ann Univ.Turku, Ser AI, No 79 (1964) 11 12 13 14 15 16 17 18 19 20 21 22 23 D R Kaprekar, Multidigitai numbers, Scripta Math., 21 (1955),27 M N Khatri, An interesting geometrical progression, Scripta Math., 20 (1954), 57 - Three consecutive integers cannot be powers, Colloq Math., (1962), 297 G P6lya, Functions not formulas for primes, Mathem Zeitschr., (1918),144 - and G Szego, Au/gohen und Lehrsatze aus der Analysis, II, Berlin, 1925 J Reiner, Zur arithmetischen Untersuchung der Polynome, Amer Math Monthly, 50 (1943), 619 O Reutter, Elemente der Math., 18 (1963),89 A Schinzel, Sur l'existence d'un cercle passant par un nombre donne de points aux coordonnees entieres, L'Enseignement Math., (1958),71-72 - Demonstration d'une consequence de l'hypothese de Goldbach, Compos Math., 14 (1959),74-76 - Remarque au travail de W Sierpmski sur les nombres azR +1, Col/oq Math., 10 (1963), 137-138 - and W Sierpitlski, Sur les sommes de quatre cubes, Acta Arithm., (1958), 20-30 - - Sur certaines hypotheses concernant les nombres premiers, Acta Arithm., (1958), 185-208 and ibidem (1960),p 259 E S Selmer, The diophantine equation ax3+by3+ cz3 = 0, Acta Math., 85 (1951), 203-362 123 124 REFERENCES 24 W Sierpmski, Sur les puissances du nombre 2, Ann Soc Polon Math., 33 (1950), 246-251 25 - On representations of rational numbers as sums of unit fractions, (in Polish) Warszawa, 1957 26 - Sur une question concernant Ie nombre de diviseurs premiers d'un nombre naturel, Colloq Math., (1958), 209-210 27 - Teoria liczb, (Number Theory; in Polish) Cz II, Monografie Matematyczne 38, Warszawa, 1958 28 - Demonstration elementaire d'un theoreme sur les sommes de trois nombres premiers distincts, Glasnik Mat.-Fiz Astronom Drustvo Mat Fiz Hrvatske, serf II, 16 (1961), 87-88 29 - Remarques sur Ie travail de M J W S Cassels "On a diophantine equation", Acta Arithm., (1961), 469-471 30 - Sur un probleme concernant les nombres k'2"+I, Elemente der Math., 15 (1961), 73-74 31 - Sur Ies sommes des chiffres de nombres premiers, Rendi Circ Mat Palermo, Serf II, 10 (1961) 229-232 32 - liczbach naturainych D, dia kt6rych okres rozwini~a yD na ulamek lancuchowy arytmetyczny rna trzy wyrazy, (On positive integers D for which the period· of expansion of D into an arithmetic continued fraction has three terms; in Polish) Roczniki Pol Tow Matem., sere II: Wiadomosci Matem., (1962), 53-55 33 - Sur quelques consequences d'une hypothese de M A Schinzel, Bull Soc Royale ~ci Liege, 31 (1962), 317-320 34 - Sur les nombres qui sont sommes et differences de deux nombres premiers, Publ Electr Facultet, Sere Math Phys., Beograd, 84 (1963), 1-2 35 - Sur les nombres an +1, Elemente der Math., 19 (1964), 106 36 - Remarques sur un probleme de M P Erdos, Publ Inst Math Beograd, (18 (1964), 125-134 37 - Elementary Theory of Numbers, Monografie Matematyczne 42, Warszawa, 1964 y NAME INDEX Anning, N 18 Anning, P 103 Aubry L 99 Kaprekar, D R 33 Khatri, M N 42, 106 KraItchik, M Banachiewicz T 35 Bindschedler, C 29 Browkin, J 69, 80, 97, 99, 103 Lebesgue, V A 17 Lejeune-Dirichlet see Lejeune Liouville, J Cantor, M 46, 47 Cassels, J W S 78 Catalan, E C 42 Chebyshev, P L 8, 56 Cipolla, M 32 Dirichlet, P G Milkowski, A 37, 42, 56, 60, 73, 99 Moessner, A 17 Mordell, L J 95 PeB 12 P61ya, G 45, 121 Demjanenko, V A 109 Dickson, L E 99 Dirichlet, P G Lejeune 6, 7, 9, 45, 46, 52, 58, 59, 69 Reiner J 62 Reutter, O 27, 28 Rotkiewicz, A 32 39, 113 Erdos, P 10, 22, 60, 73 Euler, L 12, 25, 29, 30, 33, 45, 58, 96 Fermat, P 23, 25, 27, 30, 31, 32, 35, 40 Sansone, G 78 Schinzel, A 31, 32 39,41,46-49, 51-54, 62, 67 70, 72, 74, 76, 80, 81, 92, 94, 103, 104, 109, 111, 112, 118 Seredinsky, W N 47 Sierpiilski W 9, 31, 40, 48, 51 56, 60, 67, 90, 104, lOS, 109, 110, 118, 120 Stiffel, M 66 Suranyi, M 22 Szego, G 121 Ginga, G 25 Hogatt, V E 20 109 Hyyro, S 42 125 The late Waclaw SierpiJiski, a member of the Polish Academy of Sciences, was Professor of Mathematics at the University of Warsaw from 1919 to 1960 He received the degree of Ph.D from the University of Warsaw in 1906 and served as Professor of Mathematics at the University of Lwow from 1910 to 1919 Recipient of honorary degrees from the Universities of Lwow, Amsterdam, Tartu, Sofia, Paris, Bordeaux, Prague, and Lucknow, Professor Sierpiriski was an honorary life member of the New York Academy of Sciences His mathematical research encompassed the areas of theory of sets and theory of numbers, and his more recently published works included "General Topology" (1952) and "Cardinal and Ordinal Numbers" (1958) Modern Analytic and Computational Methods in Science and Mathematics OTHER TITLES IN THE SERIES Number 21: INTEGRATION FOR ENGINEERS AND SCIENTISTS by W Squire A treatment of analytical and numerical methods for evaluating integrals with many illustrative examples and problems The last chapter is a brief introduction to integral equations and numerical procedures for their solution Number 23: FLIGHT STABILITY AND CONTROL by T Hacker Presents a contribution to the development of the theory of flight stability and control in the context of the present state of aeronautical sciences, making use of a modern mathematical apparatus Number 24: DIFFERENTIAL DYNAMIC PROGRAMMING by David H Jacobson and D Q Mayne This book describes a method for optimizing dynamic systems The behavior of the dynamic system, which is described by a set of first order differential equations, depends on a time varying input or control The book develops several algorithms which enable an initial guessed input (as a function of time) to be successively improved until a given performance index is minimized Number 25: MARKOVIAN DECISION PROCESSES by H Mine and S Osaki Deals primarily with the mathematical theory and algorithms of standard Markovian decision processes and emphasizes linear and dynamic programming algorithms Among the related topics considered are some extended models such as semi-Markovian decision processes, general sequential decision processes, and stochastic games Number 30: HEAVISIDE OPERATIONAL CALCULUS by Douglas H Moore Sets forth a clear, formal, and mathematically rigorous basis for Heaviside operational calculus, an area of applied mathematics which provides methods for solving integral-derivative equations directly and algebraically using a table of operational forms Number 32: ApPROXIMATE METHODS IN OPTIMIZATION PROBLEMS by V F Demyanov and A M Rubinov Written in the language of functional analysis, this volume considers methods of investigating and solving certain nonlinear extremal problems Number 34: MULTITYPE BRANCHING PROCESSES by C J Mode Based in large part upon the author's own mathematical research, this volume details the most recent results in that specific area of applied probability known as the multitype branching process ...ISBN ().444.()()()71·2 250 Problems, in Elementary Number Theory - WACLAW SIERPINSKI "250 Problems in Elementary Number Theory" presents problems and their solutions in five specific areas of... asserting that there exist only finitely many prime Mersenne numbers and finitely many prime Fermat numbers 8 250 PROBLEMS IN NUMBER THEORY 91 Find all numbers of the form 2n-l with positive integer... solutions in positive integers x, y, Z, t for m = and m = 3, and find all its solutions in positive integers x, y, Z, t for m = 14 250 PROBLEMS IN NUMBER THEORY 160 ' Find all solutions in positive integers