ADVANCED PROBLEMS AND SOLUTIONS PROBLEMS PROPOSED IN THIS ISSUE

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1 ADVANCED PROBLEMS AND SOLUTIONS EDITED BY LORIAN LUCA Please sen all communications concerning ADVANCED PROBLEMS AND SOLUTIONS to LORIAN LUCA, SCHOOL O MATHEMATICS, UNIVERSITY O THE WITWA- TERSRAND, PRIVATE BAG X3, WITS 2050, JOHANNESBURG, SOUTH ARICA or by at the aress florianluca@witsacza as files of the type tex, vi, ps, oc, html, pf, etc This epartment especially welcomes problems believe to be new or extening ol results Proposers shoul submit solutions or other information that will assist the eitor To facilitate their consieration, all solutions sent by regular mail shoul be submitte on separate signe sheets within two months after publication of the problems PROBLEMS PROPOSED IN THIS ISSUE H-829 Propose by Ángel Plaza an rancisco Peromo, Gran Canaria, Spain or any positive integer k, let { k,n } n0 be the sequence efine by k,0 = 0, k, =, an k,n+ = k,n + k,n for n in the limit k + k lim k 2 kk,n+ 2 arctan + k,n k,n+ 2 k,n+2 H-830 Propose by Hieyuki Ohtsuka, Saitama, Japan or an integer n, prove that n 2 k k+ k mo n n+ n+2 n+3 k= n= H-83 Propose by Prerag Terzić, Pogorica, Montenegro Let P j x = 2 j x x 2 4 j + x + x 2 4 j, where j an x are nonnegative integers Let N = k2 m + with k o, k < 2 m an m > 2 Let S 0 = P k n an S i = S 2 i 2 for i Prove the following statement: If there exists n for which S m 2 0 mo N, then N is prime H-832 Propose by Hieyuki Ohtsuka, Saitama, Japan or positive integers n an r, fin a close form expressions for i n k= rk 3 L rk; ii n k= 2 3 k 2Lk NOVEMBER

2 THE IBONACCI QUARTERLY SOLUTIONS Diophantine equations with powers of the golen section H-796 Propose by Hieyuki Ohtsuka, Saitama, Japan, an lorian Luca, Johannesburg, South Africa Vol 54, No 3, August 206 in all solutions x, y in positive integers of the equation where α is the golen section Solution by the proposers tan α x tan α y = tan x tan y, We show that x, y = 7, 5 is the only solution Applying tangent in both sies of the equation, the left sie of it becomes whereas the right sie of it becomes tan tan α x tan α y = αx α y, + αx+y tan tan x tan y = x y + xy 2 Since 2 is rational, shoul be invariant by the action of the only nontrivial Galois automorphism of K = Q 5, which sens α to β = α Applying this to, we get that is β x β y + β x+y = x α x y α y + x+y α x y = x α y y α x α x+y + x+y 3 Assume first that x + y is o Then, x = y+ an from an 3, we get α x α y α x+y + = x αy + α x α x+y Since x > y, the left sie above is positive Thus, x is even an we get α x α y α x+y + = αx + α y α x+y α x+y or, equivalently α x+y + = αx + α y α x α y, which is false since in the last equality of fractions, the left sie is < whereas the right sie is > So, x + y is even, therefore x = y So, we get from an 3 that α x α y α x+y + = x+ αx α y α x+y +, so x + is even Hence, x is o an y is o Now, the given equation is x y + xy = αx α y α x+y + = αx y/2 α y x/2 L x y/2 α x+y/2 + α x+y/2 = L if x + y 0 mo 4; x+y/2 x y/2 if x + y 2 mo 4 x+y/2 Recall that gc a, b =, where = gca, b urther, gcl a, L b equals L if an only if a/ an b/ are both o In the contrary case, gcl a, L b {, 2} So, if x+y 0 mo 4, then x y/2 is o an x+y/2 is even, therefore gcl x y/2, L x+y/2 {, 2} Thus, the enominator of the reuce fraction L x y/2 /L x+y/2 is at least L x+y/2 /2 In case x + y 2 mo 4, we have that x y/2 is even an x+y/2 is o, therefore gc x y/2, x+y/2 x y/4, so the enominator of the reuce fraction x y/2 / x+y/2 is at least x+y/2 / x y/4 374 VOLUME 56, NUMBER 4

3 Next, we recall that ADVANCED PROBLEMS AND SOLUTIONS α n 2 n α n an α n L n α n+ hol for all n Thus, in case x + y 0 mo 4, the enominator of L x y/2 /L x+y/2 is at least as large as L x+y/2 > α x/2 3, 2 2 whereas in case x + y 2 mo 4, the enominator of x y/2 / x+y/2 is at least as large as α x+y/2 2 x y/4 = α x/4+3y/4 > α x/4 αx+y/2 At any rate, this enominator is also the enominator of x y/ + xy an is therefore + xy < x 2 We thus get that { x 2 > min α x/2 3, α x/4 }, which leas to x 75 So, all solutions have y < x 75 an we finish with a computer search or other equations with the inverse tangent of powers of the golen section, see [] [] Luca an P Stănică, On Machin s formula with powers of the golen section, Int J Number Theory, , Partially solve by Dmitry leischman An ientity with ibonomial coefficients H-797 Propose by Hieyuki Ohtsuka, Saitama, Japan Vol 54, No 4, November 206 Let n k enote the ibonomial coefficient or positive integers a, b, c, an = a+b+c, prove that a 2a 2b 2c 2 2k = a b c + 2a 2b 2c 2 a+b b+c c+a a b c Solution by the proposer Let k = In [], Melham showe that That is, 2a 2a 2b 2b 2c 2c k+a+b+c k a k b k c k a b c k+a k+b k+c = k+a+b+c a+b b+c c+a 2k a+k b+k c+k + k a k b k c k ++k = a+b b+c c+a 2k 4 NOVEMBER

4 THE IBONACCI QUARTERLY We have k = a+k 2a 2a a k 2a 2a b+k 2b 2b 2b b k 2b = a+b b+c c+a 2k 2a 2b 2c 2+ 2a c+k 2c 2c 2c c k 2c 2b 2c k k by 4 Therefore, we have a 2a 2b 2c 2 2k = 2a 2b 2c 2+ a+b b+c c+a 2a = 2a 2b 2c 2+ 2b 2c 2 + a+b b+c c+a a b c 2a 2b 2c 2 + 2a b + a c + a + a + = 2a 2b 2c 2+ 2a 2b 2c 2 + a+b b+c c+a a b c = 2a 2b 2c 2+ a 2a b 2b c 2c + a+b b+c c+a 2a a 2b b 2c c 2+ = a b c a+b+c 2a 2b 2c 2 a+b b+c c+a a b c The proposer also notice that in the same manner a a + b b + c c + a 2 2k = a b c a+b+c a + b b + c c + a a+b b+c c+a a b c 2 a k [] R S Melham, On prouct ifference ibonacci ientities, INTEGERS, 200, #A0 Also partially solve by Dmitry leischman 2 An inequality with ibonacci numbers an trigonometric functions H-798 Propose by D M Bătineţu-Giurgiu, Bucharest, an Neculai Stanciu, Buzău, Romania Vol 54, No 4, November 206 If t 0, π/2 an m 0, prove that an hol for all n sin m+2 t n sin t + n+ cos t m + cos m+2 t n cos t + n+ sin t m m n+2 L n + L n+ tan t m + tan m+2 t L n tan t + L n+ m L m n+2 cos2 t 376 VOLUME 56, NUMBER 4

5 Solution by Soumitra Monal We have Again = = sin m+2 t n sin t + n+ cos t m + cos m+2 t n cos t + n+ sin t m ADVANCED PROBLEMS AND SOLUTIONS sin 2 t m+ n sin 2 t + n+ sin t cos t m + cos 2 t m+ n cos 2 t + n+ sin t cos t m sin 2 t + cos 2 t m+ n + 2 n+ sin t cos t m by Raon s inequality n + n+ sin 2 t + cos 2 t m = n + n+ m = n+2 m L n + L n+ tan t m + tan m+2 t L n tan t + L n+ m L n + L n+ tan t m + tan 2 t m+ L n tan 2 t + L n+ tan t m + tan 2 t m+ L n sec 2 t + 2L n+ tan t m by Raon s inequality sec 2m+2 t L n sec 2 t + L n+ + tan 2 t m = L n + L n+ m cos 2 t = L m n+2 cos2 t Also solve by Dmitry leischman an the proposers An inequality with ibonacci numbers H-799 Propose by D M Bătineţu-Giurgiu, Bucharest, an Neculai Stanciu, Buzău, Romania Vol 54, No 4, November 206 Prove that n n+ n n n+ + 3n 2 + n+ n 3n n n+ + n 2 4 n n+ n+2 4 an that the same inequality with all s replace by L s hols for all n Solution by Brian Braie an Note that 2 n+ + 4 n n n = n+ + n n+ + 3 n = n+2 n+ + 3 n, 3 2 n+ + 4 n n+ + 2 n = 3 n+ + n n+ + n = n+2 3 n+ + n Then, the esire inequality is equivalent to 3 n+2 2 n3 n+ + n + 2 n+3 n + n+ 4 2 n 2 n+3 n + n+ 3 n+ + n This in turn is equivalent to n n+ 2 2 n + 4 n n+ + 2 n+ 2 0, which is clearly true Moreover, inequality hols if an only if n = NOVEMBER

6 THE IBONACCI QUARTERLY Also solve by Kenneth B Davenport, Dmitry leischman, Wei Kai-Lai an John Risher jointly, Soumitra Monal, Ángel Plaza, Hieyuki Ohtsuka, an the proposers A sum involving multinomial coefficients H-800 Propose by Mehtaab Sawhney, Commack, NY Vol 54, No 4, November 206 Let k n S k = +n 2 + +n n k + n n k j + n j n, n 2,, n k n +2n 2 + +kn k =k n,n 2,,n k Z 0 Compute S, S 2 an show that S k = 0 for all k 3 Solution by Euaro H M Brietzke We generalize the argument presente in [], page 38, introucing parameters Claim : If x, x 2, is any infinite set of parameters then k + + k n! x k k! k n! xkn n = [q n ] r x j q j 5 j k +2k 2 + +nk n=n k +k 2 + +k n=r In the above, for a formal series n0 a nq n, the notation [q n ] n0 a nq n stans for the coefficient of q n equal to a n Inee, let ψq = e tq = n=0 tn n! qn an consier the infinite prouct of formal power series P := j ψx jq j Then, Also, P = t k x k j qjk = k! j k0 n0 = q n t r n0 r0 q n k +2k 2 + +nk n=n k +k 2 + +k n=r k +2k 2 + +nk n=n P = e t j x jq j = t r x j q j r! r0 j j= t k + +k n x k xkn n k! k n! x k 6 xkn n k! k n! Comparing the coefficient of q n t r in 6 an 7, we obtain 5 Applying summation on r from to infinity to both sies of 5, it follows that k + + k n! x k k! k n! xkn n = [q n ] x j q j 8 j k +2k 2 + +nk n=n k +2k 2 + +nk n=n Now, applying 8 with x j = j +, we get k +k 2 + +k n k + + k n! 2 k 3 k2 n kn = [q n ] k! k n! + 2q + 3q 2 +, 378 VOLUME 56, NUMBER 4 r 7

7 or, S n = [q n ] q 2, from which it follows that S = 2, S 2 =, an S n = 0 for n 3 ADVANCED PROBLEMS AND SOLUTIONS Remark Other choices of values for x j in 8 might yiel interesting results as well or example, for x j = j + 2 we get k +k 2 + +k n k + + k n! 2 2k q 3 2k2 n + 2kn = [q n 3 ] k! k n! + q k +2k 2 + +nk n=n 0, if n = 0 4, if n = = 7, if n = 2 n 8, if n 3 [] N J ine, Basic Hypergeometric Series an Applications, Mathematical Surveys an Monographs 27, AMS 988 Also solve by Dmitry leishman an the proposer Errata: In the statement of H-827 the last factor insie the inner limit shoul be n m / m+ instea of n m / m NOVEMBER