q-ANALOGUE OF WILSON’S THEOREM
ROBIN CHAPMAN AND HAO PAN
Abstract. We give q-analogues of Wilson’s theorem for the primes congruent 1
and 3 modulo 4 respectively. And q-analogues of two congruences due to Mordell
and Chowla are also established.
1. Introduction
For arbitrary positive integer n, let
[n]q =
1 − qn
= 1 + q + · · · + q n−1 .
1−q
Clearly limq→1 [n]q = 1, so we say that [n]q is a q-analogue of the integer n. Supposing that a ≡ b (mod n), we have
[a]q =
1 − q b + q b (1 − q a−b )
1 − qb
1 − qa
=
≡
= [b]q
1−q
1−q
1−q
(mod [n]q ).
Here the above congruence is considered over the polynomial ring in the variable q
with integral coefficients. And q-analogues of some arithmetical congruences have
been studied in [9, 1, 7, 8].
Let p be a prime. The well-known Wilson theorem states that
(p − 1)! ≡ 1
(mod p).
Unfortunately, in general,
p−1
Y
[j]q 6≡ −q n
(mod [p]q )
j=1
for any integer n. However, we have the following q-analogue of Wilson’s theorem
for a prime p ≡ 3 (mod 4).
Theorem 1.1. Suppose that p > 3 is a prime and p ≡ 3 (mod 4). Then we have
p−1
Y
[j]qj ≡ −1
(mod [p]q ).
j=1
2000 Mathematics Subject Classification. Primary 11A07; Secondary 05A30, 11R29.
1
(1.1)
2
ROBIN CHAPMAN AND HAO PAN
In [6] (or see [10, Theorem 8]), Mordell proved that if p > 3 is a prime and p ≡ 3
(mod 4) then
h(−p)+1
p−1
! ≡ (−1) 2
(mod p),
(1.2)
2
√
where h(−p) is the class number of the quadratic field Q( −p). Now we can give
a q-analogue of (1.2).
Theorem 1.2. Let p > 3 be a prime with p ≡ 3 (mod 4). Then we have
(p−1)/2
Y
[j]q16j ≡ (−1)
h(−p)+1
2
q
(mod [p]q ).
(1.3)
j=1
The case p ≡ 1 (mod 4) is a little complicated. Let p· denote the Legendre
symbol modulo p. It is well-known that for any a prime to p, ap = 1 or −1
according to whether a is a quadratic residue modulo p. Let εp and h(p) be the
√
fundamental unit and the class number of Q( p) respectively.
Theorem 1.3. Suppose that p is a prime and p ≡ 1 (mod 4). Then we have
p−1
Y
[j]qj ≡ A + B
j=1
p−1
X
j
p
j=1,(
qj
(mod [p]q ),
(1.4)
)=−1
where
2h(p)
−2h(p)
2h(p)
−2h(p)
2h(p)
−2h(p)
+ εp
εp
− εp
εp
− εp
+
and B =
.
√
√
2
2 p
p
√
Write εp = (up + vp p)/2 where up , vp are positive integers with the same parity.
Clearly u2p − pvp2 = ±4 since εp is an unit. Letting q → 1 in (1.4), we obtain that
A=
εp
2h(p)
−2h(p)
2h(p)
B(p − 1)
εp
+ εp
up
≡
≡ 2h(p) (mod p).
2
2
2
It follows that h(p) is odd and the norm of εp is always −1, i.e., u2p − pvp2 = −4.
In [4] (or see [10, Theorem 9]), Chowla extended Mordell’s result (1.2) for p ≡ 1
√
(mod 4). Let h(p) and εp = (up + vp p)/2 be defined as above. Then Chowla
proved that
h(p)+1
p−1
(−1) 2 up
!≡
(mod p).
(1.5)
2
2
Now we have the following q-analogue of Chowla’s congruence:
−1 ≡ (p − 1)! ≡ A +
Theorem 1.4. Suppose that p is a prime and p ≡ 1 (mod 4). Then
(p−1)/2
Y
j=1
[j]qj ≡ −Cq − D
p−1
X
j=1,(pj )=−1
q j+1
(mod [p]q ),
(1.6)
q-ANALOGUE OF WILSON’S THEOREM
where
h(p)
C=
εp
−h(p)
− εp
2
h(p)
+
εp
−h(p)
3
h(p)
+ εp
√
2 p
and
D=
εp
−h(p)
+ εp
√
p
.
Let us explain why (1.6) implies (1.5). Letting q → 1 in (1.6), it is derived that
h(p)
−h(p)
h(p)
−h(p)
h(p)
−h(p) p−1
εp − εp
εp + εp
p − 1 εp + εp
!≡−
+
+
·
√
√
2
2
2 p
2
p
√
√
((up + vp p)/2)h(p) − ((−up + vp p)/2)h(p)
≡−
2
h(p)−1
h(p)
h(p−1) up
up
(−1) 2 up
2
≡ − h(p) = −(up /4) 2
≡−
(mod p).
2
2
2
The proofs of Theorems 1.1-1.4 will be given in the next sections.
2. Proofs of Theorems 1.1 and 1.2
In this section we assume that p > 3 is a prime and p ≡ 3 (mod 4). Write
p−1
p−1
2
Y
Y
1 − qj
and
[j]qj =
1 − qj
j=1
j=1
(p−1)/2
Y
j=1
(p−1)/2
Y 1 − qj2
[j]qj =
.
1 − qj
j=1
Observe that
p−1
1 − qp Y
[p]q =
=
(q − ζ j )
1−q
j=1
where ζ = e2πi/p . Also we know that σs : ζ 7−→ ζ s is an automorphism over Q(ζ)
provided that p - s. Hence it suffices to show that
p−1
2
Y
1 − ζj
j=1
1 − ζj
(p−1)/2
= −1
Y 1 − ζ j2
h(−p)+1
= (−1) 2 ζ.
j
1−ζ
j=1
and
Let Q and N denote respectively the sets of quadratic residues and quadratic
non-residues of p in the interval [1, p − 1]. Then
Q p−1
p−1
2
j2 2
2
Y
1 − ζj
U2
U
j=1 (1 − ζ )
=
=
= ,
Q
p−1
j
j
1−ζ
UV
V
j=1 (1 − ζ )
j=1
where
U=
Y
(1 − ζ k ),
and
k∈Q
V =
Y
(1 − ζ k ).
k∈N
But since −1 is a quadratic non-residue modulo p,
Y
Y
Y
V =
(1 − ζ p−k ) =
(1 − ζ −k ) = U
(−ζ k ).
k∈Q
k∈Q
k∈Q
4
ROBIN CHAPMAN AND HAO PAN
Now
(p−1)/2
X
X
k≡
j2 =
j=1
k∈Q
p(p2 − 1)
≡0
24
(mod p)
2
as p is prime to 6 and so p ≡ 1 (mod 24). We conclude that
U/V = (−1)
p−1
2
ζ−
P
k∈Q
k
= −1
as desired.
Now let us begin to prove
(p−1)/2
Q(p−1)/2
2
Y 1 − ζ 16j 2
(1 − ζ 16j )
h(−p)+1
j=1
= Q(p−1)/2
= (−1) 2 ζ.
16j
1−ζ
(1 − ζ 16j )
j=1
j=1
(2.1)
Clearly the numerator of the left side of (2.1) is U . Let
(p−1)/2
W =
Y
(1 − ζ 16j )
j=1
denote its denominator. Let M = {1, 2, . . . , (p − 1)/2}. Then W = W+ W− where
Y
Y
W+ =
(1 − ζ 16j ),
and
W− =
(1 − ζ 16j ).
j∈M ∩Q
j∈M ∩N
Now
W− =
X
(1 − ζ −16k ) =
M 0 ∩Q
U
W+
Y
(−ζ −16k )
k∈M 0 ∩Q
where M 0 = {(p + 1)/2, . . . , p − 1}. We know (see [10, Section 1.3]) that
(p−1)/2 X X k
1
1
p−1
h(−p) =
−2
1
=
p
2
2 − p2 k=1
2 − p2
k∈M ∩N
≡ − 1 − 2|M ∩ N | = −1 − 2|M 0 ∩ Q| (mod 4).
Also, we have
(p−1)/2
X
X
X
X
X
p2 − 1
=
k=
k+
k=
k+
(p − k)
8
k=1
k∈M ∩Q
k∈M ∩N
k∈M ∩Q
k∈M 0 ∩Q
X
X
X
≡
k−2
k ≡ −2
k (mod p),
k∈Q
whence
P
k∈M 0 ∩Q
k∈M 0 ∩Q
k∈M 0 ∩Q
16k ≡ 1 (mod p). Thus
Y
U
U
0
=
= (−1)|M ∩Q|
ζ 16k = (−1)(1+h(−p))/2 ζ,
W
W+ W−
k∈M 0 ∩Q
which confirms (2.1).
q-ANALOGUE OF WILSON’S THEOREM
5
3. When p ≡ 1 (mod 4)
Below suppose that p is a prime congruent 1 modulo 4 and ζ = e2πi/p . Let
Q, N ⊆ [1, p − 1] be the sets of quadratic residues and quadratic non-residues of
p respectively. Let
Y
Y
U=
(1 − ζ k ) and V =
(1 − ζ k ).
k∈Q
k∈N
In order to prove Theorem 1.3, we only need to show prove that
Q
p−1
j 2
j2
Y
X
1
−
ζ
U
j∈Q (1 − ζ )
j
Q
= .
A+B
ζ =
=Q
j
j
j
1−ζ
V
j∈Q (1 − ζ )
j∈N (1 − ζ )
j=1
j∈N
(3.1)
By the analytic class number formula [2, Chapter 1, Section 4, Theorem 2]
√
√
p
and
V = εh(p)
p.
U = ε−h(p)
p
p
√
−2h(p)
Thus U/V = εp
= a − b p where
√
2a = ε2h(p)
+ ε−2h(p)
∈ Z and 2b = (ε2h(p)
− ε−2h(p)
)/ p ∈ Z.
p
p
p
p
Also, by Gauss’s formula for the quadratic Gauss sum
p−1 p−1
X
X
X
j j X j
√
p=
ζ =
ζ −2
ζ j = −1 − 2
ζj.
p
j=1
j=1
j∈N
j∈N
Hence
X U
=a+b 1+2
ζj ,
V
j∈N
which is clearly equivalent to (3.1).
Remark. The first author used products like
Q
j∈N (1
− ζ j ) extensively in [3].
Let us now consider the product
(p−1)/2
Y 1 − ζ 16j 2
U
=
1 − ζ 16j
Π16
j=1
where
(p−1)/2
Πr =
Y
(1 − ζ rj ).
j=1
When r and s are prime to p, we have Πrs = σs (Πr ) where σs is the automorphism
over Q(ζ) mapping ζ to ζ s . It turns out to be convenient to compute Π16 as σ4 (Π4 ).
6
ROBIN CHAPMAN AND HAO PAN
−h(p) √
As p ≡ 1 (mod 4) we know that U = εp
2
|Πr | = Πr Π−r
p. For each r prime to p,
p−1
Y
=
(1 − ζ j ) = p,
j=1
so |Πr | =
√
p. Now
(p−1)/2
(p−1)/2
Y 1 − ζ 4j
Y
2i sin(4πj/p)
Π4
(−ζ 2j )
=
=
4j
|Π4 |
|1 − ζ |
|2i sin(4πj/p)|
j=1
j=1
(p−1)/2
M
=(−1)
Y
2 −1)/4
(−iζ 2j ) = (−1)M (−i)(p−1)/2 ζ (p
j=1
2 −1)/4
=(−1)(p−1)/4+M ζ (p
where
M = |{1 6 j 6 (p − 1)/2 : sin(4πj/p) < 0}|.
Note that when 0 < j < p/2, sin(4πj/p) < 0 if and only if p/4 < j < p/2. So
M = (p − 1)/4, and
√
2
Π4 = ζ (p −1)/4 p.
Also,
X
p−1 p−1 p−1 X
j 4j X j j √
j j
√
σ4 ( p) = σ4
ζ =
ζ =
ζ = p
p
p
p
j=1
j=1
j=1
Thus
√
√
√
2
2
Π16 = σ4 (ζ (p −1)/4 p) = ζ p −1 p = ζ −1 p.
√
h(p)
Assume that εp = (c + d p)/2 where c, d are integers with the same parity.
√
−h(p)
Recall that the norm of εp is −1 and h(p) is odd. Hence εp
= (−c + d p)/2.
So
√
c = εh(p)
− ε−h(p)
and
d p = εh(p)
+ εp−h(p) .
p
p
p
P
√
As p = −1 − 2 j∈N ζ j , we have
√
X
−c + d p
c+d
−h(p)
εp
=
=−
−d
ζj.
2
2
j∈N
Therefore
−h(p) √
Y 1 − ζ 16j 2
X
p
εp
U
(c + d)ζ
=
=
=
−
−
d
ζ j+1 .
√
16j
−1 p
1
−
ζ
Π
ζ
2
16
j=1
j∈N
(p−1)/2
q-ANALOGUE OF WILSON’S THEOREM
This concludes that
(p−1)/2
Y
j=1
[j]q16j ≡ −
X
c+d
q−d
q j+1
2
j∈N
We are done.
7
(mod [p]q ).
Acknowledgment. The second author thanks Professor Zhi-Wei Sun for his helpful suggestions on this paper.
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Department of Mathematics, University of Exeter, Exeter, EX4 4QE, UK, 01392
263971
E-mail address: [email protected]
Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic of China
E-mail address: [email protected]