Table Of ContentProc. Amer. Math. Soc., to appear.
BINOMIAL COEFFICIENTS AND
THE RING OF p-ADIC INTEGERS
0
1
Zhi-Wei Sun* and Wei Zhang
0
2
Department of Mathematics, Nanjing University
y
a Nanjing 210093, People’s Republic of China
M [email protected], zhangwei [email protected]
9
1 Abstract. Letk >1beanintegerandletpbe aprime. Weshowthat if
pa 6k <2pa or k =paq+1 (with q <p/2) for some a=1,2,3,..., then
]
T the set {(cid:0)nk(cid:1) : n = 0,1,2,...} is dense in the ring Zp of p-adic integers,
N i.e., it contains a complete system of residues modulo any power of p.
.
h
t
a
m 1. Introduction
[
Let p be an odd prime. In section F11 of Guy [Gu, p.381] it is conjec-
4
tured that |{n! mod p : n = 1,2,3,...}| is about p(1−1/e) asympototi-
v
9 cally. [CVZ] provided certain evidence for the conjecture.
8
In [BLSS] the authors proved that for infinitely many primes p there are
0
3 at least loglogp/logloglogp distinct integers among 0,1,... ,p−1 which
2. are not congruent to n! for any n ∈ Z+ = {1,2,3,...}.
1 Garaev and Luca [GL] showed that for any ε > 0 there is a computable
8
positive constant p (ε) such that for any prime p > p (ε) and integers
0 0 0
: t > pε and s > t+p1/4+ε we have
v
i
X
{m !···m !(modp) : m +···+m = s} ⊇ {r (modp) : r = 1,... ,p−1}.
r 1 t 1 t
a
Letpbeanyprime. Asusual, wedenotebyZ theringofp-adicintegers
p
in the p-adic field Q . The reader may consult an excellent book [M] by
p
Murty for the basic knowledge of p-adic analysis. Any given p-adic integer
α has a unique representation in the form
∞
α = a pj with a ∈ [0,p−1] = {0,1,... ,p−1}.
j j
Xj=0
2010 Mathematics Subject Classification.Primary 11B65; Secondary 05A10, 11A07,
11S99.
*This author is the corresponding author. He is supported by the National Nat-
ural Science Foundation (grant 10871087) and the Overseas Cooperation Fund (grant
10928101) of China.
1
2 ZHI-WEI SUN AND WEI ZHANG
For each b ∈ N = {0,1,2,...}, we have
1
α ≡ r(b) (mod pb), i.e., |α−r(b)| 6 ,
p pb
where r(b) := a pj and |·| is the p-adic norm.
06j<b j p
In this papePr we study the following new problem (which was actually
motivated by the author’s paper [S]).
Problem 1.1. Given a prime p and a positive integer k, is the set { n :
k
n ∈ N} dense in Z ? In other words, does the set contain a comp(cid:0)le(cid:1)te
p
system of residues modulo any power of p?
Definition 1.1. Let k ∈ N and m ∈ Z+, and define
n
R (k) := (mod m) : n ∈ N . (1.1)
m
(cid:26)(cid:18)k(cid:19) (cid:27)
If R (k) = Z/mZ, then we call m a k-universal number.
m
Clearly all positive integers are 1-universal and 1 is k-universal for all
k ∈ Z+.
If p is a prime, a,k,n,n′ ∈ N and n′ ≡ n (mod pa+ordp(k!)) then
n′ (n′ −j) (n−j) n
= 06j<k ≡ 06j<k = (mod pa).
(cid:18)k(cid:19) Q k! Q k! (cid:18)k(cid:19)
Combining this observation with the Chinese Remainder Theorem we im-
mediately get the following basic proposition.
Proposition 1.1. Let k ∈ Z+ and m = pa1 ···par, where p ,... ,p are
1 r 1 r
distinct primes and a ,... ,a ∈ Z+. Then m is k-universal if and only if
1 r
pa1,... ,par are all k-universal.
1 r
In view of Proposition 1.1, we may focus on those k-universal prime
powers.
Let k > 1 be an integer. If p > k is a prime, then p is not k-universal
since
0 1 p−1
, ,... ,
(cid:26)(cid:18)k(cid:19) (cid:18)k(cid:19) (cid:18) k (cid:19)(cid:27)
is not a complete system of residues modulo p. (Note that 0 = 1 = 0.)
k k
Thus, if m ∈ Z+ is k-universal, then m has no prime diviso(cid:0)r(cid:1)grea(cid:0)te(cid:1)r than
k.
For an integer k > 1, a prime p > k and an integer r ∈ [1,p − 1] =
{1,... ,p−1}, the congruence x ≡ r (mod p) might have more than two
k
solutions. For example, (cid:0) (cid:1)
12 19 22 31
≡ ≡ ≡ ≡ 18 (mod 43)
(cid:18)5 (cid:19) (cid:18)5 (cid:19) (cid:18)5 (cid:19) (cid:18)5 (cid:19)
BINOMIAL COEFFICIENTS AND THE RING OF p-ADIC INTEGERS 3
and
15 21 25 30
≡ ≡ ≡ ≡ 14 (mod 61).
(cid:18)10(cid:19) (cid:18)10(cid:19) (cid:18)10(cid:19) (cid:18)10(cid:19)
Recall the following useful result of Lucas.
Lucas’ Theorem (cf. [Gr] and [HS]). Let p be any prime, and let
n ,k ,... ,n ,k ∈ [0,p−1]. Then we have
0 0 r r
r n pi r n
i=0 i ≡ i (mod p).
(cid:18)Pr k pi(cid:19) (cid:18)k (cid:19)
i=0 i iY=0 i
P
Clearly Lucas’ theorem implies the following proposition.
Proposition 1.2. Let p be a prime and let k = r k pi with k ∈
i=0 i i
[0,p− 1]. Then Rp(k) = ri=1Rp(ki). In particularP, when k0,... ,kr ∈
{0,p−2,p−1} we have Q
R (k) ⊆ {r(mod p) : r = 0,±1}.
p
Let p be a prime. As R (1) = Z/pZ, if the p-adic expansion of k ∈
p
Z+ has a digit 1 then R (k) = Z/pZ by Proposition 1.2. In this spirit,
p
Proposition 1.2 is helpful to study when R (k) = Z/pZ. For an integer
p
b > 1, to investigate whether pb is k-universal (i.e., R (k) = Z/pbZ) one
pb
might think that we should use extended Lucas theorem for prime powers.
However, all known generalizations of Lucas’ theorem to prime powers are
somewhat unnatural and complicated, e.g., K. Davis and W. Webb [DW]
proved that if p > 3 is a prime, a,b,n,k ∈ Z+ and n ,k ∈ [0,pa−1] then
0 0
npa+b +n np⌊b/3⌋ n
0 ≡ 0 (mod pb+1).
(cid:18)kpa+b +k (cid:19) (cid:18)kp⌊b/3⌋(cid:19)(cid:18)k (cid:19)
0 0
Therefore, weprefertoapproachProblem1.1byinductionargumentwhich
can be easily understood.
Our first result is as follows.
Theorem 1.1. Let p be a prime and let a ∈ N = {0,1,2,...}. Let k be
an integer with pa 6 k < 2pa. Then, for any b ∈ N and r ∈ Z there is an
integer n ∈ [0,pa+b−1] with n ≡ k (mod pa) such that n ≡ r (mod pb).
k
(cid:0) (cid:1)
Remark 1.1. For a prime p and a positive integer k having a digit 1 in its
p-adic expansion, p is definitely k-universal (i.e., R (k) = Z/pZ) but pb
p
might be not for some integer b > 1. For example, 21 = 4 × 5 + 1 and
ord (21!) = 4, thus
5
n n
(mod 52) : n ∈ N = (mod 52) : n ∈ [0,56 −1]
(cid:26)(cid:18)21(cid:19) (cid:27) (cid:26)(cid:18)21(cid:19) (cid:27)
={r(mod 52) : r = 0,±1,±3,±5,±10}
and hence R52(21) 6= Z/52Z.
Here is a consequence of Theorem 1.1.
4 ZHI-WEI SUN AND WEI ZHANG
Corollary 1.1. Let p be a prime and let k ∈ Z+ with log (k/2) < ⌊log k⌋.
p p
Then 1,p,p2,... are k-universal numbers and the set { n : n ∈ N} is a
k
dense subset of the ring Z of p-adic integers. (cid:0) (cid:1)
p
Proof. Set a = ⌊log k⌋. Then pa 6 k < 2pa. By Theorem 1.1, pb is
p
k-universal for every b = 0,1,2,.... Therefore { n : n ∈ N} is dense in
k
Z . This concludes the proof. (cid:3) (cid:0) (cid:1)
p
For any k ∈ Z+ there is a unique a ∈ N such that 2a 6 k < 2a+1. Thus
Theorem 1.1 or Corollary 1.1 implies the following result.
Corollary 1.2. Let k ∈ Z+. Then any power of two is k-universal and
hence the set { n : n ∈ N} is a dense subset of the 2-adic integral ring
k
Z . (cid:0) (cid:1)
2
Definition 1.2. A positive integer k is said to be universal if any power
of a prime p 6 k is k-universal, i.e., { n : n ∈ N} is a dense subset of Z
k p
for any prime p 6 k. (cid:0) (cid:1)
Theorem 1.1 implies that 1, 2, 3, 4, 5, 9 are universal numbers. To
obtain other universal numbers, we need to extend Theorem 1.1.
Theorem 1.2. Let p be a prime and let a ∈ N. Let k = k +pak with
0 1
k ∈ [0,pa −1] and k ∈ [1,p−1]. Suppose that for each r = 1,... ,p−1
0 1
there are n ∈ [k ,pa −1] and n ∈ [k ,p−1] such that
0 0 1 1
n n
1 0
≡ r (mod p) and P (n ) 6≡ 0 (mod p), (1.2)
(cid:18)k (cid:19)(cid:18)k (cid:19) k1 1
1 0
where
k1 (−1)j−1 x
P (x) = . (1.3)
k1 j (cid:18)k −j(cid:19)
Xj=1 1
Then, for any b ∈ N, the set { n : n ∈ [0,pa+b −1]} contains a complete
k
system of residues modulo pb. (cid:0) (cid:1)
Remark 1.2. Let p be an odd prime. Then
p−1
(−1)j−1 p−1
P (p−1) =
p−1
j (cid:18) j (cid:19)
Xj=1
p−1 (p−1)/2
1 1 1
≡− = − + ≡ 0 (mod p).
j (cid:18)j p−j(cid:19)
Xj=1 Xj=1
Thus, for k = p−1 there is no n ∈ [k ,p−1] with P (n ) 6≡ 0 (mod p).
1 1 1 k1 1
BINOMIAL COEFFICIENTS AND THE RING OF p-ADIC INTEGERS 5
Corollary 1.3. Let p be an odd prime and q ∈ {1,... ,(p−1)/2}. Then,
for any a ∈ Z+ and b ∈ N, the number pb is (paq +1)-universal.
Proof. Let k = q, k = 1, and k = pak + k = paq + 1. As P (x) ≡
1 0 1 0 k1
0 (mod p) cannot have more than degP (x) = k − 1 solutions (see,
k1 1
e.g., [IR, p.39]) there exists n ∈ [k ,2k − 1] ⊆ [k ,p − 1] such that
1 1 1 1
P (n ) 6≡ 0 (mod p). Note that n1 6≡ 0 (mod p). For any r ∈ [1,p−1]
k1 1 k1
there is a unique n ∈ [1,p−1] su(cid:0)ch(cid:1)that
0
n n n
1 0 1
= n ≡ r (mod p).
0
(cid:18)k (cid:19)(cid:18)k (cid:19) (cid:18)k (cid:19)
1 0 1
Applying Theorem 1.2, we immediately obtain the desired result. (cid:3)
From Theorem 1.2 we can deduce the following result.
Theorem 1.3. The integers 11, 17 and 29 are universal numbers.
We have the following conjecture based on our computation via the
software Mathematica.
Conjecture 1.2. There are no universal numbers other than 1, 2, 3, 4,
5, 9, 11, 17, 29.
In Sections 2, 3 and 4 we will prove Theorems 1.1, 1.2 and 1.3 respec-
tively.
2. Proof of Theorem 1.1
We use induction on b.
The case b = 0 is trivial, so we proceed to the induction step.
Now fix b ∈ N and r ∈ Z. Suppose that m ∈ Z, n = k + pam ∈
[0,pa+b − 1] and n ≡ r (mod pb). Let q be the smallest nonnegative
k
residue of (r − n(cid:0))/(cid:1)pb modulo p.
k
Set n′ = n+(cid:0)pa(cid:1)+bq. Then
n′ < pa+b(q +1) 6 pa+b+1 and n′ ≡ n ≡ k (mod pa).
By the Chu-Vandermonde identity (cf. (5.22) of [GKP, p.169]),
n′ n+pa+bq k pa+bq n
= = .
(cid:18)k(cid:19) (cid:18) k (cid:19) (cid:18) j (cid:19)(cid:18)k −j(cid:19)
Xj=0
If 1 6 j 6 k and j 6= pa then pa ∤ j and hence
pa+bq pa+bq pa+bq −1
= ≡ 0 (mod pb+1).
(cid:18) j (cid:19) j (cid:18) j −1 (cid:19)
6 ZHI-WEI SUN AND WEI ZHANG
Note also that
pa−1
pa+bq pa+bq −t
= pbq ≡ pbq(−1)pa−1 ≡ pbq (mod pb+1).
(cid:18) pa (cid:19) t
tY=1
Therefore
n′ n n n
≡ +pbq ≡ r −pbq +pbq (mod pb+1).
(cid:18)k(cid:19) (cid:18)k(cid:19) (cid:18)k −pa(cid:19) (cid:18)k −pa(cid:19)
So it suffices to show that
n
≡ 1 (mod p).
(cid:18)k −pa(cid:19)
As 0 6 k −pa < pa, Lucas’ theorem implies that
n (m+1)pa +(k−pa) m+1 k −pa
= ≡ = 1 (mod p).
(cid:18)k −pa(cid:19) (cid:18) 0pa +(k −pa) (cid:19) (cid:18) 0 (cid:19)(cid:18)k −pa(cid:19)
Combining the above we have completed the proof by induction.
3. Proof of Theorem 1.2
We claim that for each b = 0,1,2,... the set { n : n ∈ S(b)} contains
k
a complete system of residues modulo pb, where (cid:0) (cid:1)
{n} k1 (−1)j−1 ⌊n/pa⌋
S(b) = n ∈ [0,pa+b−1] : pa 6≡ 0 (mod p)
(cid:26) (cid:18) k (cid:19) j (cid:18)k −j(cid:19) (cid:27)
0 X 1
j=1
and {n} denotes the least nonnegative residue of n mod pa.
pa
The claim is trivial for b = 0 since
{k } k1 (−1)j−1 ⌊k /pa⌋ (−1)k1−1
0 pa 0
= 6≡ 0 (mod p).
(cid:18) k (cid:19) j (cid:18) k −j (cid:19) k
0 Xj=1 1 1
As degP (x) < k , there exists n ∈ [0,k − 1] such that P (n ) 6≡
k1 1 1 1 k1 1
0 (mod p). Combining this with the supposition in Theorem 1.2, we see
that for any r ∈ [0,p − 1] there are n ∈ [0,pa − 1] and n ∈ [0,p − 1]
0 1
satisfying (1.2) and the congruence n0 6≡ 0 (mod p). Taking n = pan +
k0 1
n ∈ [0,pa+1 −1] we find that (cid:0) (cid:1)
0
n n n
1 0
≡ ≡ r (mod p)
(cid:18)k(cid:19) (cid:18)k (cid:19)(cid:18)k (cid:19)
1 0
BINOMIAL COEFFICIENTS AND THE RING OF p-ADIC INTEGERS 7
by Lucas’ theorem. This proves the claim for b = 1.
Now let b ∈ Z+ and assume that { n : n ∈ S(b)} contains a complete
k
system of residues modulo pb. We pro(cid:0)ce(cid:1)ed to prove the claim for b+1.
Let r be any integer. By the induction hypothesis, there is an integer
n ∈ [0,pa+b −1] such that
n n k1 (−1)j−1 ⌊n/pa⌋
≡ r (mod pb) and 0 6≡ 0 (mod p),
(cid:18)k(cid:19) (cid:18)k (cid:19) j (cid:18)k −j(cid:19)
0 Xj=1 1
where n = {n} . Hence, for some q ∈ [0,p−1] we have
0 pa
n k1 (−1)j−1 ⌊n/pa⌋ r − n
q 0 ≡ k (mod p).
(cid:18)k (cid:19) j (cid:18)k −j(cid:19) pb(cid:0) (cid:1)
0 Xj=1 1
Clearly, n′ = n+pa+bq ∈ [0,pa+b+1 −1] and
{n′} k1 (−1)j−1 ⌊n′/pa⌋
pa
(cid:18) k (cid:19) j (cid:18) k −j (cid:19)
0 Xj=1 1
n k1 (−1)j−1 ⌊n/pa⌋+pbq
0
=
(cid:18)k (cid:19) j (cid:18) k −j (cid:19)
0 X 1
j=1
n k1 (−1)j−1 ⌊n/pa⌋
0
≡ 6≡ 0 (mod p).
(cid:18)k (cid:19) j (cid:18)k −j(cid:19)
0 X 1
j=1
As in the proof of Theorem 1.1, we have
n′ n k pa+bq n
− =
(cid:18)k(cid:19) (cid:18)k(cid:19) (cid:18) j (cid:19)(cid:18)k −j(cid:19)
Xj=1
⌊k/pa⌋
pa+bq n
≡ (mod pb+1).
(cid:18) paj (cid:19)(cid:18)k −paj(cid:19)
Xj=1
By Lucas’ theorem, for 1 6 j 6 ⌊k/pa⌋ = k we have
1
pa+bq pbq pbq pbq −i (−1)j−1
≡ = ≡ pbq (mod pb+1)
(cid:18) paj (cid:19) (cid:18) j (cid:19) j i j
0<Yi<j
and
n pa⌊n/pa⌋+n ⌊n/pa⌋ n
0 0
= ≡ (mod p).
(cid:18)k −paj(cid:19) (cid:18)pa(k −j)+k (cid:19) (cid:18)k −j(cid:19)(cid:18)k (cid:19)
1 0 1 0
8 ZHI-WEI SUN AND WEI ZHANG
Therefore
n′ n n k1 (−1)j−1 ⌊n/pa⌋ n
− ≡ pbq 0 ≡ r − (mod pb+1)
(cid:18)k(cid:19) (cid:18)k(cid:19) (cid:18)k (cid:19) j (cid:18)k −j(cid:19) (cid:18)k(cid:19)
0 Xj=1 1
′
and hence n ≡ r (mod pb+1). This concludes the induction step.
k
In view(cid:0)of (cid:1)the above we have proved the claim and hence the desired
result follows.
4. Proof of Theorem 1.3
(I) We first prove that 11 is universal.
Since
23 < 11 < 24, 32 < 11 < 2×32, 7 < 11 < 2×7,
and 11 = 2×5+1 with 2 6 (5−1)/2, by Theorem 1.1 and Corollary 1.3,
11 is universal.
(II) Now we want to show that 17 is universal.
Observe that
24 < 17 < 25, 32 < 17 < 2×32, 11 < 17 < 2×11,
and 13 < 17 < 3 × 13. By Theorem 1.1, pb is 17-universal for any p =
2,3,11,13 and b ∈ N.
Note that
⌊17/5⌋
(−1)j−1 x x2 −2x 1 (x−1)2 −2
= + ≡ 6≡ 0 (mod 5).
j (cid:18)⌊17/5⌋−j(cid:19) 2 3 2
Xj=1
Also, 17 = 3×5+2, and
3 2 3 3
≡ 1 (mod 5), ≡ 3 (mod 5),
(cid:18)3(cid:19)(cid:18)2(cid:19) (cid:18)3(cid:19)(cid:18)2(cid:19)
4 2 4 3
≡ 4 (mod 5), ≡ 2 (mod 5).
(cid:18)3(cid:19)(cid:18)2(cid:19) (cid:18)3(cid:19)(cid:18)2(cid:19)
So, 5b is also 17-universal for any b ∈ N.
Clearly
⌊17/7⌋
(−1)j−1 x 1
= x− ≡ x−4 (mod 7).
j (cid:18)⌊17/7⌋−j(cid:19) 2
Xj=1
BINOMIAL COEFFICIENTS AND THE RING OF p-ADIC INTEGERS 9
Also, 17 = 2×7+3, and
2 3 2 4 2 5
≡ 1 (mod 7), ≡ 4 (mod 7), ≡ 3 (mod 7),
(cid:18)2(cid:19)(cid:18)3(cid:19) (cid:18)2(cid:19)(cid:18)3(cid:19) (cid:18)2(cid:19)(cid:18)3(cid:19)
2 6 3 4 3 5
≡ 6 (mod 7), ≡ 5 (mod 7), ≡ 2 (mod 7).
(cid:18)2(cid:19)(cid:18)3(cid:19) (cid:18)2(cid:19)(cid:18)3(cid:19) (cid:18)2(cid:19)(cid:18)3(cid:19)
Thus, 7b is also 17-universal for any b ∈ N.
(III) Finally we prove that 29 is universal.
By Theorem 1.1, it remains to prove that pb is 29-universal for any
p = 7,11,13 and b ∈ N.
Note that 29 = 4×7+1. It is easy to check that
4 (−1)j−1 4
6≡ 0 (mod 7).
j (cid:18)4−j(cid:19)
X
j=1
For any r ∈ [1,6], we have 4 r ≡ r (mod 7). So, by Theorem 1.2, 7b is
4 1
29-universal for any b ∈ N.(cid:0) (cid:1)(cid:0) (cid:1)
Clearly 29 = 2×11+7, and
2 (−1)j−1 x 1
= x− ≡ x−6 (mod 11).
j (cid:18)2−j(cid:19) 2
X
j=1
Observe that
2 7 2 8
≡ 1 (mod 11), ≡ −3 (mod 11),
(cid:18)2(cid:19)(cid:18)7(cid:19) (cid:18)2(cid:19)(cid:18)7(cid:19)
2 9 2 10
≡ 3 (mod 11), ≡ −1 (mod 11),
(cid:18)2(cid:19)(cid:18)7(cid:19) (cid:18)2(cid:19)(cid:18)7 (cid:19)
3 8 3 9
≡ 2 (mod 11), ≡ −2 (mod 11),
(cid:18)2(cid:19)(cid:18)7(cid:19) (cid:18)2(cid:19)(cid:18)7(cid:19)
4 7 4 10
≡ −5 (mod 11), ≡ 5 (mod 11),
(cid:18)2(cid:19)(cid:18)7(cid:19) (cid:18)2(cid:19)(cid:18)7 (cid:19)
4 8 4 9
≡ 4 (mod 11), ≡ −4 (mod 11).
(cid:18)2(cid:19)(cid:18)7(cid:19) (cid:18)2(cid:19)(cid:18)7(cid:19)
Applying Theorem 1.2 we see that 11b is 29-universal for any b ∈ N.
Observe that 29 = 2×13+3 and
2 (−1)j−1 x 1
= x− ≡ x−7 (mod 13).
j (cid:18)2−j(cid:19) 2
Xj=1
10 ZHI-WEI SUN AND WEI ZHANG
Also,
2 3 2 4
≡ 1 (mod 13), ≡ 4 (mod 13),
(cid:18)2(cid:19)(cid:18)3(cid:19) (cid:18)2(cid:19)(cid:18)3(cid:19)
2 5 2 6
≡ −3 (mod 13), ≡ −6 (mod 13),
(cid:18)2(cid:19)(cid:18)3(cid:19) (cid:18)2(cid:19)(cid:18)3(cid:19)
2 7 2 9
≡ −4 (mod 13), ≡ 6 (mod 13),
(cid:18)2(cid:19)(cid:18)3(cid:19) (cid:18)2(cid:19)(cid:18)3(cid:19)
2 10 2 12
≡ 3 (mod 13), ≡ −1 (mod 13),
(cid:18)2(cid:19)(cid:18) 3(cid:19) (cid:18)2(cid:19)(cid:18)3 (cid:19)
3 6 3 9
≡ −5 (mod 13), ≡ 5 (mod 13),
(cid:18)2(cid:19)(cid:18)3(cid:19) (cid:18)2(cid:19)(cid:18)3(cid:19)
4 4 4 7
≡ −2 (mod 13), ≡ 2 (mod 13).
(cid:18)2(cid:19)(cid:18)3(cid:19) (cid:18)2(cid:19)(cid:18)3(cid:19)
Thus, with the help of Theorem 1.2, 13b is 29-universal for any b ∈ N.
By the above, we have completed the proof of Theorem 1.3.
Acknowledgment. The author is grateful to the referee for many helpful
comments.
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