balancing non-wieferich primes in arithmetic progressions
TRANSCRIPT
Proc. Indian Acad. Sci. (Math. Sci.) (2019) 129:21https://doi.org/10.1007/s12044-018-0459-3
© Indian Academy of Sciences 1
Balancing non-Wieferich primes in arithmetic progressions
UTKAL KESHARI DUTTA1, BIJAN KUMAR PATEL2
and PRASANTA KUMAR RAY1,∗
1Sambalpur University, Jyoti Vihar, Burla, Sambalpur 768 019, India2International Institute of Information Technology, P.O. Malipada,Bhubaneswar 751 003, India*Corresponding author.E-mail: [email protected]; [email protected]; [email protected]
MS received 5 July 2017; revised 27 February 2018; accepted 24 April 2018;published online 14 February 2019
Abstract. A prime is called a balancing non-Wieferich prime if it satisfies Bp−(8p)
�≡ 0
(mod p2), where(8
p)
and Bn denote the Jacobi symbol and the n-th balancing numberrespectively. For any positive integers k > 2 and n > 1, there are � log x/ log log xbalancing non-Wieferich primes p ≤ x such that p ≡ 1 (mod k) under the assumptionof the abc conjecture for the number field Q(
√2) (Proc. Japan Acad. Ser. A 92 (2016)
112–116). In this paper, for any fixed M , the lower bound log x/ log log x is improvedto (log x/ log log x)(log log log x)M .
Keywords. Balancing numbers; Wieferich primes; arithmetic progressions; abcconjecture.
2010 Mathematics Subject Classification. 11B25, 11B39, 11B41.
1. Introduction
In 1909, Wieferich [9] found that Fermat’s last theorem is related to the primes p satisfying2p−1 ≡ 1 (mod p2) and proved that for any odd prime p, if the Fermat equation x p +y p + z p = 0 has an integer solution with p � xyz, then 2p−1 ≡ 1 (mod p2) holds.Since then, these primes are called Wieferich primes. Conversely, an odd prime is calleda non-Wieferich prime if it does not satisfy the said congruence, in which case it followsthat the Fermat’s equation with exponent p has no integer solution. The primes 1093 and3511 are the only two Wieferich primes found till date. For any fixed integer a ≥ 2, p iscalled a Wieferich prime for base a if a p−1 ≡ 1 (mod p2). Whether there exist infinitelymany Wieferich primes or non-Wieferich primes for a given base a are still unknown.
The abc conjecture has a huge number of implications, including Fermats last theoremfor large enough exponents, as well as many important open questions in number theory. Itstates that, if a, b and c are co-prime positive integers with a + b = c, then for any ε > 0,
c ε (rad(abc))1+ε ,
where rad(n) is the product of all prime factors of n. This version of abc conjecture iscalled the abc conjecture of Masser–Oesterle–Szpiro. There is a generalization for the
21 Page 2 of 7 Proc. Indian Acad. Sci. (Math. Sci.) (2019) 129:21
number fields of the abc conjecture (see [8]). The abc conjecture for the number fieldQ[√2] is used for this purpose, as the ring Z [√2] is the ring of integers of Q[√2].
A natural number k is said to be a balancing number if it is the solution of the simplediophantine equation
1 + 2 + · · · + (k − 1) = (k + 1) + (k + 2) + · · · + (k + l),
where l is the balancer corresponding to k [1]. Balancing numbers can be derived from therecurrence relation Bn+1 = 6Bn − Bn−1 for n ≥ 1 with initial values (B0, B1) = (0, 1),
where Bn denotes the n-th balancing number. The closed-form expression of balancingnumbers is
Bn = αn − βn
α − β,
where α = 3 + 2√
2 and β = 3 − 2√
2. Panda and Rout [5] studied the periodicity ofbalancing numbers modulo any integer which helps to explore divisibility properties ofthese numbers [5]. They also conjectured that there are only three primes, 13, 31, and1546463 satisfying π(p) = π(p2), where π(m) is the period of the sequence of balancingnumbers modulo m, the least positive integer k satisfying (Bk, Bk+1) ≡ (0, 1) (mod m).Rout [7] named these numbers as balancing Wieferich primes if π(p) = π(p2), that is,analogous to the congruence
Bp−(8p)
≡ 0 (mod p2).
Otherwise, they are called balancing non-Wieferich primes. He showed that, for any arbi-trary integer k ≥ 2, there are infinitely many balancing non-Wieferich primes p withp ≡ 1 (mod k) under the assumption of the abc conjecture. He also proved that forpositive integers k > 2 and n > 1, there are � log x
log log x primes p ≤ x such that
p ≡ 1 (mod k) and Bp−(8p)
�≡ 0 (mod p2)
under the assumption of the abc conjecture.Since there are infinitely many balancing non-Wieferich primes p, it is natural to inves-
tigate their quantitative behaviour. This motivates us to improve the lower bound from(log x/ log log x) established in [7] to (log x/ log log x)(log log log x)M , where M is anypositive integer.
2. Preliminaries
The following results are found in [7].
Lemma 2.1. Balancing numbers satisfy the inequality
αn−1 < Bn < αn for n ≥ 2,
where α = 3 + 2√
2.
Lemma 2.2 . Let Bn = XnYn, where Xn and Yn are the square-free and square-full partsof Bn respectively. If p | Xn , then
Bp−(8p)
�≡ 0 (mod p2).
Proc. Indian Acad. Sci. (Math. Sci.) (2019) 129:21 Page 3 of 7 21
Lemma 2.3. If the abc conjecture for the algebraic number field Q(√
2) is true, thenYn ε B2ε
n .
We recall that the m-th cyclotomic polynomial is defined as
�m(y) =∏
1≤d<m, gcd(d,m)=1
(y − ζ dm),
where ζ dm is a primitive m-th root of unity, which gives rise to the recursive formula
�m(y) = ym − 1∏
1≤d<m, d|m �d(y).
The following result relating to this polynomial is found in [3].
Lemma 2.4 . Let n ≥ 1 be a positive integer and b ≥ 2 be an integer. If p is an odd primesuch that p | �n(b), then either p | n or p ≡ 1 (mod n).
The following result is found in [4].
Lemma 2.5. Let πn(x) denote the number of square-free integers not exceeding x andhaving exactly n distinct prime factors. Then
πn(x) � x(log log x)n−1
(n − 1)! log(x).
3. Main result
In 2017, Chen and Ding [2] proved that for any integer a ≥ 2 and any fixed integer k ≥ 2,
there are � (log x/ log log x)(log log log x)M non-Wieferich primes p ≤ x for any fixedpositive integer M such that p ≡ 1 (mod k) under the assumption of the abc conjecture.In this article, the bound for balancing non-Wieferich primes is improved using the ideasgiven in [2].
Let φ be the Euler totient function and �n(x) be the n-th cyclotomic polynomial. Letpi be the i-th prime and define
δM =M+1∏
i=1
(1 − 1
pi
).
Let τM be the set of all square-free integers with exactly M + 1 prime factors. Let Bpnk befactored into X pnkYpnk with gcd(pn, k) = 1, where X pnk is the square-free part and Ypnk
is the square-full part of Bpnk . Let
X ′pnk = gcd (X pnk,�pnk(α/β))
and
Y ′pnk = gcd (Ypnk,�pnk(α/β)).
From the definition of cyclotomic polynomials, we deduce that
�pnk(α/β) | ((α/β)pnk − 1)β pnk−1,
21 Page 4 of 7 Proc. Indian Acad. Sci. (Math. Sci.) (2019) 129:21
which implies �pnk(α/β) | ((α/β)−1)Bpnk = X pnkYpnk�1(α/β). Since gcd (X pnk, Ypnk)
= 1 = gcd (�pnk,�1), either �pnk(α/β) | X pnk or �pnk(α/β) | Ypnk . If �pnk(α/β) |X pnk , then �pnk(α/β) = X ′
pnk and Y ′pnk = 1. On the other hand, if �pnk(α/β) | Ypnk , then
�pnk(α/β) = Y ′pnk and X ′
pnk = 1. Therefore, we conclude that �pnk(α/β) = X ′pnkY ′
pnk .Before proving the main result, we will establish some lemmas that are used subse-
quently.
Lemma 3.1. Let ε be a small positive integer. Suppose that the abc conjecture is true forQ[√2], then
X ′pnk � α2(φ(pnk)−εpnk).
Proof. Using Ypnk � Y ′pnk and Lemma 2.3, Y ′
Pnk ε B2εpnk . By virtue of Lemma 2.1,
Y ′pnk ε α2εpnk . (3.1)
Using Lemma 2.10 of [7], we can write
X ′pnkY ′
pnk � α2φ(pnk). (3.2)
Now, from (3.1) and (3.2), we get
X ′pnk � α2(φ(pnk)−εpnk),
which ends the proof. �
Lemma 3.2 . If m < n, then
gcd(X ′pm k, X ′
pnk) = 1.
Proof. Let us assume that
gcd(X ′pm k, X ′
pnk) > 1,
that is, p | X ′pm k and p | X ′
pnk for some prime p. By the definitions of X ′pm k and X ′
pnk ,we have
p | �pm k(α/β) and p | �pnk(α/β).
So
p | Bpm k and p | Bpnk .
Thus, p divides gcd(Bpm k, Bpnk). In fact, since the sequence {Bn}n∈N is a strong divisibilitysequence, that is, gcd(Bm, Bn) = Bgcd(m,n) [6], p divides Bgcd(pm k,pnk) = Bk . Since
�pnk(α/β) | ((α/β) − 1)Bpnk = �1(α/β)Bpnk and Bpnk = Bpn kBk
· Bk, we have
Proc. Indian Acad. Sci. (Math. Sci.) (2019) 129:21 Page 5 of 7 21
�pnk(α/β)
∣∣∣
Bpnk
Bk
as gcd (�pnk,�1) = 1 and gcd (�pnk, Bk) = 1. Since
p | Bk and p | �pnk(α/β)
∣∣∣Bpnk
Bk,
it follows that p2 | Bpnk, which contradicts p | X ′pnk . Therefore,
gcd(X ′pm k, X ′
pnk) = 1.
�
Lemma 3.3. Suppose the abc conjecture for the number field Q(√
2) is true. Then thereexists an integer n0 depending only on α, l, M such that, if pnk ∈ τM with n ≥ n0, thenX ′
pnkl � pnkl.
Proof. Let ε = δM φ(l)2l . By Lemma 3.1, we have
X ′pnkl � α2(φ(pnkl)−εpnkl). (3.3)
Since
φ(m) = m∏
p|m
(1 − 1
p
)
and
∏
p|pnkl
(1 − 1
p
)≥
∏
p|pnk
(1 − 1
p
)∏
p|l
(1 − 1
p
),
we have φ(pnkl) ≥ φ(pnk)φ(l). Therefore, for pnk ∈ τM , we deduce that
φ(pnkl) − εpnkl ≥ φ(pnk)φ(l) − εpnkl
≥ δM pnkφ(l) − εpnkl
= 2εpnkl − εpnkl
= εpnkl.
It follows from Lemma 2.3 and (3.3) that if n ∈ τM , then
X ′pnkl � α2εpnkl � B2ε
pnkl .
This completes the proof. �
21 Page 6 of 7 Proc. Indian Acad. Sci. (Math. Sci.) (2019) 129:21
Lemma 3.4 . Let n0 be as in Lemma 3.3. If pnk ∈ τM with n ≥ n0, then there exists aprime qn such that qn divides X ′
pnk and
Bqn−( 8qn)
�≡ 0 (mod q2n ), qn ≡ 1 (mod pnk).
Proof. Let pnk ∈ τM with n ≥ n0. Since X ′pnk is a product of distinct primes, it follows
from Lemma 3.3 that there is a prime qn such that
qn | X ′pnk and qn � pnk.
Since
X ′pnkY ′
pnk = �pnk(α/β) and qn � pnk,
it follows from Lemma 2.4 that qn is congruent to 1 modulo pnk. Also from Lemma 2.2,if qn | X pnk , then Bqn−( 8
qn)�≡ 0 (mod q2
n ), which implies the result. �
Theorem 3.5. Let k > 2 and n > 1 be positive integers, and also assume the abcconjecture for the number field Q(
√2). Then, for any positive integer M , we have
#{primes p ≤ x : p ≡ 1 (mod k),
Bp−(8p)
�≡ 0 (mod p2)}�(log x/ log log x)(log log log x)M .
Proof. By Lemma 2.1, Bpnk ≤ x if
pnk ≤ log x
log α.
Thus, Bpnk ≤ x with pnk ∈ τM if
pnk ≤ log x
log α, pnk ∈ τM .
Hence by Lemma 2.5, the number of integers n with Bpnk ≤ x, pnk ∈ τM and n > n0 is
� (log x)(log log log x)M
log log x.
By virtue of Lemma 3.4, as qn ≤ X ′pnk ≤ Bpnk , it follows that the number of qn with
qn ≤ x, pnk ∈ τM and n ≥ n0 is
� (log x)(log log log x)M
log log x,
and hence
#{primes p ≤ x : p ≡ 1 (mod k),
Proc. Indian Acad. Sci. (Math. Sci.) (2019) 129:21 Page 7 of 7 21
Bp−(8p)
�≡ 0 (mod p2)}� (log x/ log log x)(log log log x)M ,
which completes the proof. �
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Communicating Editor: S D Adhikari