Here, we look at the Fibonacci and Lucas numbers whose Euler function is a factorial, as well as Lucas numbers whose Euler function is a product of power of two and power of thre

Luca, Florian

Stănică, Pantelimon

Calhoun, Institutional Archive of the Naval Postgraduate School

Calhoun: The NPS Institutional ArchiveFaculty and Researcher Publications Faculty and Researcher Publications2013The Euler function of Fibonacci andLucas  numbers and factorialsLuca, FlorianAnnales Univ. Sci. Budapest., Sect. Comp  41 (2013), 119-124.http://hdl.handle.net/10945/38823Annales Univ. Sci. Budapest., Sect. Comp. 40 (2013) nn–nnnTHE EULER FUNCTION OF FIBONACCI ANDLUCAS NUMBERS AND FACTORIALSFlorian Luca (Morelia, Mexico)Pantelimon Sta˘nica˘ (Monterey, USA)Dedicated to Professors Zolta´n Daro´czy and Imre Ka´taion the occasion of their L9 − 1’st birthdayCommunicated by Bui Minh Phong(Received December 22, 2012; accepted July 18, 2013)Abstract. Here, we look at the Fibonacci and Lucas numbers whose Eulerfunction is a factorial, as well as Lucas numbers whose Euler function is aproduct of power of two and power of three.1. IntroductionLet (Fn)n≥0 be the Fibonacci sequence given by F0 = 0, F1 = 1 andFn+2 = Fn+1+Fn for all n ≥ 0. Let (Ln)n≥0 be the companion Lucas sequencesatisfying the same recurrence with initial conditions, L0 = 2, L1 = 1. In ourprevious paper [2], we noticed the relationF1F2F3F4F5F6F7F8F10F12 = 11!and proved that it is the largest solution of the Diophantine equationFn1Fn2 · · ·Fnk = m1!m2! · · ·m`!Key words and phrases: Fibonacci and Lucas numbers, factorials, diophantine equations.2010 Mathematics Subject Classification: 11B39, 11D45.This paper was written during a visit of F. L. to the Applied Mathematics Department inDecember, 2012. During the preparation of this paper, F. L. was supported in part byProject PAPIIT IN104512 (UNAM), VSP N62909-12-1-4046 (Department of the US Navy,ONR–Global) and a Marcos Moshinsky fellowship.2 F. Luca and P. Sta˘nica˘in positive integers n1 < n2 < · · · < nk and m1 ≤ m2 ≤ · · · ≤ m` where by“largest” we mean that the number appearing in the left (or right) hand sideof the above equation is largest among all solutions. Here, we note thatφ(F21) = 7! and φ(L6) = 3!and conjecture that the above solutions are the largest solutions of the equationφ(Fn) = m!, respectively, φ(Ln) = m!but have no idea how to attack this problem. Instead, we putN = {n : φ(Fn) = m! for some positive integer m}and prove the following properties of the set N . Put N (x) = N ∩ [1, x]. For apositive real number x we write log x for the natural logarithm of x.Theorem 1.1. The following hold:(i) #N (x) x log log xlog x, and so N is of asymptotic density zero.(ii) The only primes in N are 2 and 3.In [1] it was shown that F9 = 34 and L3 = 4 are the largest Fibonacci andLucas numbers, respectively, whose Euler function is a power of 2. Here, weshow the following result.Theorem 1.2. The only solutions in nonnegative integers of the equationφ(Ln) = 2a3b are(n, a, b) = (0, 0, 0), (1, 0, 0), (2, 1, 0), (3, 1, 0), (4, 1, 1), (6, 1, 1), (9, 2, 2).We do not know how to find all the nonnegative solutions (n, a, b) of theDiophantine equationφ(Fn) = 2a3b.Also, we noted that φ(L30) = 5!7!, but we do not even know how to prove thatthe set of positive integers n such thatφ(Fn) = m1! · · ·m`! or φ(Ln) = m1! · · ·m`!for some integers 1 ≤ m1 ≤ · · · ≤ m` is of asymptotic density zero. We leavesuch questions for the reader.The Euler function of Fibonacci and Lucas numbers 32. The proofs2.1. The proof of Theorem 1.1(i) Let x be a large real number and γ = (1 +√5)/2 be the golden section.Let n ∈ N (x). Since(me)m< m! = φ(Fn) < Fn < γn ≤ γx,it follows that for large x we have m ≤ x/ log x. Let us denote K = bx/ log xc.For k = 1, . . . ,K, putNk(x) = {n ≤ x : φ(Fn) = k!}}.Fix k and let n1 < n2 < . . . < nt be all elements in Nk(x). Since1 ≤ Fnφ(Fn) log logFn  log x,we get thatFntFn1=(Fntk!)(k!Fn1)=(Fntφ(Fnt))(φ(Fn1)Fn1) log x.Since γn−2 ≤ Fn ≤ γn−1 holds for all n, we get that Fnt/Fn1 ≥ γnt−n1−1.Hence,γnt−n1−1  log x yielding #Nk(x) ≤ nt − n1  log log x.Since certainlyN (x) =⋃1≤k≤KNk(x),it follows that#N (x) ≤K∑k=1#Nk(x) K log log x x log log xlog x ,which completes the proof of (i).(ii) Assume that p > 12 is in N . Then all prime factors q of Fp satisfythe relation q ≡ (5|q) (mod p), where (a|q) is the Legendre symbol of a with4 F. Luca and P. Sta˘nica˘respect to q. If q ≡ 1, 4 (mod p), then p | (q− 1) | φ(Fp). Since φ(Fp) = m! forsome integer m, we get that m ≥ p. Thus,γp > Fp ≥ φ(Fp) ≥ p! ≥ (p/e)p,an inequality which is false for any p > 12. A similar argument proves that Fpis square free. Indeed, if q2 | Fp, then q | φ(Fp), therefore m ≥ q. Since q ≡ ±1(mod p), we get that q ≥ 2p − 1 > p, and we get again that φ(Fp) ≥ q! > p!,a contradiction. Thus, Fp is square free and q ≡ 2, 3 (mod 5) for all primefactors q of Fp. Since the above congruence is true for all prime factors q of Fp,we get that 5 - φ(Fp), so that m ≤ 4. Hence, φ(Fp) ≤ 4! = 24. This is false ifFp is a prime, or if Fp has at least one prime factor > 23, or if Fp has at leastfour distinct prime factors because (2 − 1)(3 − 1)(5 − 1)(7 − 1) > 24. Hence,Fp < 233, leading to p ≤ 19. A quick search now completes the proof of (ii).Remark 1. The argument used to prove (ii) shows that for each fixed positiveinteger a, there are only finitely many primes p such that ap ∈ N . To see why,assume that p > 12 and ap ∈ N . Then every prime factor q of Fap either is aprime factor of Fa, or is a primitive prime factor of Fdp for some divisor d ofa. In the second case, either q ≡ 1 (mod p), and we getγap > Fap > φ(Fap) ≥ p! ≥ (p/e)p therefore p < eaγ,or q ≡ 2, 3 (mod 5). If this last scenario happens for all prime factors q of Fapwhich are not prime factors of Fa, we then deduce that ν5(m!) = ν5(φ(Fa)),where ν5(m) is the exponent of 5 in the factorization of m. Since certainlyν5(m!) ≥ bm/5c, we get that bm/5c ≤ ν5(φ(Fa)), so that m ≤ 5ν5(φ(Fa)) + 4.This in turn puts an upper bound on ap. For example, for a ∈ {2, 3, 4}, weget that either p < e4γ, therefore p ≤ 19, or m ≤ 4ν5(φ(Fa)) + 4 = 4, soφ(Fap) ≤ 4!, which again gives that p ≤ 19, and a quick search reveals that theonly such values of ap in N are 4 and 21.Remark 2. The conclusions of the above theorem (with the same boundsand primes membership in N ) as well as the above Remark 1 still hold if wereplace the Fibonacci numbers by Lucas numbers. One just uses the inequalitiesγn−1 ≤ Ln ≤ γn+1 valid for all n ≥ 1.2.2. The proof of Theorem 1.2Assume that n = 2αm for some odd positive integer m. We start by showingthat α ≤ 2. Assume that α ≥ 4. SinceL2α = L22α−1 − 2,The Euler function of Fibonacci and Lucas numbers 5it follows that L2α ≡ 3 (mod 4). In particular, there exists a prime factor q ofL2α such that q ≡ 3 (mod 4). Reducing the relation L22α − 5F 22α = 4 moduloq, we get that (−5|q) = 1. Since q ≡ 3 (mod 4), we deduce that (−1|q) = −1,therefore (5|q) = −1. It follows that q ≡ −1 (mod 2α). Write q = 2a3b + 1.Then since q ≡ 3 (mod 4), we get that a = 1. Thus, 2α | (q+1), or 2α−1 | 3b+1,and this is impossible for α ≥ 4 because ν2(3b+1) = 1, 2 according to whetherb is even or odd. This shows that α ≤ 3. The case α = 3 is not possible since itwould lead to L8 | Ln, hence 23 | φ(L8) | φ(Ln), a contradiction. We now lookat the prime factors of m. Since 107 | φ(L27), 41 | φ(L18) and 11 | φ(L36), itfollows that 33 - m. In fact, if α ∈ {1, 2}, then 32 - m.Now assume that p > 3 is a prime factor of m. Then L2αp has the sameproperty that its Euler function is divisible only by primes which are at most3. Let q > 2 be any prime factor of L2αp which is not a prime factor of L2α .If α = 0, then reducing the formula L2p − 5F 2p = −4 modulo q, we get that(5|q) = 1. This shows that q ≡ 1 (mod p), therefore p | φ(Lp), which is acontradiction because p > 3. This shows that the only acceptable solutionswhen α = 0 are n = 3, 9. Assume now that α ≥ 1. Reducing the formulaL22αp − 5F 22αp = 4 modulo q we get (−5|q) = 1. If q ≡ 1 (mod 4), then weget q ≡ 1 (mod p), leading to p | φ(L2αp), which is a contradiction for p > 3.So, we get that n ∈ {2, 4, 6, 12} and the solution n = 12 is not convenient.So, we need to treat the case when q ≡ −1 (mod 4) for all prime factors q ofL2αp/L2α , which leads to the conclusion that q = 2 · 3bq + 1. Moreover, q ≡ −1(mod p), therefore 2 · 3bq + 1 = aqp− 1 for some even integer aq. Further, it isclear that L2αp/L2α is square free. Thus, we get thatL2αp = L2αq1q2 · · · q`,where qi = 2 ·3bqi +1 for i = 1, . . . , `. We may assume that 1 ≤ bq1 < · · · < bq` .We thus get that3b1 | L2αp − L2α = 5F2α−1(p−1)F2α−1(p+1).Now Fm is a multiple of 3 if and only if 4 | m. Moreover, in this case, ν3(Fm) == ν3(m)+1. Since exactly one of p−1 and p+1 is a multiple of 3, and exactlyone of these two numbers is a multiple of 4, it follows thatmin{ν3(F2α−1(p−1), F2α−1(p+1)} ≤ 1,max{ν3(F2α−1(p−1), F2α−1(p+1)} ≤ 1 + max{ν3(p− 1), ν3(p+ 1)}.In particular, we deduce that if bq1 ≥ 2, then 3bq1−2 | (p − 1)/2 or 3bq1−2 || (p+ 1)/2. On the one hand, writingp =2 · 3bq1 + 2aq1, we get that 3bq1−2 | aq1 + 1, or 3bq1−2 | aq1 − 1.6 F. Luca and P. Sta˘nica˘Since (p+ 1)/2 ≥ 3bq1−2, we get that3bq1 + 1a1=p2> 3bq1−2 − 1.On the one hand, if aq1 ≥ 10, then 3bq1 + 1 > 10(3bq1−2 − 1), or 11 ≥ 3bq1−2,or bq1 ≤ 4. On the other hand, if aq1 ≤ 8, then 3bq1−2 divides one of aq1 − 1 oraq1 + 1, a number which is at most 9, so again bq1 ≤ 4. Thus, bq1 ∈ {1, 2, 3, 4},so the only possibilities are q1 ∈ {7, 19, 163}. The case q1 = 7 leads to α = 2,then p = 7, which is false because 72 cannot divide L2αp. The case q1 = 19leads to p | q1 − 1, which is false because p > 3. The case q1 = 163 leads top | 164, so p = 41. However, in this case since q = 163 divides L2ap, we getthat α = 1. In this case, 31 | φ(L82), and we get a contradiction. So, we indeedconclude that n cannot be divisible by any prime p > 3, which completes theproof of the theorem. References[1] Luca, F., Arithmetic functions of Fibonacci numbers, Fibonacci Quart.,37 (1999), 265–268.[2] Luca F. and P. Sta˘nica˘, F1F2F3F4F5F6F8F10F12 = 11!, Port. Math.(N.S.), 63 (2006), 251–260.F. LucaCentro de Ciencias Matema´ticas,Universidad Nacional Autonoma de Me´xico,C.P. 58089, Morelia, Michoaca´n, Me´xicofluca@matmor.unam.mxP. Sta˘nica˘Naval Postgraduate SchoolApplied Mathematics DepartmentMonterey, CA 93943, USApstanica@nps.edu

The Euler function of Fibonacci and Lucas  numbers and factorials

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The Euler function of Fibonacci and Lucas numbers and factorials

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