Torsten Sillke, Nov. 1996 Update: June 1998 I am interessted in the prime divisors and the period length of the linear recursion A[n+1] = 4 A[n] - A[n-1]. This recurence appears in the following problems: - Lucas-Lehmer test - Consecutively Integral Triangles on the Grid - Heronian Triangles with Consecutive Integral Sides I have stated a conjecture connecting the period length of Lucas-Lehmes's V series modulo a prime p with the factor p in series V-1. Lucas-Lehmer test ================= Knuth gives a proof of the Lucas-Lehmer test in Knuth II, 4.5.4 Theorem L: "Lucas-Lehmer Test" The proof analyzes two series: U and V. He give some relations: (25) U(0) = 0, U(1) = 1, U(n+1) = 4 U(n) - U(n-1) V(0) = 2, V(1) = 4, V(n+1) = 4 V(n) - V(n-1) (27) U(n) = ((2 + sqrt(3))^n - (2 - sqrt(3))^n)/sqrt(12) (28) V(n) = (2 + sqrt(3))^n + (2 - sqrt(3))^n Consecutively Integral Triangles on the Grid ============================================ Karl Scherer analyzes in his book "Nutts and other Cracker" in Chapter 5: Consecutively Integral Triangles on the Square Grid and their Strange Sequence of Solutions (E6) Y^2 = 3*X^2 - 3. All positive solutions of (E6) are given by: (E7) X(i) = 2*X(i-1) + Y(i-1) and X(0) = 1 (E8) Y(i) = 3*X(i-1) + 2*Y(i-1) and Y(0) = 0 or X(0) = 1, X(1) = 2, X(n+1) = 4 X(n) - X(n-1) So we find 2 * X(n) = V(n). Karl Scherer shows that triangles on the square grid with consecutive integral side length (L1,L2,L3) are L1 = 2 * X(n) - 1 L2 = 2 * X(n) L3 = 2 * X(n) + 1 Heronian Triangles with Consecutive Integral Sides ================================================== See the article of Kevin Brown below. I. Period length of A[n+1] = 4 A[n] - A[n-1] (modulo p): -------------------------------------------------------- We shall indicate the behavior of this Lucas sequences modulo a prime p. p = 2 period_length(A,2) | 2 period_length(U,2) = 2 period_length(V/2,2) = 2 p = 3 period_length(A,3) | 6 period_length(U,3) = 3 period_length(V,3) = 2 p >= 5 period_length(A,p) | p - (3/p) where (3/p) is the Legendre symbol which is 1 if p = 1, -1 (modulo 12) -1 if p = 5, -5 (modulo 12) This can be found in (Ribenboim, 1996, Chap 2, Equations IV.30 and IV.31) where the general 2nd order recurence A[n+1] = P A[n] - Q A[n-1] is analyzed. II. Prime divisors of V(n)+c: ----------------------------- n V(n)-1 prime factors 0 1 - 1 3 prime 2 13 prime 3 51 3 * 17 4 193 prime 5 723 3 * 241 6 2701 37 * 73 7 10083 3 * 3361 8 37633 prime 9 140451 3 * 46817 10 524173 13 * 61 * 661 11 1956243 3 * 652081 12 7300801 prime 13 27246963 3 * 313 * 29017 14 101687053 13 * 757 * 10333 15 379501251 3 * 17 * 89 * 83609 16 1416317953 prime 17 5285770563 3 * 7753 * 227257 18 19726764301 109 * 1297 * 139537 19 73621286643 3 * 113 * 569 * 381673 20 274758382273 193 * 1201 * 1185361 . 26 ... 13*13*R (not square free) n V(n)/2 prime factors 0 1 - 1 2 prime 2 7 prime 3 26 2 * 13 4 97 prime 5 362 2 * 181 6 1351 7 * 193 7 5042 2 * 2521 8 18817 31 * 607 9 70226 2 * 13 * 37 * 73 10 262087 7 * 37441 11 978122 2 * 489061 12 3650401 97 * 37633 13 13623482 2 * 6811741 14 50843527 7 * 7 * 337 * 3079 (not square free) 15 189750626 2 * 13 * 61 * 181 * 661 16 708158977 prime 17 2642885282 2 * 1321442641 18 9863382151 7 * 193 * 7300801 19 36810643322 2 * 18405321661 20 137379191137 79 * 97 * 17927599 n V(n)+1 prime factors 0 3 prime 1 5 prime 2 15 3 * 5 3 53 prime 4 195 3 * 5 * 13 5 725 5 * 5 * 29 (not square free) 6 2703 3 * 17 * 53 7 10085 5 * 2017 8 37635 3 * 5 * 13 * 193 9 140453 prime 10 524175 3 * 5 * 5 * 29 * 241 (not square free) 11 1956245 5 * 391249 12 7300803 3 * 17 * 37 * 53 * 73 13 27246965 5 * 5449393 14 101687055 3 * 5 * 2017 * 3361 15 379501253 53 * 7160401 16 1416317955 3 * 5 * 13 * 193 * 37633 17 5285770565 5 * 2957 * 357509 18 19726764303 3 * 46817 * 140453 19 73621286645 5 * 6269 * 2348741 20 274758382275 3 * 5 * 5 * 13 * 29 * 61 * 241 * 661 Karl Scherer (karl@kiwi.gen.nz) is interessted in the prime factors of V(n)-1. Theorem1: (prime divisors) (Kevin Brown 1998) --------- (-1) Is p a prime divisor of V(n) - 1 then we have is n even then p = 1, 13 (modulo 24) is n odd then p = 1, 17 (modulo 24) or p = 3. ( 0) Is p a prime divisor of V(n) / 2 then we have is n even then p = 1, 7 (modulo 24) is n odd then p = 1, 13 (modulo 24) or p = 2. Conjecture 1: moreover every prime p = 7, 13 (modulo 24) is divisor of some element of the sequence V(n). ( 1) Is p a prime divisor of V(n) + 1 then we have is n even then p = 1, 5, 13, 17 (modulo 24) or p = 3. is n odd then p = 1, 5 (modulo 24) Proof: (-1) Kevin Brown, June 1998 even case: V(2n)-1 = V(n)^2 - 3 (a) V(2n)-1 = 3 (2U(n))^2 + 1 (b) These relations follow from the additive rule V(m+n) = V(m) V(n) - V(m-n) = 12 U(m) U(n) + V(m-n). Equations (a) and (b) are special cases of the quadratic form x^2 + Ny^2 where gcd(x,Ny)=1 with N = -3 and N = 3. But by a theorem of Euler (ref ???) we have for primes p>3 p | x^2 - 3y^2 <=> ( 3 / p) = 1 <=> p = 1, 11 (modulo 12) p | x^2 + 3y^2 <=> (-3 / p) = 1 <=> p = 1, 7 (modulo 12) As both relations must hold we conclude p = 1, 13 (modulo 24). odd case: V(2n+1)-1 = 3 ( 2 (U(n+1) - U(n))^2 - 1 ) (c) V(2n+1)-1 = 2 (U(n+1) + U(n))^2 + 1 (d) After factoring out 3 (c) and (d) are special cases of the quadratic form x^2 + Ny^2 where gcd(x,Ny)=1 with N = -2 and N = 2. But by a theorem of Euler (ref ???) we have for primes p>3 p | x^2 - 2y^2 <=> ( 2 / p) = 1 <=> p = 1, 7 (modulo 8) p | x^2 + 2y^2 <=> (-2 / p) = 1 <=> p = 1, 3 (modulo 8) As both relations must hold and for every prime p>3 we have p = 1, 5 (modulo 6) we conclude p = 1, 17 (modulo 24). Proof: (0) Kevin Brown, June 1998 even case: V(2n)/2 = 2 (V(n)/2)^2 - 1 (a) V(2n)/2 = (V(n)/2)^2 + 3 U(n)^2 (b) These relations follow from the additive rule V(m+n) = V(m) V(n) - V(m-n) and 2 V(m+n) = V(m) V(n) + 12 U(m) U(n). Equations (a) and (b) are special cases of the quadratic form x^2 + Ny^2 where gcd(x,Ny)=1 with N = -2 and N = 3. But by a theorem of Euler (ref ???) we have for primes p>3 p | x^2 - 2y^2 <=> ( 2 / p) = 1 <=> p = 1, 7 (modulo 8) p | x^2 + 3y^2 <=> (-3 / p) = 1 <=> p = 1, 7 (modulo 12) As both relations must hold we conclude p = 1, 7 (modulo 24). odd case: V(2n+1)/2 = 3 (U(n+1) - U(n))^2 - 1 (c) V(2n+1)/2 = (U(n+1) + U(n))^2 + 1 (d) Equations (c) and (d) are special cases of the quadratic form x^2 + Ny^2 where gcd(x,Ny)=1 with N = -3 and N = 1. But by a theorem of Euler (ref ???) we have for primes p>3 p | x^2 - 3y^2 <=> ( 3 / p) = 1 <=> p = 1, 11 (modulo 12) p | x^2 + y^2 <=> (-1 / p) = 1 <=> p = 1 (modulo 4) As both relations must hold we conclude p = 1, 13 (modulo 24). Proof: ( 1) Kevin Brown, June 1998 even case: V(2n)+1 = V(n)^2 - 1 (a) V(2n)+1 = 3((2U(n))^2 + 1) (b) These relations follow from the additive rule V(m+n) = V(m) V(n) - V(m-n) = 12 U(m) U(n) + V(m-n). After factoring out 3 (c) and (d) are special cases of the quadratic form x^2 + Ny^2 where gcd(x,Ny)=1 with N = -1 and N = 1. But by a theorem of Euler (ref ???) we have for primes p>3 p | x^2 + y^2 <=> (-1 / p) = 1 <=> p = 1 (modulo 4) As this relation must hold and for every prime p>3 we have or p = 1, 5 (modulo 6) we conclude p = 1, 5, 13, 17 (modulo 24). odd case: V(2n+1)+1 = 6 (U(n+1) - U(n))^2 - 1 (c) V(2n+1)+1 = 2 (U(n+1) + U(n))^2 + 3 (d) Equations (c) and (d) are special cases of the quadratic form Mx^2 + Ny^2 where gcd(Mx,Ny)=1 with (M,N) = (6,-1) and (M,N) = (2,3). But by a theorem of Euler (ref ???) we have for primes p>3 p | 6x^2 - y^2 <=> ( 6 / p) = 1 <=> p = 1, 5, 19, 23 (modulo 24) p | 2x^2 + 3y^2 <=> (-6 / p) = 1 <=> p = 1, 5, 7, 11 (modulo 24) As both relations must hold we conclude p = 1, 5 (modulo 24). The theorem of Kevin Brown is based on the reduction theorem of Euler Theorem: Quadatic Form (Euler, ???) Let p be a prime with does not divide N*M then Mx^2 + My^2 = 0 (modulo p) has a solution with (Mx,Ny) = 1 iff (-N*M / p) = 1. Conjecture2: (prime divisors and period length) (Sillke) ------------ Is p prime and p>3 then it is equivalent: -- 6 | period_length(V,p) -- p | V(n)-1 for some integer n. moreover if p | V(n)-1 with 0 < n < period_length(V,p) then n = period_length(V,p)/6 or n = 5 period_length(V,p)/6 The distribution of prime divisors mod 24 (1oo} #Primes(n,V-1, 1)/#Primes(n, 1) = 5/8 lim_{n->oo} #Primes(n,V-1,13)/#Primes(n,13) = 3/4 lim_{n->oo} #Primes(n,V-1,17)/#Primes(n,17) = 3/4 alltogether we get the density lim_{n->oo} #Primes(n,V-1)/#Primes(n) = 17/64 = 0.265625 lim_{n->oo} #Primes(n,V, 1)/#Primes(n, 1) = 2/3 Primes(n,V, 7) = Primes(n, 7) Primes(n,V,13) = Primes(n,13) alltogether we get the density lim_{n->oo} #Primes(n, V )/#Primes(n) = 1/3 = 0.333333 The set Primes(10000, V-1) \ {3} 13 17 37 61 73 89 109 113 137 157 193 229 233 241 257 281 313 397 401 409 421 433 449 457 541 569 577 593 601 613 617 641 661 673 709 757 761 769 809 829 857 877 881 929 937 953 997 1009 1021 1069 1093 1117 1153 1201 1217 1249 1297 1361 1381 1409 1429 1453 1481 1489 1549 1597 1601 1609 1621 1657 1669 1721 1741 1777 1789 1801 1861 1889 1913 1933 2053 2081 2113 2161 2269 2273 2293 2341 2377 2389 2393 2437 2441 2473 2557 2593 2633 2657 2677 2689 2713 2729 2749 2753 2777 2797 2801 2833 2857 2897 2917 2953 2969 3001 3041 3061 3109 3121 3137 3169 3209 3253 3257 3301 3313 3329 3361 3373 3449 3457 3517 3529 3613 3617 3637 3709 3733 3769 3793 3833 3853 3881 3889 3929 4021 4049 4057 4073 4129 4153 4201 4217 4241 4261 4273 4289 4297 4337 4357 4409 4481 4513 4561 4621 4657 4673 4721 4789 4801 4817 4861 4889 4909 4933 4937 4957 4969 5009 5077 5101 5113 5197 5273 5281 5297 5417 5437 5441 5449 5581 5641 5653 5657 5689 5701 5737 5749 5821 5849 5869 5881 5897 5953 6037 6073 6133 6217 6229 6257 6277 6301 6329 6353 6361 6373 6421 6449 6473 6481 6521 6529 6569 6709 6733 6781 6793 6829 6833 6841 6857 6949 6997 7001 7069 7129 7177 7193 7237 7297 7309 7321 7369 7417 7433 7481 7489 7573 7577 7621 7649 7669 7673 7681 7741 7753 7789 7793 7817 7841 7873 7933 7937 7993 8009 8017 8053 8081 8089 8101 8209 8233 8273 8293 8297 8317 8329 8353 8369 8377 8389 8513 8537 8581 8609 8641 8677 8681 8689 8713 8737 8753 8761 8821 8849 8893 8929 8969 9001 9013 9109 9133 9161 9181 9241 9277 9281 9337 9377 9397 9433 9473 9497 9521 9601 9613 9649 9661 9689 9721 9829 9833 9929 9973 Table of primes p with period_length(V,p) = p+1 (p<10000) 7 17 31 79 89 103 113 127 137 151 199 223 233 257 271 281 367 401 439 463 487 593 617 631 641 727 751 761 809 823 857 929 953 967 991 1039 1063 1087 1217 1231 1279 1303 1327 1361 1399 1409 1423 1447 1471 1481 1543 1601 1663 1759 1783 1831 1879 1889 1951 1999 2081 2143 2239 2273 2311 2383 2441 2503 2551 2633 2647 2657 2671 2729 2767 2777 2791 2801 2887 2897 2969 3041 3137 3209 3271 3319 3343 3391 3449 3463 3511 3559 3583 3607 3617 3631 3727 3823 3833 3847 3881 3929 3967 4049 4073 4111 4159 4217 4231 4241 4327 4337 4409 4423 4447 4519 4567 4591 4639 4673 4721 4759 4783 4817 4831 4889 4903 4937 4951 4999 5023 5119 5167 5273 5297 5407 5417 5431 5441 5503 5527 5623 5647 5657 5743 5791 5839 5849 5897 6007 6079 6151 6199 6247 6257 6271 6329 6343 6353 6367 6449 6473 6521 6569 6607 6703 6823 6833 6871 6967 6991 7001 7159 7193 7207 7351 7433 7481 7577 7591 7639 7649 7673 7687 7759 7817 7841 7879 7927 7937 7951 8081 8167 8191 8263 8273 8287 8297 8311 8431 8527 8537 8599 8609 8623 8647 8681 8753 8839 8849 8863 8887 8969 9007 9103 9127 9151 9161 9199 9281 9319 9343 9391 9439 9463 9473 9497 9511 9521 9631 9689 9833 9871 9929 9967 Table of primes p with period_length(V,p) = p-1 (30 with p | U[n] then p^2 | U[n]. For the Fibonacci numbers Fib[n] = U[n](1,-1) there is no prime p < 10^9 with p^2 | Fib[p - (5/p)]. For the Pell numbers Pell[n] = U[n](2,-1) the primes p < 10^8 which satisfies p^2 | Pell[p - (2/p)] are p = 13 or 31 or 1546463. This has been done by Williams (1982). References: ----------- - Christian Ballot; Density of Prime Divisors of Linear Recurrences. Vol 115, No. 551 of the Memoirs of the AMS, 1995. ISBN 0-8218-2610-7 MR 95i:11110 - Brillhart, Montgomery, & Silverman Tables of Fibonacci & Lucas Factorizations Math. Comp. 50 (1988) - R. D. Carmichael; On the numerical factors of the arithmetic forms a^n +- b^n. Annals of Math., 15 (1913) 30-70 - H. Hasse; "Uber die Dichte der Primzahlen p, f"ur die eine vorgegebene ganzrationale Zahl a != 0 von gerader bzw. ungerader Ordnung mod p ist. Math. Nann., 168 (1966) 19-23 - Donald E. Knuth; The Art of Computer Programming, Vol. II - J. C. Lagarias; The set of primes dividing the Lucas number has density 2/3. Pacific J. Math., 118 (1985) 19-23 - Rudolf Lidl, Harald Niederreiter; Introduction to finite fields and their applications. Cambridge University Press, Cambridge, 1994 ISBN 0-521-46094-8 - Paulo Ribenboim; The New Book of Prime Number Records. Springer, New York, 1996, 3rd ed. ISBN 0-387-94457-5 (Chap 2 Sect IV: Lucas Sequences) (Chap 5 Sect VIII: Primes and Second-Order Linear Recurrence Sequences) - Paulo Ribenboim; Gauss and the class number problem. Proceedings of the International Symposium on Mathematics and Theoretical Physics (Guaruja, 1989), 3--63, Sympos. Gaussiana Ser. A Math. Theoret. Phys., 1, Inst. Gaussianum, Toronto, ON, 1990. MR 92h:11002 (Reviewer: Duncan A. Buell) 11-03 (01A55 11E16 11R11 11R29) - Karl Scherer; Nutts and other Cracker. (A book on geometrical problems with shapes for beginners and advanced puzzlists with strong emphasis on dissections) Self-published and distributed (New Zealand), 1994 - H. C. Williams; A note on the Fibonacci quotient F[p-e]/p. Can. Math. Bull., 25 (1982) 366-370. MR 84a:10013 ------------------------------------------------------------------------------ From: Kevin Brown Subject: Re: Heronian Triangles with Consecutive Integer Sides Date: 29 May 1998 The existence of infinitely many triangles with integer sides and areas is well known. (Of course, ordinary Pythagorean triples lead immediately to such triangles, so it's usually stipulated that the triangle must not be right-angled.) For example, in 598 AD Brahmegupta noted that for any integers a,b,c we have an oblique triangle with edges of length L1 = a^2 c + b^2 c L2 = a^2 b + c^2 b L3 = (a^2 - bc)(b + c) whose area is A = abc(b+c)(bc-a^2). Another reference that might be of interest is the result found by the Japanese mathematician Nakane Genkei in 1722. He gave a recurrence for an infinite family of triangles with consecutive integer sides and integer areas. This family of solutions is L1 L2 L3 Area ---- ---- ---- -------- 3 4 5 6 (The original Pythagorean triple) 13 14 15 84 (Heron's original example) 51 52 53 1170 193 194 195 16296 723 724 725 226974 2701 2702 2703 3161340 10083 10084 10085 44031786 etc. where each sequence of edge lengths satisfies the recurrence L1[n] = 4 L1[n-1] - L1[n-2] + 2 L2[n] = 4 L2[n-1] - L2[n-2] L3[n] = 4 L3[n-1] - L3[n-2] - 2 and the areas satisfy the recurrence A[n] = 14 A[n-1] - A[n-2] which is exactly the kind of recurrence one normally encounters in the solution of a Pell equation. See Dickson's History of the Theory of Numbers (vol 2) for much more about rational triangles. _________________________________________________________ | MathPages /*\ http://www.seanet.com/~ksbrown/ | | / \ | |___________/"The thing which hath been is that which ____| shall be, and that which is done is that which shall be done." ------------------------------------------------------------------------------ 95f:11098 11Txx 94A60 94B27 Lidl, Rudolf (5-TAS); Niederreiter, Harald (A-OAW) Introduction to finite fields and their applications. Revision of the 1986 first edition. Cambridge University Press, Cambridge, 1994. xii+416 pp. $44.95. ISBN 0-521-46094-8 ------------------------------------------------------------------------------ The first edition has been reviewed [MR 88c:11073]. From the preface: "The subject of finite fields has undergone a spectacular development in the last 10 years. Great strides have been made, especially in the computational and algorithmic aspects of finite fields which are important in the rapidly developing areas of computer algebra and symbolic computation. The numerous applications of finite fields to combinatorics, cryptology, algebraic coding theory, pseudorandom number generation and electrical engineering, to name but a few, have provided a steady impetus for further research. Thus, the subject grows inexorably. Nevertheless, we hope that this book will continue to serve its purpose as an introduction for students, since it is devoted mainly to those parts that have a certain quality of timelessness, namely the classical theory and the standard applications of finite fields. "This book has been out of print for some time, but because of the continuing demand we want to make it available again. We have taken this opportunity to revise the book slightly. Historical and bibliographical notes have been added to each chapter, the bibliography has become more detailed, and the misprints in the original edition have of course been corrected." ------------------------------------------------------------------------------ 95i:11110 11N64 11B37 Ballot, Christian(1-AZ) Density of prime divisors of linear recurrences. (English. English summary) Mem. Amer. Math. Soc. 115 (1995), no. 551, viii+102 pp. If $(u\sb h)$ is a binary integer linear recurrence sequence one may define the set $P(u)$ of its "prime divisors" as those primes $p$ dividing some $u\sb h$, and the density of $P$ (if it exists) as $\delta(u) = \lim\sb {N\to \infty} {\vert P(u) \cap [1, N]\,\vert /\pi(N)}$. This memoir develops a method proposed by Hasse and refined and extended by Lagarias and provides the precise value of $\delta(u)$ for many interesting sequences. The method is based on two crucial points: certain group properties of linear recurrence sequences variously noticed by M. Ward and by R. R. Laxton and here studied in detail and generalised by the author, and Chebotarev's density theorem. The author then turns to integer linear recurrence sequences of arbitrary order $n$. In order to apply the ideas and techniques of the binary case, it seems appropriate in the general case to consider "maximal prime divisors" in place of prime divisors, namely those primes $p$ with $u\sb h \equiv \cdots \equiv u\sb {h+n-2} \bmod p$ for some integer $h$ (of course this is just the definition of "prime divisor" if $n=2$). Although a wide range of particular sequences are successfully studied there is as yet no general result on the existence of the density $\delta(u)$. Reviewed by A. J. van der Poorten ------------------------------------------------------------------------------ Chap 2 Sect IV: Lucas Sequences (P. Ribenboim, 1996) Definition of the general Lucas sequences: Let P, Q be nonzero integers. Consider the polynomial X^2 - PX + Q; its discriminant is D = P^2 - 4Q and its roots are a = (P + sqrt(D))/2, b = (P - sqrt(D))/2. So a + b = P, a b = Q, a - b = sqrt(D). I shall assume that D != 0. Define the sequenes of numbers U[n](P,Q) = (a^n - b^n)/(a - b) and V[n](P,Q) = a^n + b^n for integral n. In particular, U[0](P,Q) = 0, U[1](P,Q) = 1, while V[0](P,Q) = 2, V[1](P,Q) = P. The sequences U(P,Q) and V(P,Q) are called the 'Lucas sequences associated to the pair (P,Q)'. The second sequence is also called the 'companion Lucas sequence', with the given parameters. Algebraic relations: -------------------- To simplify notations, let's write U[n]=U[n](P,Q) and V[n]=V[n](P,Q). (0) Negations: U[-n] = - U[n]/Q^n, V[-n] = V[n]/Q^n, Q^n U[m-n] = - Q^m U[n-m], Q^n V[m-n] = Q^m V[n-m]. (1) Quadratic relations: V[n]^2 - D U[n]^2 = 4 Q^n, U[n]^2 - U[n-1] U[n+1] = Q^(n-1). (2) Conversion formulars: for n integral D U[n] = V[n+1] - Q V[n-1] V[n] = U[n+1] - Q U[n-1] (3) Addition formulars: for n, m integral U[m+n] = U[m] V[n] - Q^n U[m-n] = U[m] V[n] + Q^m U[n-m] V[m+n] = V[m] V[n] - Q^n V[m-n] = D U[m] U[n] + Q^n V[m-n] = V[m] V[n] - Q^m V[n-m] = D U[m] U[n] + Q^m V[n-m] (m) Multisection: V[n](V[k],Q^k) = V[k*n] Algebraic relations: special case Q=1 -------------------- V[-n] = V[n] U[-n] = -U[n] V[m+n] = V[m] V[n] - V[m-n] = D U[m] U[n] + V[m-n] V[2n] = V[n]^2 - 2 V[2n] = D U[n]^2 + 2 V[2n+1] = (P+2) (U(n+1) - U(n))^2 - 2 V[2n+1] = (P-2) (U(n+1) + U(n))^2 + 2 V[n] = a^n + 1/a^n U[n] = (a^n - 1/a^n)/(a - 1/a) (m) Multisection: V[n](V[k](P,1),1) := V[k*n](P,1) P' = V[k](P,1) = a^k + 1/a^k Q' = 1 ------------------------------------------------------------------------------ 97c:11021 11B39 Ribenboim, Paulo(3-QEN); McDaniel, Wayne L.(1-MOSL) The square terms in Lucas sequences. (English. English summary) J. Number Theory 58 (1996), no. 1, 104--123. Let $P$ and $Q$ be relatively prime odd integers such that $P\sp 2-4Q>0$. The Lucas sequences $\{U\sb n\}$ and $\{V\sb n\}$ are given by $U\sb 0=0$, $U\sb 1=1$, $U\sb n=PU\sb {n-1}-QU\sb {n-2}$ and $V\sb 0=2$, $V\sb 1=P$, $V\sb n=PV\sb {n-1}-QV\sb {n-2}$. Then the authors prove that $U\sb n$ [resp. $V\sb n$] is a square only if $n=0,1,2,3,6,12$ [resp. $n=1,3,5$]. It is also shown that $U\sb n$ [resp. $V\sb n$] is twice a square only if $n=0,3,6$ [resp. $n=0,3,6$]. Further, the authors give explicit information on $P$ and $Q$ whenever a term of the above Lucas sequences is a square or twice a square. For this information, we refer to the paper. The proofs depend on an elementary method which is an extension of that used by Cohn. If you have any suggestions on this problems or any comments please mail me. mailto:Sillke@mathematik.uni-bielefeld.de