[PDF] [PDF] There are No Multiply-Perfect Fibonacci Numbers - Mathematics

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1 Proof We prove the result for the case when n is multiplicatively e-perfect, the other prime number p, 2p-multiplicatively e-superperfect numbers which have the Te((p1 · p2)2p)=(p1 · p2)σ(2p)·d(2p) = (p1 · p2)3·(p+1)·4 = (p1 · p2)12·(p+1)

[PDF] On Perfect Numbers - Semantic Scholar

17 mai 2016 · Proof Let p be an integer such that 2p − 1 is a prime number We aim to show that all perfect numbers of the form 2p-1(2p − 1)? The next significant step in the answering of σ(6) = 1 + 2 + 3 + 6 = 12 σ(18) = 1 + 2 + 3 + 6 + 

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Roots that a perfect number other than 6 has digital root 1 We now provide a proof of this claim 12 is abundant, because 1 + 2 + 3 + 4 + 6 + 12 > 24 For, if p is a prime number, then 12p is necessarily an abundant number, because

[PDF] Perfect numbers - Irish Mathematical Society

Theorem 1 1f the number (2"-1) 18 prime, then the number 2n-1/2"-1) is Proof Let p = (2"-1) and suppose p 18 prime Then the 1 e (2"-1)(1+p) This equals 12"-1)2", One could say that m is perfect if the sum of all the factors of in, which is  

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p − 1 is prime, and discuss an application to even perfect numbers The proof requires us to study the field Z/qZ[ √

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previous paragraph This shows that a = 1,2 have an unique balanced ternary expansion that p > n In particular, given a positive integer n we can always find a prime larger than n; by growing n Since (12,18) = 6 50 there are no solutions

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e-superperfect numbers and prove some results on them We also p = 1 It is impossible since p is a prime number If at least an αi is equal to 1, say α1 = 1, then from (2 2), we 2 be the prime factorisation of an integer n > 1 where p1 and p2 are r = 1: (3 3) becomes 2 · (α1 + 1) = 12 · (2p − 1); it gives α1 = 12p − 5

[PDF] There are No Multiply-Perfect Fibonacci Numbers - Mathematics

We show that no Fibonacci number (larger than 1) divides the sum of its divi- sors 2 Notation For a positive integer a and a prime p we write p a/ for the exact [ 3] and [21] we get that n0 2 ¹1; 2; 3; 4; 6; 12º, so n 2 ¹3; 6; 9; 12; 18; 36º, and the

SOLUTIONS

greater than 10 contains a prime number greater than or equal to 11, which is impossible And from 1,2, ,10 we cannot select nine consecutive numbers with the required We need the fact that every integer p ≥ 2 can be represented as a2 + b2 Methods of Proof 335 1 = 12, 2 = −12 − 22 − 32 + 42, 3 = −12 + 22,

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Integers11(2011), 363-397

DOI 10.1515/INTEG.2011.027 © de Gruyter 2011There are No Multiply-Perfect Fibonacci Numbers Kevin A. Broughan, Marcos J. González, Ryan H. Lewis, Florian Luca, V. Janitzio Mejía Huguet and Alain Togbé Abstract.We show that no Fibonacci number (larger than1) divides the sum of its divi- sors. Keywords.Fibonacci Numbers, Multiply-Perfect Numbers.

2010 Mathematics Subject Classification.11B39.

- To Professor Carl Pomerance on his 65th birthday

1 Introduction

For a positive integernwe put.n/for the sum of its divisors. Given an inte- gerk, the numbernis said to be multiperfect, multiply-perfect, ork-fold perfect if.n/Dkn. Of course, ordinary perfect numbers are2-fold perfect. The single

1-fold perfect is the trivial casenD1. The3-fold perfect numbers are also called

triperfect, and only six of them are known: they are120,672,523776,459818240,

1476304896,51001180160. All of them were already known by the seventeenth

century. Several multiperfect numbers are also known for everyk11. Their number vary from thousands forkD8;9;10, to only one forkD11which has more than a thousand decimal digits and was discovered in 2001. Descartes dis- covered the first4-fold number, and Fermat the first5-fold number. Dickson"sHis- tory of the Theory of Numbers[5, p. 33-38] records a long interest in such num- bers. Seealso[7,SectionB2], orthewebpage[23]formoredetailsandreferences. Except for the well-known Euclid-Euler rule forkD2, no formula to generate multiperfect numbers is known. Lehmer [8] proved that ifnis odd, thennis per- fect just if2nis3-fold perfect. Moreover, no odd multiperfect number is known. There are several conjectures on the size ofkin what relates to the size ofn. For

example, from the maximal order of the sum of divisors function, it is known thatThis work was done during a visit of M. G. and V. J. M. H. at the Mathematical Institute of the

UNAM in Morelia, Mexico. During the preparation of this paper, F. L. was supported in part by Grants SEP-CONACyT 79685 and PAPIIT 100508, V. J. M. H. was supported in part by Grant UAM-A 2232508 and a postdoctoral position at the IFM of the UMSNH, and A. T. was partially supported by Purdue University North Central. Computation for this research was supported in part

by the Research Computing group of the Rochester Institute of Technology.Brought to you by | University of Waikato (University of Waikato)

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364Broughan, González, Lewis, Luca, Mejía Huguet and Togbéthere exists a positive constantcsuch that the inequality.n/=n > cloglogn

holds for infinitely many positive integersn, where here and from now on we use log for the natural logarithm. Contrary to this inequality, Erd os conjectured that if there were infinitely many multiperfect numbers, thenkDo.loglogn/as n! 1through multiperfect numbers. It has even been suggested there may be only finitely manyk-fold perfect numbers altogether withk3, and it is further believed that all multiperfect numbers withkD3,4,5,6, and7are known. There are several results in the literature addressing perfect and multiperfect numbers of various shapes. For example, Pomerance [20] proposed as a problem to find all positive integers such thatnŠis multiperfect. In the solution [6] to the above problem, it is shown that this happens only fornD1,3,5. In [10], it is shown that there is no Fibonacci number which is perfect and in [11], it was shown that there are at most finitely many Fibonacci numbers which are multiperfect. In [9], it is shown that no Fermat number, i.e., number of the form22nC1for some nonnegative integern, is perfect, and it is remarked that the method of proof also yieldsthatsuchnumbersarenotmultiperfecteither. Whetherbinomialcoefficients can be multiply-perfect numbers is a problem which was studied in [13], where it is shown that any fixed line through the Pascal triangle contains at most finitely many multiply-perfect numbers. In this paper, we look at multiply-perfect numbers in the Fibonacci sequence. Recall that the Fibonacci sequence.Fn/n0is given byF0D0; F1D1and F nC2DFnC1CFnfor alln0. As we have mentioned before, in [10], it was shown that there is no perfect Fibonacci number, while the main result in [11] is that there are at most finitely many Fibonacci numbers which are multiply-perfect and, at least in theory, they are all effectively computable.

Here, we prove the following result.

Theorem 1.No Fibonacci number (larger than1) is multiply-perfect. En route to the proof of Theorem 1, we prove certain results concerning the number of odd prime factors appearing at odd exponents in the factorization ofFn, as well as some estimates involving the primitive prime factors ofFn. Such results might have some interest in their own and could be useful for other Diophantine questions involving Fibonacci numbers.

2 Notation

For a positive integeraand a primepwe writep.a/for the exact exponent ofp in the factorization ofa. We write nD2tm; mDp1p;wherep1p2 p3Brought to you by | University of Waikato (University of Waikato)

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There are No Multiply-Perfect Fibonacci Numbers365areprimesnotnecessarilydistinct. WealsoputsWD2.Fn/. Inparticular,wehave:

(i) if 3n, thensD0; (ii) if 3jnbut2n, thensD1; (iii) if 6jn, thensDtC2. We write.Ln/n0for the companion Lucas sequence of the Fibonacci sequence given byL0D2; L1D1andLnC2DLnC1CLnfor alln0. There are many formulas relating the Fibonacci and Lucas numbers and which are valid for all n0, such asF2nDFnLnandL2n5F2nD4.1/n. Throughout the paper, we shallmakeuseofseveralofsuchformulas. Theycanallbeprovedimmediatelyus- ing the Binet formulas F nD nn andLnD nCnfornD0;1;:::;(1) where. ;/D..1Cp5/=2;.1p5/=2/. Throughout, we shall also use several well-known divisibility properties of the Fibonacci numbers. For example,Fa dividesFbifadividesb. Furthermore, ifajbandpis a prime number, then pjgcd.Fa;Fb=Fa/if and only ifpjb=a. These divisibility properties will be used in Section 3. We also write.n/,!.n/,.n/,.n/, andP.n/for the Euler function ofn, the number of distinct prime factors ofn, the total number of prime factors ofn, the numberofdivisorsofn, andthelargestprimefactorofnrespectively. Inparticular, with our notations, we have that.n/DtC,P.n/Dp1, and.n/.tC1/2.

3 The Exponent of2in.Fn/

IfFn> 1is multiply-perfect, then

.F n/DkFn(2) holds for some positive integerk > 1. The casekD2is impossible by the result from [10]. A short computation with MAPLE[14] reveals that there is no suchn inŒ3;200. We writeD2.k/. Clearly,2.kFn/DCs. To proceed further, weneedalowerboundfor2..Fn//. Thisisachievedbygivingalowerboundon ofFn. This is the aim of the current section. Before stating the main result of this section, let us make some observations. If

2..Fn//D0, thenFnD, or2. Theonlypossibilitiesaren2 ¹1;2;3;6;12º

(see [3] for a more general result). If2..Fn//D1, thenFnDp, or2p,Brought to you by | University of Waikato (University of Waikato)

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366Broughan, González, Lewis, Luca, Mejía Huguet and Togbéwherepis some odd prime. In the former case, eithern2 ¹4;25º, ornis prime.

Indeed, this is a result of Robbins from [21]. Observe that the casenD4is not convenient for us, sinceF4D3, so2..F4// > 1. In the latter case, suppose first thatpD3. We then getFnD6, and sinceL2n5F2nD 4, we are led to an integer point.X;Y/with positive coordinates, whereYWDLn, on one of the two curves Y

2180X4D 4:(3)

A short computation with MAGMA[2] reveals that the Diophantine equation (3) Clearly,3jn, so we can writenD3n0. SinceF3n0DFn0.5F2n0C3.1/n0/and the greatest common divisor of the two factors above is1or3, we get that either F n0D; 2; 3; 6;or5F2n

0C3.1/n0D; 2; 3; 6:

If one of the equalities from the first set of equalities holds, then by the results from [3] and [21] we get thatn02 ¹1;2;3;4;6;12º, son2 ¹3;6;9;12;18;36º, and the only suchnfor whichFnis of the form2pwith some odd primepisnD9 for whichF9D217. If one of the equalities from the second set of equalities holds, then since5F2n0C4.1/n0DL2n0, we are led to an integer point.X;Y/ with positive coordinates, whereXWDFn0, on one of the curves .5X

23/.5X24/DY2; 2Y2; 3Y2; 6Y2:(4)

With MAGMA, wegetthatthetotalityofthesixteenDiophantineequations(4)lead only ton02 ¹1;3;5º, so thatn2 ¹3;9;15º. However, sinceF15D610D2561, we get that2..F15//D4, which is not convenient for us. So far, we have seen that if2..Fn//D0, thenn2 ¹1;2;3;6;12º, while if

2..Fn//D1, thennD9;25, ornis prime. In the following result, we prove a

lower bound for2..Fn//in terms of the number of prime factors ofn.

Lemma 2.LetnD2tp1p, wherep1p2 p3are primes. The

following inequalities hold: (i)IftD0;1;2, then

2..Fn//.tC1/.1/:

(ii)Ift3, then

2..Fn//3.1/C.C1/C3.t3/:

In both instances, the factor (or term)1can be replaced byifp1> 5.

Proof.Write

F nDFmLmL2mL2t1m:(5)Brought to you by | University of Waikato (University of Waikato)

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There are No Multiply-Perfect Fibonacci Numbers367It is clear that the greatest common divisor of any two of the above numbers is in

¹1;2º. Indeed, for0, we have thatFmjF2mand

L

22m5F22mD 4;(6)

where the sign on the right depends on whetherD0or > 0. Formula (6) shows that gcd.Fm;L2m/D1; 2. It is2exactly when3jm, otherwise it is1. Similarly, if < , thenL2mjF2C1mjF2m, and by the above formula we get again that gcd.L2m;L2m/D1; 2. Again, it is2exactly when3jm.

Let us deal withFm. We assume that1. Write

F mDFp1Fp1p2F p1

Fp1piF

p1pi1 Fp1pF p1p1

We put

F p1piF p1pi1DdiforiD1;:::;; wherediis squarefree. Here, and in subsequent occasions, we putp0WD1. Since p

1;:::;pare all odd, it follows that all the numbersdiare odd except if3jm.

In this last case, withibeing the smallest index such thatpiD3, we have thatdi prime factor except whenmD3uis a power of3. Indeed, for if not, then either d iD1, ordiD2. In the first case, we get that F pF D; wherepDpi, andDp1pi1. However, this is not possible by the results of McDaniel and Ribenboim [16]. WhendiD2, we have that F 3F D2; whereDp1pi1,pi1> 3, andpiD3. Sinceis odd, we have that F

3=FDL2C1. Thus, puttingxWDLandyWDF, we are then led to the

simultaneous Pell equations x

22v2D 1;

x

25y2D 4:

In turn, any solution of the above system of simultaneous Pell equations leads to an integer point.X;Y/WD.x;vy/with positive coordinates, on the curve .X

2C1/.X2C4/D10Y2:(7)Brought to you by | University of Waikato (University of Waikato)

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