PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 101(115) (2017), 231–245 https://doi.org/10.2298/PIM1715231F 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 1 and 2. 1. Introduction Let φ(n) be the Euler function of the positive integer n. Recall that if n has the prime factorization n = pa1 pak 1 · · · k with distinct primes p1,...,pk and positive integers a1,...,ak, then φ(n)= pa1−1(p 1) pak−1(p 1). 1 1 − · · · k k − There are many papers in the literature dealing with diophantine equations involv- ing the Euler function in members of a binary recurrent sequence. For example, in [11], it is shown that 1, 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 φ(5n 1)=5m 1 has no positive integer solutions (m,n). Furthermore, the divisibility− relation φ(−n) n 1 when n is a Fibonacci number, or a Lucas num- ber, or a Cullen number (that| is,− a number of the form n2n + 1 for some positive integer n), or a rep-digit (gm 1)/(g 1) in some integer base g [2, 1000] have been investigated in [10, 5, 7,− 3], respectively.− ∈ Here we look for a similar equation with members of the Pell sequence. The Pell sequence (Pn)n>0 is given by P0 = 0, P1 = 1 and Pn+1 = 2Pn + Pn−1 for all n > 0. Its first terms are 0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832,... We have the following result. 2010 Mathematics Subject Classification: 11B39. 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. 231 232 FAYE AND LUCA Theorem 1.1. The only solutions in positive integers (n,m) of the equation (1.1) φ(Pn)= Pm are (n,m)=(1, 1), (2, 1). For the proof, we begin by following the method from [11], but we add to it some ingredients from [10]. 2. Preliminary results Let (α, β)=(1+ √2, 1 √2) be the roots of the characteristic equation x2 − − 2x 1 = 0 of the Pell sequence P > . The Binet formula for P is − { n}n 0 n αn βn P = − for all n > 0. n α β − This implies easily that the inequalities n−2 n−1 (2.1) α 6 Pn 6 α hold for all positive integers n. We let Q > be the companion Lucas sequence of the Pell sequence given { n}n 0 by Q0 = 2, Q1 = 2 and Qn+2 =2Qn+1 + Qn for all n > 0. Its first few terms are 2, 2, 6, 14, 34, 82, 198, 478, 1154, 2786, 6726, 16238, 39202, 94642, 228486, 551614,... The Binet formula for Qn is n n (2.2) Qn = α + β for all n > 0. We use the well-known result. Lemma . 2 2 n 2.1 The relations (i) P2n = PnQn and (ii) Qn 8Pn = 4( 1) hold for all n > 0. − − For a prime p and a nonzero integer m let νp(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 2.2. The relations (i) ν2(Qn)=1 and (ii) ν2(Pn)= ν2(n) hold for all positive integers n. The following divisibility relations among the Pell numbers are well known. Lemma 2.3. Let m and n be positive integers. We have: (i) If m n, then P P , (ii) gcd(P , P )= P . | m | n m n gcd(m,n) For each positive integer n, let z(n) be the smallest positive integer k such that n Pk. It is known that this exists and n Pm if and only if z(n) m. This number is| referred to as the order of appearance of|n in the Pell sequence.| Clearly, z(2) = 2. 2 Further, putting for an odd prime p, ep = p , where the above notation stands for the Legendre symbol of 2 with respect to p, we have that z(p) p e . A prime | − p factor p of Pn such that z(p) = n is called primitive for Pn. It is known that Pn ep has a primitive divisor for all n > 2 (see [2] or [1]). Write Pz(p) = p mp, where PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 233 k mp is coprime to p. It is known that if p Pn for some k>ep, then pz(p) n. In particular, | | (2.3) νp(Pn) 6 ep whenever p ∤ n. We need a bound on ep. We have the following result. Lemma 2.4. The inequality (p + 1)log α (2.4) e 6 . p 2log p holds for all primes p. Proof. Since e2 = 1, the inequality holds for the prime 2. Assume that p is odd. Then z(p) p + ε for some ε 1 . Furthermore, by Lemmas 2.1 and 2.3, | ∈ {± } we have pep P P = P Q . By Lemma 2.1, it follows easily that | z(p) | p+ε (p+ε)/2 (p+ε)/2 p cannot divide both Pn and Qn for n = (p + ε)/2 since otherwise p will also divide Q2 8P 2 = 4, n − n ± ep a contradiction since p is odd. Hence, p divides one of P(p+ε)/2 or Q(p+ε)/2. If ep ep (p+1)/2 p divides P(p+ε)/2, we have, by (2.1), that p 6 P(p+ε)/2 6 P(p+1)/2 < α , which leads to the desired inequality (2.4) upon taking logarithms of both sides. In ep case p divides Q(p+ε)/2, we use the fact that Q(p+ε)/2 is even by Lemma 2.2 (i). ep Hence, p divides Q(p+ε)/2/2, therefore, by formula (2.2), we have (p+1)/2 Q(p+ε)/2 Q(p+1)/2 α +1 pep 6 6 < < α(p+1)/2, 2 2 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 ω(n) for the number of distinct prime factors of n. The following inequalities (i), (ii) and (iii) are inequalities (3.13), (3.29) and (3.41) in [15], while (iv) is Théoréme 13 from [6]. Lemma 2.5. Let p <p < be the sequence of all prime numbers. We have: 1 2 · · · (i) The inequality pn <n(log n + loglog n) holds for all n > 6. (ii) The inequality 1 1 1+ < 1.79 log x 1+ p 1 2(log x)2 6 pYx − holds for all x > 286. (iii) The inequality n φ(n) > 1.79loglog n +2.5/ log log n holds for all n > 3. 234 FAYE AND LUCA (iv) The inequality log n ω(n) < log log n 1.1714 − holds for all n > 26. For a positive integer n, we put n = p : z(p) = n . We need the following result. P { } Lemma 2.6. Put S := 1 . For n> 2, we have n p∈Pn p−1 P 2log n 4+4loglog n (2.5) S < min , . n n φ(n) n o Proof. Since n > 2, it follows that every prime factor p n is odd and satisfies the congruence p 1 (mod n). Further, putting ℓ :=∈ # P , we have ≡± n Pn (n 1)ℓn 6 p 6 P < αn−1 − n pY∈Pn (by inequality (2.1)), giving (n 1)log α (2.6) ℓ 6 − . n log(n 1) − Thus, the inequality n log α (2.7) ℓ < n log n holds for all n > 3, since it follows from (2.6) 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 prove7→ the first bound, we use (2.7) to deduce that 1 1 2 1 1 1 (2.8) S 6 + 6 + n nℓ 2 nℓ n ℓ m 2 − m 6 6 6 6 > 1 Xℓ ℓn − 1 Xℓ ℓn mXn − 2 ℓn dt 1 1 2 n 6 +1 + + 6 log ℓn +1+ n 1 t n 2 n 1 n n 2 Z − − − 2 (log α)e2+2/(n−2) 6 log n . n log n Since the inequality log n > (log α)e2+2/(n−2) holds for all n > 800, (2.8) implies 2 > that Sn < n log n for n 800. The remaining range for n can be checked on an individual basis. For the second bound on Sn, we follow the argument from [10] and split the primes in in three groups: Pn (i) p< 3n; (ii) p (3n,n2); (iii) p>n2; ∈ We have 1 1 + 1 + 1 + 1 + 1 < 10.1 , n even, 6 n−2 n 2n−2 2n 3n−2 3n (2.9) T1 = 1 1 7.1 p 1 ( + < , n odd, p∈Pn − 2n−2 2n 3n p<X3n PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 235 where the last inequalities above hold for all n > 84.