Download On integers of the forms k ± 2n and k2 n ± 1

Journal of Number Theory 125 (2007) 14–25
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On integers of the forms k ± 2n and k2n ± 1 ✩
Yong-Gao Chen
Department of Mathematics, Nanjing Normal University, Nanjing 210097, Jiangsu, PR China
Received 14 June 2005; revised 27 October 2005
Available online 29 November 2006
Communicated by David Goss
Dedicated to Professor Yuan Wang on his 75th birthday
Abstract
In this paper we consider the integers of the forms k ± 2n and k2n ± 1, which are ever focused by
F. Cohen, P. Erdős, J.L. Selfridge, W. Sierpiński, etc. We establish a general theorem. As corollaries, we
prove that (i) there exists an infinite arithmetic progression of positive odd numbers for each term k of
which and any nonnegative integer n, each of four integers k − 2n , k + 2n , k2n + 1 and k2n − 1 has at least
two distinct odd prime factors; (ii) there exists an infinite arithmetic progression of positive odd numbers
for each term k of which and any nonnegative integer n, each of ten integers k + 2n , k + 1 + 2n , k + 2 + 2n ,
k + 3 + 2n , k + 4 + 2n , k2n + 1, (k + 1)2n + 1, (k + 2)2n + 1, (k + 3)2n + 1 and (k + 4)2n + 1 has at least
two distinct odd prime factors; (iii) there exists an infinite arithmetic progression of positive odd numbers
for each term k of which and any nonnegative integer n, each of ten integers k + 2n , k + 2 + 2n , k + 4 + 2n ,
k + 6 + 2n , k + 8 + 2n , k2n + 1, (k + 2)2n + 1, (k + 4)2n + 1, (k + 6)2n + 1 and (k + 8)2n + 1 has at least
two distinct odd prime factors. Furthermore, we pose several related open problems in the introduction and
three conjectures in the last section.
© 2006 Elsevier Inc. All rights reserved.
MSC: 11A07; 11B25
Keywords: Covering systems; Odd numbers; Sums of prime powers
✩
Supported by the National Natural Science Foundation of China, Grant No. 10471064.
E-mail address: [email protected].
0022-314X/$ – see front matter © 2006 Elsevier Inc. All rights reserved.
doi:10.1016/j.jnt.2006.10.005
Y.-G. Chen / Journal of Number Theory 125 (2007) 14–25
15
1. Introduction
Let N denote the set of all positive integers and P denote the set of all positive odd primes.
Is there any integer k
Is there any integer k
Is there any integer k
Is there any integer k
such that all integers k − 2n (n = 1, 2, . . .) are composite?
such that all integers k + 2n (n = 1, 2, . . .) are composite?
such that all integers k2n + 1 (n = 1, 2, . . .) are composite?
such that all integers k2n − 1 (n = 1, 2, . . .) are composite?
In this section, we observe some facts and recall the related history.
(A) Write a positive odd number k as the form 2n + p, n ∈ N, p ∈ P. This is equivalent to find a
positive integer n such that k − 2n is a positive prime.
If we try to write odd numbers as the form 2n + p, n ∈ N, p ∈ P, then we find that almost all
positive odd numbers can be done. First counterexample is 127. There are only 19 counterexamples from 5 to 1087. In 1934, Romanoff [32] proved that the set of positive odd numbers which
can be expressed in the form 2n + p has positive lower asymptotic density in the set of all positive odd numbers, where n is a nonnegative integer and p is a positive prime. Chen and Sun [13]
proved that the lower asymptotic density is more than 0.1736. For a positive integer n and an
integer a, let a (mod n) = {a + nk: k ∈ Z}. {ai (mod mi )}ki=1 is called a covering system if every
integer b satisfies b ≡ ai (mod mi ) for at least one value of i. By employing a covering system,
P. Erdős [18] proved that there is an infinite arithmetic progression of positive odd numbers each
of which has no representation of the form 2n + p.
(B) Write a positive odd number k as the form 2n − p, n ∈ N, p ∈ P. This is equivalent to find a
positive integer n such that 2n − k is a positive prime.
If we try to write odd numbers as the form 2n − p, n ∈ N, p ∈ P, then we find no counterexample up to 219. We do not know the least positive odd number which cannot be represented
as 2n − p. For a given odd number k, we have no effective method to determine whether k can
be represented as 2n − p. Erdős’ proof implies that there is an infinite arithmetic progression of
positive odd numbers each of which has no representation of the form 2n − p. Up to my knowledge, there is no proof or disproof that the set of positive odd numbers which can be expressed
in the form 2n − p has positive lower asymptotic density in the set of all positive odd numbers,
where n is a nonnegative integer and p is a positive prime.
(C) Write a positive odd number k as p − 2n , p ∈ P, n ∈ N. This is equivalent to find a positive
integer n such that k + 2n is a positive prime.
If we try to write odd numbers as the form p − 2n , n ∈ N, p ∈ P, then we find no counterexample up to 503. We have the similar problems as in paragraph (B).
(D) Odd numbers k for which all k2n + 1 (n = 1, 2, . . .) are composite.
n
It is well known that for n 2, the prime factors of Fermat number 22 + 1 have the form
n+2
+ 1. Primes of the form k2n + 1 have been studied extensively [1,2,4,6,7,9,16,24,25,31,
k2
33–36]. For each odd number k, if we try to find an integer n such that k2n + 1 is prime, then we
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Y.-G. Chen / Journal of Number Theory 125 (2007) 14–25
find no counterexample up to k < 3061 (see Jaeschke [24], Baillie, Cormack and Williams [2]).
Erdős and Odlyzko [19] proved that the set of odd numbers k for which there exists a positive
integer n with k2n + 1 being prime has positive lower asymptotic density in the set of all positive
odd integers. On the other hand, Sierpiński [34] proved that there are infinitely many positive odd
numbers k for which all k2n + 1 (n = 1, 2, . . .) are composite. In 1962, J.L. Selfridge (unpublished) discovered that for any positive integer n, the integer 78 557 · 2n + 1 is divisible by one of
the primes 3, 5, 7, 13, 19, 37, 73. It is still an open question whether 78 557 is the least positive
odd number k for which all k2n + 1 (n = 1, 2, . . .) are composite. For a given odd number k, we
have no effective method to determine whether there is an integer n such that k2n + 1 is prime.
(E) Odd numbers k for which all k2n − 1 (n = 1, 2, . . .) are composite.
Comparing with k2n + 1, people do not pay much attention to k2n − 1. Meng [29] verified
that for each positive odd number k < 106 , there exists a positive integer n 6000 such that
k2n − 1 is a prime except for k = 659, 1211, 2293, 3817, 8681, 9221, 10 391, 10 451, 11 519,
13 451. Similarly to Sierpiński’s proof, one may prove that there are infinitely many positive odd
numbers k for which all k2n − 1 (n = 1, 2, . . .) are composite. For a given odd number k, we
have no effective method to determine whether there is an integer n such that k2n − 1 is prime.
In 1975, Cohen and Selfridge [15] proved that there exist infinitely many odd numbers which
are neither the sum nor the difference of two prime powers. In [8] Chen proved the following
result: the set of positive integers which have no representation of the form 2n ± p a q b , where
p, q are distinct odd primes and n, a, b are nonnegative integers, has positive lower asymptotic
density in the set of all positive odd integers. That is, the lower asymptotic density of the set of
positive odd integers k such that k − 2n has at least three distinct prime factors for all positive
integers n is positive. Chen [10] proved that the set of positive odd integers k such that k − 2n
has at least three distinct prime factors for all positive integers n contains an infinite arithmetic
progression. Chen [10] also showed that the set of positive odd integers k such that k2n + 1
has at least three distinct prime factors for all positive integers n contains an infinite arithmetic
progression. Similarly, one may prove that the set of positive odd integers k such that k + 2n
has at least three distinct prime factors for all positive integers n contains an infinite arithmetic
progression, and the set of positive odd integers k such that k2n − 1 has at least three distinct
prime factors for all positive integers n contains an infinite arithmetic progression. Chen [11]
studied the integers of the forms k r − 2n and k r 2n + 1. Recently, Chen [12] proved that there
is an arithmetic progression of positive odd numbers, for each term M of which, none of five
consecutive odd numbers M, M − 2, M − 4, M − 6 and M − 8 can be expressed in the form
2n ± p α , where p is a prime and n, α are nonnegative integers. That is, for every nonnegative
integer n, each of M − 2n , M − 2 − 2n , M − 4 − 2n , M − 6 − 2n and M − 8 − 2n has at least
two odd prime factors. For further related references, one may refer to Guy [22], Crocker [17],
Granville and Soundararajan [21].
A similar problem is to represent large even integers as a sum of two primes and k powers of 2.
Under GRH, D.R. Heath-Brown and J.C. Puchta [23], J. Pintz and I.Z. Ruzsa [30] independently
proved that k = 7 is admissible. The uncondition value k = 13 is due to D.R. Heath-Brown and
J.C. Puchta [23]. J. Pintz and I.Z. Ruzsa [30] announced that it will be shown that k = 8 is
admissible unconditionally. For the related research, one may refer to [20,26–28].
Y.-G. Chen / Journal of Number Theory 125 (2007) 14–25
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2. Main results
With the observations and arguments in the previous section, we need only to concern integers
of four forms k − 2n , k + 2n , k2n + 1, k2n − 1.
Theorem. Let k1 , k2 , . . . , ku+v be integers, {aij (mod mij )}tji=1 (i = 1, 2, . . . , u + v) be covering
systems with aij 0, and pij (j = 1, 2, . . . , ti , i = 1, 2, . . . , u + v) be positive primes with
pij | 2mij − 1,
for all i, j.
Let rij be integers with 0 rij < pij and
rij ≡ 2aij − ki
rij ≡ −2aij − ki
(mod pij ),
(mod pij ),
j = 1, 2, . . . , ti , 1 i u,
j = 1, 2, . . . , ti , u + 1 i u + v.
Suppose that if pij = puv , then rij = ruv . Then there exist two positive integers M and M0 with
2 | M and 2 M0 such that if k ≡ M0 (mod M), then for any nonnegative integer n, each of
k + ki − 2n
k + ki + 2 n
(1 i u),
(u + 1 i u + v),
(k + ki )2n − 1 (1 i u),
(k + ki )2n + 1 (u + 1 i u + v)
has at least two distinct odd prime factors.
Remark. If each covering system {aij (mod mij )}tji=1 covers every integer at least li times and
pi1 , pi2 , . . . , piti are distinct, suppose that the conditions in the theorem hold, then there exist
two positive integers M and M0 with 2 | M and 2 M0 such that if k ≡ M0 (mod M), then for
any nonnegative integer n, each of
k + ki − 2 n
k + ki + 2 n
(1 i u),
(u + 1 i u + v),
(k + ki )2n − 1 (1 i u),
(k + ki )2n + 1 (u + 1 i u + v)
has at least li + 1 (correspondingly) distinct odd prime factors. Similarly to the proof of the
theorem, a proof can be given with minor changes.
Corollary 1. Let the conditions be as in the theorem. If a is an integer such that all ki 2a are
integers, then there exist two positive integers M and M0 with 2 | M and 2 M0 such that if
k ≡ M0 (mod M), then for any nonnegative integer n, each of
k + ki 2 a − 2 n
k + ki 2 a + 2 n
(1 i u),
(u + 1 i u + v),
k + ki 2a 2n − 1 (1 i u),
k + ki 2a 2n + 1 (u + 1 i u + v)
has at least two distinct odd prime factors.
Corollary 2. There exists an infinite arithmetic progression of positive odd numbers for each
term k of which and any nonnegative integer n, each of four integers k − 2n , k + 2n , k2n + 1 and
k2n − 1 has at least two distinct odd prime factors.
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Y.-G. Chen / Journal of Number Theory 125 (2007) 14–25
Corollary 3. Let a be a nonnegative integer. Then there exists an infinite arithmetic progression
of positive odd numbers for each term k of which and any nonnegative integer n, each of ten
integers
k − 2n ,
k2n − 1,
k − 3 · 2a − 2n ,
k − 3 · 2a 2n − 1,
k − 2a − 2n ,
k − 2a 2n − 1,
k − 4 · 2a − 2n ,
k − 4 · 2a 2n − 1
k − 2 · 2a − 2n ,
k − 2 · 2a 2n − 1,
has at least two distinct odd prime factors.
Corollary 4. Let a be a nonnegative integer. Then there exists an infinite arithmetic progression
of positive odd numbers for each term k of which and any nonnegative integer n, each of ten
integers
k + 2n ,
k2n + 1,
k + 3 · 2a + 2n ,
k + 3 · 2a 2n + 1,
k + 2a + 2n ,
k + 2a 2n + 1,
k + 4 · 2a + 2n ,
k + 4 · 2a 2n + 1
k + 2 · 2a + 2n ,
k + 2 · 2a 2n + 1,
has at least two distinct odd prime factors.
3. Proofs
Proof of the theorem. Without loss of generality, we may assume that 0 aij < mij
for all i, j . For a prime p and a positive integer l, let n(l) be the order of 2 (mod p l ).
Then n(l) → +∞ as l → +∞. By this fact there exist positive integers lij , lij , nij , nij
(j = 1, 2, . . . , ti , i = 1, 2, . . . , u + v) such that all nij , nij are distinct, nij , nij > maxs,t {mst } + 6
l
l
(j = 1, 2, . . . , ti , i = 1, 2, . . . , u + v) and nij , nij are the orders of 2 (mod pijij ) and 2 (mod pijij )
respectively. By [5] (see also [3,37]) there exist primes qij , qij such that nij , nij are the orders of
2 (mod qij ) and 2 (mod qij ) respectively (j = 1, 2, . . . , ti , i = 1, 2, . . . , u + v). Then all qij , qij
are distinct and qij = pfg , qij = pfg for j = 1, 2, . . . , ti ; g = 1, 2, . . . , tf ; i, f = 1, 2, . . . , u + v.
Let bij be the least nonnegative integers n with
l rij + ki − 2n ≡ 0 mod pijij , j = 1, 2, . . . , ti , 1 i u,
l rij + ki + 2n ≡ 0 mod pijij , j = 1, 2, . . . , ti , u + 1 i u + v.
be the least nonnegative integers n
If for some i, j such n does not exist, then let bij = aij . Let bij
with
l (rij + ki )2n − 1 ≡ 0 mod pijij , j = 1, 2, . . . , ti , 1 i u,
(rij + ki )2n + 1 ≡ 0
l mod pijij ,
j = 1, 2, . . . , ti , u + 1 i u + v.
=m −a .
If for some i, j such n does not exist, then let bij
ij
ij
Y.-G. Chen / Journal of Number Theory 125 (2007) 14–25
19
Fix an integer m with 2m−1 > (m + 1)(u + v). Since the number of integers −ki + 2n
(0 n m, 1 i u), −ki − 2n (0 n m, u + 1 i u + v) is at most (m + 1)(u + v) <
2m−1 , there exists an odd number a with 0 < a < 2m such that for any nonnegative integer n,
a ≡ −ki + 2n mod 2m , 1 i u,
a ≡ −ki − 2n mod 2m , u + 1 i u + v.
Since nij , nij > maxs,t {mst } + 6 (j = 1, 2, . . . , ti , i = 1, 2, . . . , u + v) and nij , nij are the orders
l
l
n for all i, j . Let
of 2 (mod pijij ) and 2 (mod pijij ) respectively, we have bij nij and bij
ij
T = max lij , lij .
i,j
Fix an integer l with
T
2l > 2m max pij + a + max |ki |.
i,j
i
Noting that rij = rst if pij = pst , by the Chinese Remainder Theorem, the congruent system
k ≡ rij
T
mod pij
,
j = 1, . . . , ti , 1 i u + v,
k ≡ 2bij − ki
k≡2
nij −bij
(mod qij ), j = 1, . . . , ti , 1 i u,
− ki mod qij , j = 1, . . . , ti , 1 i u,
k ≡ −2bij − ki
(mod qij ), j = 1, . . . , ti , u + 1 i u + v,
k ≡ −2
− ki mod qij , j = 1, . . . , ti , u + 1 i u + v,
k ≡ a + 2l + 2l+1 mod 2l+3
nij −bij
has solutions in integers. Let k be such a solution and n be any nonnegative integer. Suppose that
at least one of
k + ki − 2n
k + ki + 2 n
(1 i u),
(u + 1 i u + v),
(k + ki )2n − 1 (1 i u),
(k + ki )2n + 1 (u + 1 i u + v),
has at most one odd prime factor.
Case 1. k + ki − 2n has at most one odd prime factor (1 i u).
Since {aij (mod mij )}tji=1 is a covering system, there exists a j with
n ≡ aij
(mod mij ).
By pij | 2mij − 1 we have
k + ki − 2n ≡ rij + ki − 2aij ≡ 0 (mod pij ).
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Y.-G. Chen / Journal of Number Theory 125 (2007) 14–25
Since k + ki − 2n has at most one odd prime factor, we may assume that
β
k + ki − 2n = ±2αij pijij .
(1)
By k ≡ a + 2l + 2l+1 (mod 2l+3 ) and l > m, we have k ≡ a (mod 2m ). Thus, by the definition
of a, we have
k + ki − 2n ≡ a + ki − 2n ≡ 0
mod 2m .
So αij < m. Since
mod 2l+3 ,
k ≡ a + 2l + 2l+1
we may write
k = a + 2l + 2l+1 + 2l+3 t,
t ∈ Z.
Thus
k + ki − 2n = a + 2l + 2l+1 + 2l+3 t + ki − 2n 2l + 2l+1 + 2l+3 t − 2n − a − |ki |
2l − a − |ki |
T
.
2m pij
(2)
By αij < m, (1) and (2) we have βij T . So
T
mod pij
.
k + ki − 2 n ≡ 0
Thus
rij + ki − 2n ≡ k + ki − 2n ≡ 0
T
.
mod pij
Hence
rij + ki − 2n ≡ 0
l mod pijij .
(3)
rij + ki − 2bij ≡ 0
l mod pijij .
(4)
On the other hand, we have
By (3) and (4) we have
2n − 2bij ≡ 0
l mod pijij .
Y.-G. Chen / Journal of Number Theory 125 (2007) 14–25
21
l
Since nij is the order of 2 (mod pijij ), we have
n ≡ bij
(5)
(mod nij ).
By (5) and the order of 2 (mod qij ) being nij , we have
2n − 2bij ≡ 0
(mod qij ).
Thus
k + ki − 2n ≡ k + ki − 2bij ≡ 0 (mod qij ),
a contradiction with (1).
Case 2. (k + ki )2n − 1 has at most one odd prime factor (1 i u).
Since {aij (mod mij )}tji=1 is a covering system, we have {(mij − aij ) (mod mij )}tji=1 is a
covering system. Hence there exists a j with
n ≡ (mij − aij )
(mod mij ).
By pij | 2mij − 1 we have
2aij (k + ki )2n − 1 ≡ (k + ki )2n+aij − 2aij
≡ k + ki − 2aij
≡ rij + ki − 2aij
≡ 0 (mod pij ).
Since (k + ki )2n − 1 has at most one odd prime factor, we may assume that
β
(k + ki )2n − 1 = ±2αij pijij .
If n > 0, then αij = 0. If n = 0, then, by the proof of Case 1 and the definition of a, we have
k + ki − 1 ≡ a + ki − 1 ≡ 0
mod 2m .
So αij < m. Since k ≡ a + 2l + 2l+1 (mod 2l+3 ), we have
(k + ki )2n − 1 |k + ki |2n − 1
|k + ki | − 1
|k − a| − a − |ki | − 1
2l − a − |ki | − 1
T
.
2m pij
(6)
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Y.-G. Chen / Journal of Number Theory 125 (2007) 14–25
Hence βij T and
T
.
mod pij
(k + ki )2n − 1 ≡ 0
Thus
(rij + ki )2n − 1 ≡ (k + ki )2n − 1 ≡ 0
T
.
mod pij
So
(rij + ki )2n − 1 ≡ 0
l mod pijij .
(7)
l mod pijij .
(8)
On the other hand, we have
(rij + ki )2bij − 1 ≡ 0
By (7) and (8) we have pij rij + ki and
l (rij + ki ) 2n − 2bij ≡ 0 mod pijij .
So
2n − 2bij ≡ 0
l mod pijij .
l
Since nij is the order of 2 (mod pijij ), we have
n ≡ bij
mod nij .
(9)
By (9) and the order of 2 (mod qij ) being nij , we have
2n − 2bij ≡ 0
mod qij .
Thus
(k + ki )2n − 1 ≡ (k + ki )2bij − 1 ≡ 2nij −bij 2bij − 1 ≡ 0 (mod qij ),
a contradiction with (6).
Case 3. k + ki + 2n has at most one odd prime factor (u + 1 i u + v). Similarly to Case 1,
we obtain a contradiction.
Case 4. (k + ki )2n + 1 has at most one odd prime factor (u + 1 i u + v). Similarly to
Case 2, we obtain a contradiction.
This completes the proof of the theorem. 2
Proof of Corollary 1. Since
aij
t
(mod mij ) ji=1
(i = 1, 2, . . . , u + v)
Y.-G. Chen / Journal of Number Theory 125 (2007) 14–25
are all covering systems, we have
aij + a + |a|mij
t
(mod mij ) ji=1
23
(i = 1, 2, . . . , u + v)
are all covering systems. Let ki = 2a ki (i = 1, 2, . . .). Then, for all i, j , rij ≡ 2a rij (mod pij ) if
a 0 and 2−a rij ≡ rij (mod pij ) if a < 0. By the theorem we obtain a proof of Corollary 1. 2
Proof of Corollary 2. Let u = v = 1 and k1 = k2 = 0. Two covering systems in Cohen and
Selfridge [15] satisfy the conditions of the theorem. By the theorem, there exists an infinite
arithmetic progression of positive odd numbers for each term k of which and any nonnegative
integer n, each of four integers k − 2n , k + 2n , k2n + 1 and k2n − 1 has at least two distinct odd
prime factors. This completes the proof of Corollary 2. 2
Proofs of Corollaries 3 and 4. According to the proof of [12, Theorem 2], there exist five
covering systems {aij (mod mij )}tji=1 (i = 1, 2, 3, 4, 5) and primes pij with pij | 2mij − 1 for all
i, j such that if pij = puv , then
2aij − ki ≡ 2auv − ku
(mod pij ),
where k1 = 0, k2 = −2, k3 = −4, k4 = −6 and k5 = −8. Since 2a−1 k1 , 2a−1 k2 , 2a−1 k3 , 2a−1 k4
and 2a−1 k5 are all integers, by Corollary 1 there exist two positive integers M0 and M with 2 | M
and 2 M0 such that if k ≡ M0 (mod M), then for any nonnegative integer n, each of ten integers
k − 2n ,
k2n − 1,
k − 3 · 2a − 2n ,
k − 3 · 2a 2n − 1,
k − 2 a − 2n ,
k − 2a 2n − 1,
k − 4 · 2a − 2n ,
k − 4 · 2a 2n − 1
k − 2 · 2a − 2n ,
k − 2 · 2a 2n − 1,
has at least two distinct odd prime factors.
If k ≡ M −M0 (mod M), then −k ≡ M0 (mod M). Thus, for any nonnegative integer n, each
of ten integers −k − 2n , −k − 2a − 2n , −k − 2 · 2a − 2n , −k − 3 · 2a − 2n , −k − 4 · 2a − 2n ,
−k 2n − 1, (−k − 2a )2n − 1, (−k − 2 · 2a )2n − 1, (−k − 3 · 2a )2n − 1 and (−k − 4 · 2a )2n − 1
has at least two distinct odd prime factors. Hence, for any nonnegative integer n, each of ten
integers
k + 2n ,
k 2n + 1,
k + 2a + 2n ,
k + 2a 2n + 1,
k + 3 · 2a + 2n ,
k + 3 · 2a 2n + 1,
k + 4 · 2a + 2 n ,
k + 4 · 2a 2n + 1
k + 2 · 2a + 2n ,
k + 2 · 2a 2n + 1,
has at least two distinct odd prime factors. Then Corollaries 3 and 4 follow easily.
2
4. Unsolved problems
In this section, I pose the following problem and conjectures. These are mentioned in the first
section. To emphasis these, I formulate here.
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Y.-G. Chen / Journal of Number Theory 125 (2007) 14–25
Problem. For a given odd number k, how do we determine whether k can be represented as
2n − p? What is the least positive odd number which cannot be represented as 2n − p, where n
is a positive integer and p a positive prime? We have the similar problem for p − 2n .
Conjecture 1. The set of positive odd numbers which can be expressed in the form 2n − p has
positive lower asymptotic density in the set of all positive odd numbers, where n is a positive
integer and p a positive prime.
Conjecture 2. The set of positive odd numbers which can be expressed in the form p − 2n has
positive lower asymptotic density in the set of all positive odd numbers, where n is a positive
integer and p a positive prime.
Conjecture 3. For any positive integer r, there exists an infinitely arithmetic progression of
positive odd numbers, for each term k of which and any positive integer n, each of
k + 2n ,
k − 2n ,
k2n + 1,
k2n − 1,
k + 1 + 2n ,
k + 1 − 2n ,
(k + 1)2n + 1,
(k + 1)2n − 1,
...,
...,
...,
...,
k + r − 1 + 2n ,
k + r − 1 − 2n ,
(k + r − 1)2n + 1,
(k + r − 1)2n − 1
has at least two distinct prime factors.
P. Erdős (see [22]) conjectured that for every c > 0, there exists a congruent covering system
{ai (mod mi )}ti=1 with c m1 < m2 < · · · < mt . S.L.G. Choi [14] proved that Erdős conjecture
is true for c = 20. By the theorem, it is clear that if Erdős conjecture is true, then Conjecture 3 is
true.
Acknowledgment
We are grateful to the referee for his/her kindly remarks.
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