PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER. Bernadette Faye and Florian Luca

Transcription

1 PUBLICATIONS DE L INSTITUT MATHÉMATIQUE Nouvelle série tome PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER Bernadette Faye and Florian Luca Abstract We show that the only Pell numbers whose Euler function is also a Pell number are and 2 Introduction Let φn be the Euler function of the positive integer n Recall that if n has the prime factorization n = p a pa k k with distinct primes p p k and positive integers a a k then φn = p a p p a k k p k There are many papers in the literature dealing with diophantine equations involving the Euler function in members of a binary recurrent sequence For example in [] it is shown that 2 and 3 are the only Fibonacci numbers whose Euler function is also a Fibonacci number while in [4] it is shown that the Diophantine equation φ5 n = 5 m has no positive integer solutions m n Furthermore the divisibility relation φn n when n is a Fibonacci number or a Lucas number or a Cullen number that is a number of the form n2 n + for some positive integer n or a rep-digit g m /g in some integer base g [2 000] have been investigated in [ ] respectively Here we look for a similar equation with members of the Pell sequence The Pell sequence n 0 is given by P 0 = 0 P = and + = 2 + for all n 0 Its first terms are We have the following result 200 Mathematics Subject Classification: B39 Key words and phrases: Pell numbers; Euler function B Faye was supported by a grant from OWSD and SIDA F Luca was supported by an NRF A-rated researcher grant Communicated by Žarko Mijajlović 23

2 232 FAYE AND LUCA Theorem The only solutions in positive integers n m of the equation φ = P m are n m = 2 For the proof we begin by following the method from [] but we add to it some ingredients from [0] 2 Preliminary results Let α β = be the roots of the characteristic equation x 2 2x = 0 of the Pell sequence { } n 0 The Binet formula for is = αn β n α β This implies easily that the inequalities for all n 0 2 α n 2 α n hold for all positive integers n We let {Q n } n 0 be the companion Lucas sequence of the Pell sequence given by Q 0 = 2 Q = 2 and Q n+2 = 2Q n+ + Q n for all n 0 Its first few terms are The Binet formula for Q n is 22 Q n = α n + β n for all n 0 We use the well-known result Lemma 2 The relations i P 2n = Q n and ii Q 2 n 8P 2 n = 4 n hold for all n 0 For a prime p and a nonzero integer m let ν p m be the exponent with which p appears in the prime factorization of m The following result is well known and easy to prove Lemma 22 The relations i ν 2 Q n = and ii ν 2 = ν 2 n hold for all positive integers n The following divisibility relations among the Pell numbers are well known Lemma 23 Let m and n be positive integers We have: i If m n then P m ii gcdp m = P gcdmn For each positive integer n let zn be the smallest positive integer k such that n P k It is known that this exists and n P m if and only if zn m This number is referred to as the order of appearance of n in the Pell sequence Clearly z2 = 2 Further putting for an odd prime p e p = 2 p where the above notation stands for the Legendre symbol of 2 with respect to p we have that zp p e p A prime factor p of such that zp = n is called primitive for It is known that has a primitive divisor for all n 2 see [2] or [] Write P zp = p ep m p where

3 PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 233 m p is coprime to p It is known that if p k for some k > e p then pzp n In particular 23 ν p e p whenever p n We need a bound on e p We have the following result Lemma 24 The inequality 24 e p holds for all primes p p + log α 2 log p Proof Since e 2 = the inequality holds for the prime 2 Assume that p is odd Then zp p + ε for some ε {±} Furthermore by Lemmas 2 and 23 we have p ep P zp P p+ε = P p+ε/2 Q p+ε/2 By Lemma 2 it follows easily that p cannot divide both and Q n for n = p + ε/2 since otherwise p will also divide Q 2 n 8P 2 n = ±4 a contradiction since p is odd Hence p ep divides one of P p+ε/2 or Q p+ε/2 If p ep divides P p+ε/2 we have by 2 that p ep P p+ε/2 P p+/2 < α p+/2 which leads to the desired inequality 24 upon taking logarithms of both sides In case p ep divides Q p+ε/2 we use the fact that Q p+ε/2 is even by Lemma 22 i Hence p ep divides Q p+ε/2 /2 therefore by formula 22 we have p ep Q p+ε/2 2 Q p+/2 2 < αp+/2 + 2 < α p+/2 which leads again to the desired conclusion by taking logarithms of both sides For a positive real number x we use log x for the natural logarithm of x We need some inequalities from the prime number theory For a positive integer n we write ωn for the number of distinct prime factors of n The following inequalities i ii and iii are inequalities and 34 in [5] while iv is Théoréme 3 from [6] Lemma 25 Let p < p 2 < be the sequence of all prime numbers We have: i The inequality p n < nlog n + log log n holds for all n 6 ii The inequality + < 79 log x + p 2log x 2 p x holds for all x 286 iii The inequality n φn > 79 log log n + 25/ log log n holds for all n 3

4 234 FAYE AND LUCA iv The inequality holds for all n 26 ωn < log n log log n 74 For a positive integer n we put = {p : zp = n} We need the following result Lemma 26 Put S n := p p { 2 log n 25 S n < min n For n > 2 we have } log log n φn Proof Since n > 2 it follows that every prime factor p is odd and satisfies the congruence p ± mod n Further putting l n := # we have n ln p p < α n by inequality 2 giving 26 l n Thus the inequality n log α logn 27 l n < n log α log n holds for all n 3 since it follows from 26 for n 4 via the fact that the function x x/ log x is increasing for x 3 while for n = 3 it can be checked directly To prove the first bound we use 27 to deduce that 28 S n nl nl n l l n 2 ln dt n t + + n 2 + n 2 n 2 log αe 2+2/n 2 n n log log n l l n l + m n m 2 m log l n + + n n 2 Since the inequality log n > log αe 2+2/n 2 holds for all n implies that S n < 2 n log n for n 800 The remaining range for n can be checked on an individual basis For the second bound on S n we follow the argument from [0] and split the primes in in three groups: We have 29 T = p p<3n i p < 3n; ii p 3n n 2 ; iii p > n 2 ; p { n 2 + n + 2n 2 + 2n + 3n 2 < 0 3n 2n 2 + 2n < 7 3n n even n odd

5 PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 235 where the last inequalities above hold for all n 84 For the remaining primes in we have 20 p < p + m = T 2 + T 3 + m 3n p p>3n p p>3n m 3n+ where T 2 and T 3 denote the sums of the reciprocals of the primes in satisfying ii and iii respectively The sum T 2 was estimated in [0] using the large sieve inequality of Montgomery and Vaughan [3] see also page 397 in [] and the bound on it is 2 T 2 = p < 4 4 log log n + < φn log n φn φn 3n<p<n 2 4 log log n + φn where the last inequality holds for n 55 Finally for T 3 we use estimate 27 on l n to deduce that 22 T 3 < l n n 2 < log α n log n < 09 3n where the last bound holds for all n 9 To summarize for n 84 we have by and 22 S n < 0 3n + 3n n + 4 log log n + = 4 φn φn n + 4 log log n log log n + φn φn φn for n even which is stronger that the desired inequality Here we used that φn n/2 for even n For odd n we use the same argument except that the first fraction 0/3n on the right-hand side above gets replaced by 7/3n by 29 and we only have φn n for odd n This was for n 84 For n [3 83] the desired inequality can be checked on an individual basis The next lemma from [9] gives an upper bound on the sum appearing in the right-hand side of 25 Lemma 27 We have d n log d d < p n log p n p φn Throughout the rest of this paper we use p q r with or without subscripts to denote prime numbers 3 Proof of the Theorem 3 A bird e eye view of the proof of the Theorem In this section we explain the plan of attack for the proof of the Theorem We assume n > 2 We put k for the number of distinct prime factors of and l = n m We first show that 2 k m and that any putative solution must be large This only uses the fact that p φ = P m for all prime factors p of and all such primes with at most one exception are odd We show that k 46 and n > m 2 46 This is Lemma 3 We next bound l in terms of n by showing that l < log log log n/ log α +

6 236 FAYE AND LUCA Lemma 32 Next we show that k is large by proving that 3 k > n/6 Lemma 33 When n is odd then every prime factor of is congruent to modulo 4 This implies that 4 k m Thus 3 k > n/6 and n > m 4 k a contradiction in our range for n This is done in Subsection 35 When n is even we write n = 2 s n with an odd integer n and bound s and the smallest prime factor r of n We first show that s 3 that if n and m have a common divisor larger than then r {3 5 7} Lemma 34 A lot of effort is spend into finding a small bound on r As we saw r 7 if n and m are not coprime When n and m are coprime we succeed in proving that r < 0 6 Putting e r for the exponent of r in the factorization of P zr it turns out that our argument works well when e r = and we get a contradiction but when e r = 2 then we need some additional information about the prime factors of Q r It is always the case that e r = for all primes r < 0 6 except for r {3 3} for which e r = 2 but lucky for us both Q 3 and Q 3 have two suitable prime factors each which allows us to obtain a contradiction Our efforts in obtaining r < 0 6 involve quite a complicated argument roughly the entire argument after Lemma 34 until the end which we believe it is justified by the existence of the mighty prime r = for which e r = 2 Should we have only obtained say r < we would have had to say something nontrivial about the prime factors of Q a nuisance which we succeeded in avoiding simply by proving that r cannot get that large! 32 Some lower bounds on m and ω We start with a computation showing that there are no other solutions than n = 2 when n 00 So from now on n > 00 We write = q α qα k k where q < < q k are primes and α α k are positive integers Clearly m < n McDaniel [2] proved that has a prime factor q mod 4 for all n > 4 Thus McDaniel s result applies for us showing that 4 q φ P m so 4 m by Lemma 22 Further it follows from a the result of the second author [5] that φ P φn Hence m φn Thus n 3 m φn 79 log log n + 25/ log log n by Lemma 25 iii The function x x 79 log log x + 25/ log log x is increasing for x 00 Since n 00 inequality 3 together with the fact that 4 m show that m 24 Put l = n m Since m is even we have β m > 0 therefore 32 = αn β n P m α m β m > αn β n α m α l α m+n > αl 0 40 where we used the fact that αm+n α 24 < 0 40 We now are ready to provide a large lower bound on n We distinguish the following cases

7 PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 237 Case : n is odd Here we have l So /P m > α 0 40 > 2442 Since n is odd it follows that is divisible only by primes q such that zq is odd Among the first 0000 primes there are precisely 2907 of them with this property They are F = { } Since p F < 963 < 2442 < = p P m k i= q i we get that k > 2907 Since 2 k φ P m we get by Lemma 22 that 33 n > m > Case 2: n 2 mod 4 Since both m and n are even we get l 2 Thus P m > α > If q is a prime factor of as in Case we have that zq is not divisible by 4 Among the first 0000 primes there are precisely 585 of them with this property They are F 2 = { } Writing p j as the j th prime number in F 2 we check with Mathematica that = = p i p i i= which via inequality 34 shows that k 46 Of the k prime factors of we have that only k of them are odd q = 2 because n is even but one of those is congruent to modulo 4 by McDaniel s result Hence 2 k φ P m which shows via Lemma 22 that i= 35 n > m 2 46 Case 3: 4 n In this case since both m and n are multiples of 4 we get that l 4 Therefore /P m > α > 3397 Letting p < p 2 < be the sequence of all primes we have that 2000 i= p i < 74 < 3397 < P m = showing that k > 2000 Since 2 k φ = P m we get 36 n > m k i= q i To summarize from and 36 we get the following results Lemma 3 If n > 2 then i 2 k m; ii k 46; iii n > m 2 46

8 238 FAYE AND LUCA 33 Bounding l in term of n We saw in the preceding section that k 46 Since n > m 2 k we have 37 k < kn := log n log 2 Let p j be the j th prime number Lemma 25 shows that p k p kn knlog kn + log log kn := qn We then have using Lemma 25 ii that P m = k q i i= 2 p qn > p 79 log qn + /2log qn 2 Inequality ii of Lemma 25 requires that x 286 which holds for us with x = qn because kn 46 Hence we get 79 log qn + 2log qn 2 > > α l 0 40 > α l P m 0 40 Since k 46 we have qn > 3256 Hence we get log qn log > α l which yields after taking logarithms to 38 l The inequality log log qn log α qn < log n 45 holds in our range for n in fact it holds for all n > 0 83 which is our case since for us n > 2 46 > 0 25 Inserting inequality 39 into 38 we get l < log loglog n45 log α Thus we proved the following result Lemma 32 If n > 2 then l < < log log log n log α log log log n log α Bounding the primes q i for i = k Write = q q k B where B = q α q α k k Clearly B φ therefore B P m Since also B we have by Lemma 23 that B gcd P m = P gcdnm P l where the last relation follows again by Lemma 23 because gcdn m l Using inequality 2 and Lemma 32 we get 30 B m α n m α 0 log log n

9 PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 239 To bound the primes q i for all i = k we use the inductive argument from Section 33 in [] We write Therefore k q i i= Using the inequality k q i i= = φ = P m = P m = P m > x x s x + + x s valid for all x i [0 ] for i = s we get < k q i i= k i= q i < k q therefore q < k / < 3k Using the method of the proof of inequality 3 in [] one proves by induction on the index i { k} that if we put u i := i j= q j then u i < 2α 2 k log log n 3i /2 In particular q q k = u k < 2α 2 k log log n 3k /2 which together with formula 38 and 30 gives = q q k B < 2α 2 k log log n +3k /2 = 2α 2 k log log n 3k +/2 Since > α n 2 by inequality 2 we get n 2 log α < 3k + 2 Since k < log n/ log 2 see 37 we get 3 k > n 2 2 log α log2α 2 log nlog log nlog 2 log2α 2 k log log n where the last two inequalities above hold because n > 2 46 So we proved the following result Lemma 33 If n > 2 then 3 k > n/6 > 07n 2 > n 6 35 The case when n is odd Assume that n > 2 is odd and let q be any prime factor of Reducing relation Q 2 n 8P 2 n = 4 n of Lemma 2 ii modulo q we get Q 2 n 4 mod q Since q is odd because n is odd we get that q mod 4 This is true for all prime factors q of Hence 4 k k q i φ P m i=

10 240 FAYE AND LUCA which by Lemma 22 ii gives 4 k m Thus n > m 4 k inequality which together with Lemma 33 gives n > 3 k log 4/ log 3 > n 6 log 4/ log 3 so in contradiction with Lemma 3 n < 6 log 4/ log4/3 < Bounding n From now on n > 2 is even We write it as n = 2 s r λ rλt t =: 2 s n where s t 0 and 3 r < < r t are odd primes Thus by inequality 32 we have α l 0 40 < α l 0 40 < φ = + = 2 + p p p d 3 p P d d n and taking logarithms we get 3 l log α 0 39 < log α l 0 40 < log 2 + d 3 d n p P d log + p < log 2 + d 3 d n In the above we used the inequality log x > 0x valid for all x 0 /2 with x = /0 40 and the inequality log + x x valid for all real numbers x with x = p for all p P d and all divisors d n with d 3 Let us deduce that the case t = 0 is impossible Indeed if this were so then n is a power of 2 and so by Lemma 3 both m and n are divisible by 2 46 Thus l 2 46 Inserting this into 3 and using Lemma 26 we get 2 46 log α 0 39 < 2 log2 a 2 a = 4 log 2 a a contradiction Thus t so n > We now put I := {i : r i m} and J = { t} I We put M = i I r i We also let j be minimal in J We split the sum appearing in 3 in two parts: S d = L + L 2 where L := d n r d r 2M d n S d and L 2 := d n r u d for some u J To bound L we note that all divisors involved divide n where n = 2 s i I r λi i S d S d

11 PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 24 Using Lemmas 26 and 27 we get 32 L 2 log d < 2 d d n r n log r n = 2 r φn r 2M log r 2M r φ2m We now bound L 2 If J = then L 2 = 0 and there is nothing to bound So assume that J We argue as follows Note that since s by Lemma 2 i we have = Q n Q 2n Q 2 s n Let q be any odd prime factor of Q n By reducing relation ii of Lemma 2 modulo q and using the fact that n and q are both odd we get 2Pn 2 mod q therefore 2 q = Hence zq q for such primes q Now let d be any divisor of n which is a multiple of r j The number of them is τn /r j where τu is the number of divisors of the positive integer u For each such d there is a primitive prime factor q d of Q d Q n Thus r j d q d This shows that 33 ν rj φ ν rj φq n τn /r j τn /2 where the last inequality follows from the fact that τn /r j τn = λ j λ j + 2 Since r j does not divide m it follows from 23 that 34 ν rj P m e rj Hence and imply that Invoking Lemma 24 we get τn 2e rj 35 τn r j + log α log r j Now every divisor d participating in L 2 is of the form d = 2 a d where 0 a s and d is a divisor of n divisible by r u for some u J Thus { } L 2 τn min S 2a d := gn s r 0 a s d n r u d for some u J In particular d 3 and since the function x log x/x is decreasing for x 3 we have that 36 gn s r 2τn log2 a r j 2 a r j 0 a s Putting also s := min{s 46} we get by Lemma 3 that 2 s l Thus inserting this as well as 32 and 36 all into 3 we get 37 l log α 0 39 < 2 log r 2M + gn s r r φ2m r 2M

12 242 FAYE AND LUCA Since 38 0 a s log2 a r j 2 a r j < 4 log log r j r j inequalities and 36 give us that gn s r log 2 log α := gr j r j log r j The function gx is decreasing for x 3 Thus gr j g3 < 064 For a positive integer N put fn := N log α log r N r φn r N Then inequality 37 implies that both inequalities 39 fl < gr j l M log α + fm < gr j hold Assuming that l 26 we get by Lemma 25 that l log α log log l + 25/ log log l log l 2log log log l 74 Mathematica confirmed that the above inequality implies l 500 Another calculation with Mathematica showed that the inequality fl < 064 for even values of l [ 500] Z implies that l [2 8] The minimum of the function f2n for N [ 250] Z is at N = 3 and f6 > 22 For the remaining positive integers N we have f2n > 0 Hence inequality 39 implies 2 s 2 log α < 064 and 2 s 23 log α < = 276 according to whether M 3 or M = 3 and either one of the above inequalities implies that s 3 Thus s = s { 2 3} Since 2M l 2M is square-free and l 8 we have that M { 3 5 7} Assume M > and let i be such that M = r i Let us show that λ i = Indeed if λ i 2 then 99 Q P Q 49 according to whether r i = respectively and Thus we get that divide φ = P m showing that divide l Since l 8 only the case l = 8 is possible In this case r j 5 and inequality 39 gives 84 < f8 g5 < 79 a contradiction Let us record what we have deduced so far Lemma 34 If n > 2 is even then s { 2 3} Further if I then I = {i} r i {3 5 7} and λ i = We now deal with J For this we return to 3 and use the better inequality namely 2 s M log α l log α log Pn + + L 2 φ p log d 2 s M p P d

13 PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 243 so L 2 2 s M log α 0 39 log d 2 s M p P d + p In the right-hand side above M { 3 5 7} and s { 2 3} The values of the right-hand side above are in fact hu := u log α 0 39 logp u/φp u for u = 2 s M { } Computing we get: where hu H sm M φm for M { 3 5 7} s { 2 3} H > 069 H M > 28 for M > H 2M > 2426 H 3M > 5897 We now exploit the relation M 320 H sm φm d n r u d for sume u J < L 2 Our goal is to prove that r j < 0 6 Assume this is not so We use the bound log log d L 2 < φd of Lemma 26 Each divisor d participating in L 2 is of the form 2 a d where a [0 s] Z and d is a multiple of a prime at least as large as r j Thus and log log d φd Using 35 we get log log 8d φ2 a φd for a {0 s} d φd n φn M + ωn φm r j 2 ωn τn r j + log α log r j < r j where the last inequality holds because r j is large Thus 32 ωn < log r j log 2 < 2 log r j Hence 322 n φn M + ωn M < φm r j φm < M 2 log φm exp rj < M r j φm + r j + 4 log r j r j 2 log rj

14 244 FAYE AND LUCA where we used the inequalities + x < e x valid for all real numbers x as well as e x < + 2x which is valid for x 0 /2 with x = 2 log r j /r j which belongs to 0 /2 because r j is large Thus the inequality log log d φd log log 8d + 4 log r j d r j M φ2 a φm holds for d = 2 a d participating in L 2 The function x log log8x/x is decreasing for x 3 Hence log log8rj 323 L 2 τn + 4 log r j M r j r j φ2 a φm 0 a s Inserting inequality 35 into 323 and using 320 we get 324 log r j < log r j + log log8r j log α r j r j where G s = 0 a s φ2 a Gs H sm For s = 2 3 inequality 324 implies r j < and r j < 300 respectively For s = and M > inequality 324 implies r j < 5000 When M = and s = we get n = 2n and j = Here inequality 324 implies that r < This is too big so we use the bound S d < 2 log d d of Lemma 26 instead for the divisors d of participating in L 2 which in this case are all the divisors of n larger than 2 We deduce that 06 < L 2 < 2 < 4 d 2n d>2 log d d log d d d n The last inequality above follows from the fact that all divisors d > 2 of n are either of the form d or 2d for some divisor d 3 of n and the function x log x/x is decreasing for x 3 Using Lemma 27 and inequalities 32 and 322 we get log r n 4 log r 06 < 4 < ωn + 4 log r r φn r r < r n 4 log r r 2 log r + 4 log r r which gives r < 59 So in all cases r j < 0 6 Here we checked that e r = for all such r except r {3 3} for which e r = 2 If e rj = we then get τn /r j so n = r j Thus n in contradiction with Lemma 3 Assume now that r j {3 3} Say r j = 3 In this case 79 and 599 divide Q 3 which divides therefore φ = P m Thus if there is some other prime factor r of n /3 then 3r n and Q 3r has a primitive prime factor q mod 3r In particular 3 q Thus ν 3 φ 3 showing that

15 PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER P m Hence 3 m therefore 3 M a contradiction A similar contradiction is obtained if r j = 3 since Q 3 has two primitive prime factors namely and so 3 M This finishes the proof Acknowledgement We thank the referee for comments which improved the quality of this paper References Y F Bilu G Hanrot P M Voutier Existence of Primitive Divisors of Lucas and Lehmer Numbers With an appendix by M Mignotte J Reine Angew Math R D Carmichael On the numerical factors of the arithmetic forms α n ± β n Ann Math J Cilleruelo F Luca Repunit Lehmer numbers Proc Edinb Math Soc II Ser B Faye F Luca A Tall On the equation φ5 m = 5 n Bull Korean Math Soc B Faye F Luca Lucas numbers with the Lehmer property Math Rep Buchar to appear 6 R Guy Estimation de la fonction de Tchebychef θ sur le k-ieme nombre premier et grandes valeurs de la fonction ωn nombre de diviseurs premiers de n Acta Arith J M Grau Ribas F Luca Cullen numbers with the Lehmer property Proc Am Math Soc : Corrigendum F Luca Arithmetic functions of Fibonacci numbers Fibonacci Q Multiply perfect numbers in Lucas sequences with odd parameters Publ Math Fibonacci numbers with the Lehmer property Bull Pol Acad Sci Math F Luca F Nicolae φf m = F n Integers A30 2 W McDaniel On Fibonacci and Pell Numbers of the Form kx 2 Fibonacci Q H L Montgomery R C Vaughan The large sieve Mathematika A Pethő The Pell sequence contains only trivial perfect powers in Sets graphs and numbers Budapest 99 Colloq Math Soc János Bolyai 60 North-Holland Amsterdam J B Rosser L Schoenfeld Approximate formulas for some functions of prime numbers Ill J Math School of Mathematics Received University of the Witwatersrand Revised Wits South Africa Ecole Doctorale de Mathematiques et d Informatique Université Cheikh Anta Diop de Dakar Dakar Fann Senegal bernadette@aims-senegalorg School of Mathematics University of the Witwatersrand Wits South Africa florianluca@witsacza