Heath-Brown, D. R.

English

Mathematical Institute Eprints Archive

Carmichael Numbers With Three Prime Factors
D.R. Heath-Brown
Mathematical Institute, Oxford
A Carmichael number (or absolute pseudo-prime) is a composite positive
integer n such that n|an − a for every integer a. It is not difficult to prove that
such an integer must be square-free, with at least 3 prime factors. Moreover
if the numbers p = 6m + 1, q = 12m + 1 and r = 18m + 1 are all prime,
then n = pqr will be a Carmichael number. However it is not currently known
whether there are infinitely many prime triplets of this form. Indeed it is not
known whether or not there are infinitely many Carmichael numbers with 3
prime factors. None the less it was proved by Alford, Granville and Pomerance
[1] that there are infinitely many Carmichael numbers.
Let C3(x) denote the number of Carmichael numbers n ≤ x having ω(n) = 3.
It has been conjectured by Granville and Pomerance [3] that
C3(x) ∼ c x
1/3
(log x)3
,
with an explicit positive constant c. Various upper bounds approximating this
have been given, the best available in the literature being the estimate
C3(x)¿ε x5/14+ε,
for any fixed ε > 0, due to Balasubramanian and Nagaraj [2]. The goal of the
present paper is to improve this as follows.
Theorem For any fixed ε > 0 we have
C3(x)¿ε x7/20+ε.
Note that 1/3 < 7/20 < 5/14. Indeed
7
20 − 13
5
14 − 13
=
7
10
.
Thus our results reduces the previous excess in the exponent by 30%.
Suppose that n = pqr is counted by C3(x), where p < q < r. We write
m = (p − 1, q − 1, r − 1) and p = 1 + am, q = 1 + bm, r = 1 + cm. Since
p− 1 = am divides n− 1 = abcm3 + (ab+ ac+ bc)m2 + (a+ b+ c)m, we must
have
a|(ab+ ac+ bc)m+ a+ b+ c. (1)
Thus (a, b)|c, and since we know that (a, b, c) = 1 we deduce that a and b are
coprime. Similarly a and c are coprime, and also b and c. In analogy to (1) we
see likewise that both b and c divide (ab + ac + bc)m + a + b + c. Since a, b
1
and c have been shown to be coprime in pairs we conclude that the product abc
divides (ab+ ac+ bc)m+ a+ b+ c. We may therefore write
(ab+ ac+ bc)m+ a+ b+ c = abck, (2)
say, for some integer k. We proceed to divide the available ranges for a, b, c,m
and k into O((log x)5) dyadic ranges (A, 2A], (B, 2B], (C, 2C], (M, 2M ] and
(K, 2K]. We shall write C3(x;A,B,C,M,K) for the corresponding contribution
to C3(x). Since 2 ≤ p < q < r we may suppose that
1 ≤ A ≤ B ≤ C. (3)
Moreover, since pqr ≤ x we will have
M3ABC ≤ x. (4)
Finally we observe from (2) that
ABCK < abck = (ab+ ac+ bc)m+ a+ b+ c¿ BCM,
whence
K ¿M/A. (5)
To prove our result we give three different bounds for C3(x;A,B,C,M,K).
The first two are completely elementary, and appear in a slightly different form
in the work of Balasubramanian and Nagaraj [2].
For our first estimate we count the 5-tuples (a, b, c,m, k) according to the
values of a, b and k. We multiply (2) by a+ b and re-write the answer as
{abk − 1− (a+ b)m}{(a+ b)c+ ab} = N1, with N1 = a2b2k + a2 + b2 + abk.
One readily sees that N1 = O(x) by (3), (4) and (5). Hence the number of
possible divisors of N1 is Oε(xε). Moreover, given a, b and k, each pair of
divisors determines c and m uniquely. We therefore conclude that
C3(x;A,B,C,M,K)¿ε xεABK ¿ε xεBM, (6)
by (5).
For our second estimate we substitute h = ak −m into (2), to deduce that
hbc = ma(b+ c) + a+ b+ c. (7)
It follows that
h ≥ 1 and h¿ (BC)−1.MAC =MAB−1. (8)
We proceed to count solutions according to the values of h and m, there being
O(M2AB−1) such pairs. Given h and m we have ak = h + m. However
h + m has Oε(xε) divisors, since h + m ¿ x, by (3), (4) and (8). Thus we
have Oε(xεM2AB−1) possible 4-tuples (a,m, k, h). For each such 4-tuple we
multiply (7) by h to obtain
{hb−ma− 1}{hc−ma− 1} = N2 with N2 = (ma+ 1)2 − ha. (9)
2
If N2 = 0 then a = 1, since a|N2 − 1. We would then have h = (m + 1)2.
However if N2 = 0 then (9) would imply hb−ma− 1 = 0 or hc−ma− 1 = 0.
However, since we must have a = 1 and h = (m+ 1)2 these would entail either
(m+1)2b = m+1 or (m+1)2c = m+1, which are impossible for m ≥ 1. Thus
we must have N2 6= 0 in (9). However it is also clear that N2 ¿ x, by (3), (4)
and (8). Thus N2 has Oε(xε) pairs of divisors, and each such pair determines b
and c, once the 4-tuple (a,m, k, h) has been specified. It therefore follows that
C3(x;A,B,C,M,K)¿ε x2εM2AB−1. (10)
We proceed to present a new estimate.
Lemma For any fixed ε > 0 we have
C3(x;A,B,C,M,K)¿ε MA+ xεA1/2BC + xεA2B1/2C1/2. (11)
Before proving this we show how the theorem follows. We consider three
cases. If
max{MA, A1/2BC , A2B1/2C1/2} =MA
in (11), then it suffices to observe that MA ≤ (M3ABC)1/3 ¿ x1/3, by (3) and
(4), so that C3(x;A,B,C,M,K)¿ x1/3. If
max{MA, A1/2BC , A2B1/2C1/2} = A1/2BC,
then (11), in conjunction with (6) and (10), yields
C3(x;A,B,C,M,K) ¿ε x2εmin{BM ,M2AB−1 , A1/2BC}
¿ε x2ε(BM)11/20(M2AB−1)1/4(A1/2BC)1/5
= x2ε(M3ABC)7/20(BC−1)3/20
≤ x2ε+7/20,
by (3) and (4). Finally, if
max{MA, A1/2BC , A2B1/2C1/2} = A2B1/2C1/2,
then (6), (10) and (11) yield
C3(x;A,B,C,M,K) ¿ε x2εmin{BM ,M2AB−1 , A2B1/2C1/2}
¿ε x2ε(BM)3/4(M2AB−1)3/20(A2B1/2C1/2)1/10
= x2ε(M3ABC)7/20(BC−1)3/10
≤ x2ε+7/20,
by (3) and (4). Thus C3(x;A,B,C,M,K) ¿ε x2ε+7/20 in every case. The
theorem then follows on replacing ε by ε/3 and summing over the O(log x)5 sets
of dyadic ranges for A,B,C,M and K.
We turn now to the proof of the lemma. From (2) we have
a(b+ c)m+ a+ b+ c ≡ 0 (mod bc). (12)
For coprime positive integers u and v we now define u(v) by the conditions
0 ≤ u(v) < v and uu(v) ≡ 1 (mod v). Where the context is clear we shall write
3
u for u(v). Since a, b and c are coprime in pairs we may now re-write (12) as
m ≡ −(b+ c) − a (mod bc), where the inverses are taken modulo bc. For a
given triple (a, b, c), each solution m of this congruence, with m in the range
M < m ≤ 2M , corresponds to at most value of k. We now write r := −(b+ c)−a
for brevity, and observe, using the familiar transformation formula for the theta-
function, that
#{m ∈ (M, 2M ] : m ≡ r (mod bc)}
≤ e4
∑
m∈Z
m≡r (mod bc)
exp{−m2M−2}
= e4
∑
n∈Z
exp{−(r + nbc)2M−2}
= e4
√
pi
M
bc
∑
n∈Z
e(
rn
bc
) exp{−(pinM/bc)2}
= e4
√
pi
M
bc
∑
n∈Z
ebc(−n(b+ c)− na) exp{−(pinM/bc)2}.
Here we use the standard notations e(t) := exp(2piit) and eq(t) := exp(2piit/q).
We may now deduce that
C3(x;A,B,C,M,K)
≤ e4√pi
∑
a,b,c
M
bc
∑
n∈Z
ebc(−n(b+ c)− na) exp{−(pinM/bc)2}
= e4
√
pi
∑
b,c
M
bc
∑
n∈Z
ebc(−n(b+ c)) exp{−(pinM/bc)2}
∑
a
ebc(−na)
¿ M
BC
∑
b,c
∑
n∈Z
exp{−(pinM/bc)2}
∣∣∣∑
a
ebc(−na)
∣∣∣
¿ M
BC
∑
b,c
∑
n∈Z
exp{−(pinM/4BC)2}
∣∣∣∑
a
ebc(−na)
∣∣∣
¿ MA+ M
BC
∑
b,c
∞∑
n=1
exp{−(pinM/4BC)2}
∣∣∣∑
a
ebc(−na)
∣∣∣,
where the variables a, b, c are restricted to be coprime in pairs, and to lie in
the intervals A < a ≤ 2A, B < b ≤ 2B and C < c ≤ 2C. At this point we
combine the variables b and c, writing bc = f , say. Thus f lies in the range
BC < f ≤ 4BC, and is coprime to a. Moreover each available value of f arises
at most d(f)¿ε xε times, so that
C3(x;A,B,C,M,K)
¿ε MA+ xε M
BC
∑
BC<f≤4BC
∞∑
n=1
exp{−(pinM/4BC)2}
∣∣∣ ∑
A<a≤2A
(a,f)=1
ef (−na)
∣∣∣.
We proceed to apply Cauchy’s inequality, whence
C3(x;A,B,C,M,K)¿ε MA+ xε M
BC
S
1/2
1 S
1/2
2 , (13)
4
where
S1 :=
∑
BC<f≤4BC
∞∑
n=1
exp{−(pinM/4BC)2}
and
S2 :=
∑
BC<f≤4BC
∞∑
n=1
exp{−(pinM/4BC)2}
∣∣∣ ∑
A<a≤2A
(a,f)=1
ef (−na)
∣∣∣2.
Since ∞∑
n=1
exp{−n2t−2} ¿ t (14)
uniformly for t > 0, we trivially have
S1 ¿ B2C2M−1. (15)
To handle S2 we expand the square, and rearrange the sum to produce
S2 =
∞∑
n=1
exp{−(pinM/4BC)2}
∑
A<a1,a2≤2A
S3(a1, a2;n), (16)
with
S3(a1, a2;n) :=
∑
BC<f≤4BC
(f,a1a2)=1
ef (n(a1 − a2)).
To deal with this last sum we observe that
a1 − a2 ≡ (a2 − a1)a1a2 (mod f).
Moreover we have
a1a2a1a2
(f) + ff
(a1a2) ≡ 1 (mod a1a2f),
whence
a1a2
(f)
f
= −f
(a1a2)
a1a2
+
1
a1a2f
+ v
for some integer v. It follows that
ef (n(a1 − a2)) = ef (n(a2 − a1)a1a2) = ea1a2(n(a1 − a2)f
(a1a2))e
(n(a2 − a1)
a1a2f
)
.
We may therefore write
S3(a1, a2;n) =
∑
BC<f≤4BC
(f,g)=1
eg(rf
(g)
)e(δ/f),
where we have written g := a1a2, r := n(a1 − a2) and δ := n(a2 − a1)/a1a2 for
brevity. An elementary estimate for S3(a1, a2;n) will suffice for our purposes.
By partial summation we have
S3(a1, a2;n)¿
(
1 +
|δ|
BC
)
max
BC<F≤4BC
∣∣∣ ∑
BC<f≤F
(f,g)=1
eg(rf
(g)
)
∣∣∣.
5
Moreover, on writing
cg(r) =
∑
0<f≤g
(f,g)=1
eg(rf
(g)
) =
∑
d|g, d|r
µ(g/d)d¿ε (g, r)gε
for the Ramanujan sum, we have∑
BC<f≤F
(f,g)=1
eg(rf
(g)
) =
[F −BC
g
]
cg(r) +O(g)
¿ε BC
g
(g, r)gε + g.
Thus, on recalling the definitions of g, r and δ, we deduce from (3) that
S3(a1, a2;n) ¿
(
1 +
n
ABC
){BC
A2
(
a1a2, n(a1 − a2)
)
A2ε +A2}
¿ BC
A2
(
a1a2, n(a1 − a2)
)
A2ε +A2 +
n
ABC
{BCA2ε +A2}
¿ BC
A2
(
a1a2, n(a1 − a2)
)
A2ε +A2 + nA−1+2ε,
in view of the fact that
(
a1a2, n(a1 − a2)
) ≤ a1a2.
We are now ready to perform the summation over n in (16).We begin by
noting that
(
a1a2, n(a1 − a2)
) ≤ (a1a2, (a1 − a2))(a1a2, n). If we now classify
integers n according to the value of (a1a2, n) we find, using (14), that
∞∑
n=1
(
a1a2, n(a1 − a2)
)
exp{−(pinM/4BC)2}
≤ (a1a2, (a1 − a2)) ∑
d|a1a2
d
∑
n:d|n
exp{−(pinM/4BC)2}
=
(
a1a2, (a1 − a2)
) ∑
d|a1a2
d
∞∑
s=1
exp{−(pisdM/4BC)2}
¿ε
(
a1a2, (a1 − a2)
) ∑
d|a1a2
d
BC
dM
¿ε
(
a1a2, (a1 − a2)
)BC
M
xε,
since the number of divisors of a1a2 is ¿ε (a1a2)ε ¿ε xε. Thus
∞∑
n=1
BC
A2
(
a1a2, n(a1 − a2)
)
A2ε exp{−(pinM/4BC)2}
¿ε
(
a1a2, (a1 − a2)
)
B2C2A−2M−1x2ε.
A second application of (14) shows that
∞∑
n=1
A2 exp{−(pinM/4BC)2} ¿ A2BCM−1,
6
while the bound ∞∑
n=1
n exp{−n2t−2} ¿ t2,
which holds uniformly for t > 0, implies that
∞∑
n=1
nA−1+2ε exp{−(pinM/4BC)2} ¿ A−1B2C2M−2xε.
On combining these bounds we find that
∞∑
n=1
exp{−(pinM/4BC)2}S3(a1, a2;n)
¿ε
(
a1a2, (a1 − a2)
)
B2C2A−2M−1x2ε
+A2BCM−1 +A−1B2C2M−2xε.
To complete our bound for S2 we must investigate∑
A<a1,a2≤2A
(
a1a2, (a1 − a2)
)
.
When a1 = a2 we have
(
a1a2, (a1 − a2)
)
= a1a2 ¿ A2, while for a1 6= a2 we
have
(
a1a2, (a1 − a2)
) ≤ |a1 − a2| ≤ A. It follows that the sum above is O(A3).
We therefore deduce from (16) and (3) that
S2 ¿ε AB2C2M−1x2ε +A4BCM−1 +AB2C2M−2xε
¿ε AB2C2M−1x2ε +A4BCM−1.
If we combine this with (13) and (15), we deduce that
C3(x;A,B,C,M,K)¿ε MA+ x2εA1/2BC + xεA2B1/2C1/2,
and the lemma follows on replacing ε by ε/2.
References
[1] W.R. Alford, A. Granville and C. Pomerance, There are infinitely many
Carmichael numbers, Ann. Math., 140 (1994), 703–722.
[2] R. Balasubramanian and S.V. Nagaraj, Density of Carmichael numbers with
three prime factors, Math. Comp., 66 (1997), 1705–1708.
[3] A. Granville and C. Pomerance, Two contradictory conjectures concerning
Carmichael numbers, Math. Comp., 71 (2002), 883–90.
Mathematical Institute,
24–29, St. Giles’,
Oxford
OX1 3LB
UK
rhb@maths.ox.ac.uk
7


Carmichael numbers with three prime factors

Abstract

Similar works

Full text

Available Versions

Mathematical Institute Eprints Archive

Oxford University Research Archive