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Repdigits in Euler Functions of Associated Pell Numbers

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Repdigits in Euler Functions of Associated Pell Numbers Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:25 https://doi.org/10.1007/s12044-019-0551-3 Repdigits in Euler functions of associated pell numbers G K PANDA and M K SAHUKAR∗ Department of Mathematics, National Institute of Technology, Rourkela 769 008, India *Corresponding author. E-mail: [email protected]; [email protected] MS received 28 January 2019; revised 16 September 2019; accepted 22 October 2019 Abstract. In this paper, we discuss some properties of Euler totient function of associated Pell numbers which are repdigits in base 10. Keywords. Pell numbers; associated Pell numbers; Euler totient function; repdigits. Mathematics Subject Classification. Primary: 11A25 Secondary: 11B39. 1. Introduction The Euler totient function φ(n) of a positive integer n is the number of positive integers less than or equal to n and relatively prime to n.Ifn has the canonical decomposition = a1 ··· ar , ,..., , ,..., n p1 pr , where p1 p2 pr are distinct primes and a1 a2 ar are positive integers, then it is well-known that − − φ( ) = a1 1( − ) ··· ar 1( − ). n p1 p1 1 pr pr 1 The Pell sequence {Pn} is defined by means of a binary recurrence relation Pn+1 = 2Pn + Pn−1 with initial terms P0 = 0, P1 = 1. The associated Pell sequence {Qn} satisfies an identical recurrence relation with initial terms Q0 = 1, Q1 = 1. The sequence {Rn}, where Rn = 2Qn, is known as the Pell–Lucas sequence. The Binet formulas for Pell and associated Pell numbers are αn − βn αn + βn Pn = and Qn = , α − β 2 √ √ where α = 1 + 2 and β = 1 − 2. The Lucas and associated Lucas sequences are defined as u0 = 0, u1 = 1, un+1 = Aun + Bun−1 and u0 = 2, u1 = A,vn+1 = Avn + Bvn−1 respectively, where A and B are positive integers. Many well-known binary recurrence sequences are special cases of these sequences. In the last two decades, Diophantine equations involving Euler functions of binary recur- rence sequences have been extensively studied. Luca [6] explored all Fibonacci numbers whose Euler function is a power of 2. Luca and St˘anic˘a[8] proved that the Euler function of the n-th Pell number Pn or the n-th associated Pell number Qn is a power of 2 only © Indian Academy of Sciences 0123456789().: V,-vol 25 Page 2 of 8 Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:25 if n ≤ 8. Damir et al. [3] extended their search for powers of 2 in Euler functions of the terms of the Lucas sequence corresponding to B =±1 and proved that there are finitely many such n. A repdigit is a positive integer with only one distinct digit in its decimal expansion. In particular, numbers of the form d(10m − 1)/9forsomem ≥ 1 and 1 ≤ d ≤ 9 are called repdigits in base 10. In the year 2006, Luca [7] explored the repdigits associated with the Euler functions of Fibonacci numbers. In 2015, Faye and Luca [4] proved that there is no Pell or Pell–Lucas number larger than 10 with only one distinct digit. Subsequently, Bravo et al. [2] in 2016 investigated the presence of repdigits in the Euler functions of Lucas numbers. Sahukar and Panda [12] proved that Euler function of no Pell number is a repdigit having at least two digits. In this paper, we study the presence of repdigits in the sequence {φ(Qn)}. In particular, we prove the following theorem on the solvability of the equation: m φ(Qn) = d(10 − 1)/9 (1.1) for d ∈{1, 2,...,9}. Theorem 1.1. Let (1.1) hold for some n > 16 and for some d ∈{1, 2,...,9}. Then (a) d ∈{4, 8}. (b) n is odd. (c) n is an odd prime and n2|10m − 1 if d = 4. a Throughout this paper, we use p with or without subscripts as a prime number, b as Jacobi symbol of a and b and (a, b) as the greatest common divisor of a and b. 2. Preliminaries In this section, we present some results about Pell and associated Pell numbers. The results of this section are necessary to prove the theorems stated in Section 1. Lemma 2.1 [5]. If m and n are natural numbers, then (1) (Pn, Qn) = 1, 2 − 2 = (− )n (2) Qn 2Pn 1 , | | n (3) Qm Qn if and only if m n and m is odd, (4) v2(Pn) = v2(n) and v2(Qn) = 0, where v2(n) is the exponent of 2 in the canonical decomposition of n, (5) 3|Qn if and only if n ≡ 2(mod 4), (6) 5 Qn for any n, = ( 2 − ) (7) Q3·2t Q2t 4Q2t 3 . Lemma 2.2 ([1], Theorem A).Ifn, y, m are positive integers with m ≥ 2, then the equation m Qn = y has the only solution (n, y) = (1, 1). 2 Lemma 2.3 ([9], Theorem 3). The solutions of the Diophantine equation Qm Qn = x with 0 ≤ m < naren= 3m,3 m, m is odd. Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:25 Page 3 of 8 25 Table 1. Periods of Qn. kQn(mod k) Period 4 1,1,3,3 4 5 1,1,3,2,2,1,4,4,2,3,3,4 12 8 1,1,3,7 4 20 1, 1, 3, 7, 17, 1, 19, 19, 17, 13, 3, 19 12 Lemma 2.4 ([1], Lemma 2.1). A prime p is a primitive prime divisor of Lucas number un 2 if p divides un but does not divide (α − β) u2 ···un−1 and p is always congruent to ±1 modulo n. Lemma 2.5 ([1], Lemma 2.1). A primitive prime divisor of n-th associated Pell number Qn exists only if n ≥ 2. Lemma 2.6 ([13], Pell–Lucas numbers). If the associated Pell number Qn is a prime, then n is either a prime or a power of 2. 3. Main results In this section, we prove Theorem 1.1 stated at the end of Section 1. To prove this theorem, we need the least residues and periods of associated Pell sequence {Qn}n≥0 modulo 4, 5, 8 and 20. We list them in Table 1. If n ≤ 16, it is easy to see that 1, 3 and 7 are the only numbers in the associated Pell sequence {Qn}n≥0 such that φ(Qn) is a repdigit. If φ(Qn) is a repdigit for some n > 16, then by Lemma 2.6, 2 2 5 φ(Qn) ≥ √ Qn ≥ √ Q17 > 450390 > 10 , 3 3 and thus m > 5. Since Qn is odd for all n, the canonical decomposition of Qn is = a1 a2 ··· ar , ≥ , Qn p1 p2 pr r 0 (3.1) where p1, p2,...,pr are distinct odd primes and ai ≥ 1 for all i and − − − φ( ) = a1 1 a2 1 ··· ar 1( − )( − ) ···( − ). Qn p1 p2 pr p1 1 p2 1 pr 1 (3.2) m If φ(Qn) = d(10 − 1)/9forsomen, then r v2(φ(Qn)) = v2(pi − 1) = v2(d), where v2(d) ∈{0, 1, 2, 3}. (3.3) i=1 With these ideas, we now proceed with the proof of Theorem 1.1. 25 Page 4 of 8 Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:25 3.1 Proof of Theorem 1.1(a) In view of (3.3), if v2(d) = 0, then d ∈{1, 3, 5, 7, 9} and consequently, φ(Qn) is odd, which is possible only when n = 0, 1. If v2(d) = 1, then d = 2, 6 and because of (3.2), = = ( ) = a1 r 1 and p1 3 mod 4 . In particular, Qn p1 and an application of Lemma 2.2 gives a1 = 1 and hence 10m − 1 φ(Qn) = Qn − 1 = d · . 9 If d = 2, then Qn = 3 (mod 20) and from Table 1, n ≡ 2, 10 (mod 12). Because of Lemma 2.6, n is either a prime or a power of 2. If n ≡ 2 (mod 12), we can write n as n = 2(6k + 1) for some k ≥ 0, which is a power of 2 only if k = 0 and consequently n = 2, which contradicts our assumption that n > 16. If n ≡ 10 (mod 12), then we can write n as n = 2(6l + 5) for some l ≥ 0, which is not a power of 2 for any l.Ifd = 6, then Qn = 7 (mod 20) and from Table 1, it follows that n ≡ 3 (mod 12). Thus n = 3(4k +1), which is a prime only if k = 0. Hence, the only possibility left is n = 3, a contradiction to n > 16. 3.2 Proof of Theorem 1.1(b) From Theorem 1.1(a), it is clear that if (1.1) holds for n > 16, then d ∈{4, 8}. To prove part (b) of Theorem 1.1, we need the following lemma. Lemma 3.1.Ifn> 16 is even and (1.1) holds for d ∈{4, 8}, then all prime factors in the canonical decomposition of Qn are congruent to 3 modulo 4. t Proof. Since n is even, we can write n as n = 2 ·n1, where t, n1 ∈ N and n1 is odd. Assume on the contrary that there exist a prime p ≡ 1(mod 4) in the canonical decomposition of Qn.Inviewof(3.3), the number of prime factors of Qn is at most 2 and therefore, r ≤ 2. = = = a1 = If d 4, then r 1 and Qn p1 . From Lemma 2.2, it follows that Qn p1 and hence 10m − 1 φ(Qn) = Qn − 1 = 4 · ≡ 4(mod 10).
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