Download An inequality for 3-factor Carmichael numbers due to J.M. Chick Let

An inequality for 3-factor Carmichael numbers due to J.M. Chick
Let n = pqr be a Carmichael number, with p, q, r prime and p < q < r. Let g be the
gcd of p − 1, q − 1 and r − 1, and
p = ag + 1,
q = bg + 1,
r = cg + 1.
Then g is an even integer, a, b, c are pairwise coprime, and there is an integer k such that
E =: (bc + ca + ab)g + a + b + c = kabc
[e.g. Jam, 2.6]. It follows that ka ≤ 3g − 1 (so, in particular, a ≤ 3g − 1), since
ab
a b
kab = g a + b +
+ 1+ +
c
c c
< g(2a + b) + 3 ≤ g(3b − 2) + 3 < 3bg.
The following stronger inequality was established by J.M. Chick [Ch, Theorem 4.1]. Here
we present a slightly simplified version of Chick’s proof.
THEOREM. For all 3-factor Carmichael numbers, we have a < 3g − (g/2)1/2 .
To start the proof, note that if k ≥ 2, then a < 32 g, so we assume that k = 1. Now
1 1 1
1
1
1
k=
+ +
+ .
(1)
g+ +
a b c
bc ca ab
Write
a = 3g + α,
b = 3g + β,
c = 3g + γ.
(2)
Then α < β < γ, and since 1 ≤ a < 3g, we have −3g < α < 0. Our aim is to show that
−α > (g/2)1/2 , which equates to g < 2α2 . Note that
k=
E
m
=1−
,
abc
abc
where m = abc − E, so that k = 1 if m = 0. We write
S1 = α + β + γ,
S2 = βγ + γα + αβ.
LEMMA 1. We have
m = m(g) =: 3S1 g 2 + (2S2 − 9)g + αβγ − S1 .
Proof. This follows from the identities
abc = 27g 3 + 9S1 g 2 + 3S2 g + αβγ,
1
(3)
E = (27g 2 + 6S1 g + S2 )g + 9g + S1 .
LEMMA 2. If k = 1, then S1 > 0.
Proof. If x + y > 0, then x/(x + y) > (x − y)/x, since x2 > x2 − y 2 . Hence
g
g
1
α
=
>
1−
,
a
3(g + α3 )
3
3g
so by (1),
1=k>
1 1 1
+ +
a b c
g >1−
S1
,
9g
hence S1 > 0.
Call a triple of integers (α, β, γ) admissible if α < 0, α ≤ β < γ and S1 > 0. Given
such a triple, define m(g) by (3): this is a quadratic function of g with positive coefficient
of g 2 . Also, for g > −α/3, define a, b, c by (2)) and k(g) by (1). Then a > 0 and
k(g) = 1 − m(g)/(abc). For (a, b, c, g) to generate a Carmichael number with k = 1, we
require g to be an even positive integer satisfying g > −α/3 and m(g) = 0..
LEMMA 3. If (α, β, γ) is admissible, then m(−α/3) < 0.
Proof. We have
α
1 2
α
m −
=
α (α + β + γ) − (2βγ + 2γα + 2αβ − 9) + αβγ − α − β − γ
3
3
3
α 2
=
(α − αβ − αγ + βγ) + 2α − β − γ
3
α
=
(β − α)(γ − α) + (α − β) + (α − γ)
3
< 0.
Since m(g) is a quadratic with positive leading coefficient, it follows that there is a
unique g ∗ = g ∗ (α, β, γ) > −α/3 with m(g ∗ ) = 0. Further, m(g) < 0 for −α/3 < g < g ∗ and
m(g) > 0 for g > g ∗ .
LEMMA 4. Suppose that (α1 , β1 , γ1 ) and (α2 , β2 , γ2 ) are admissible and α1 = α2 = α,
β1 ≤ β2 and γ1 ≤ γ2 . Let mj (g) be derived from (αj , βj , γj ) for j = 1, 2. Suppose that
g > −α/3 and m1 (g) > 0. Then m2 (g) > m1 (g).
Proof. Then (with obvious notation), a2 = a1 > 0, b2 ≥ b1 and c2 ≥ c1 . By (1),
k2 (g) < k1 (g), hence m2 (g)/(a2 b2 c2 ) > m1 (g)/(a1 b1 c1 ), so m2 (g) > m1 (g).
LEMMA 5. Suppose that (α1 , β1 , γ1 ) and α2 , β2 , γ2 ) are admissible and α1 = α2 = α,
β2 = β1 +δ and γ2 ≤ γ1 −δ for some δ > 0. If g > −α/3 and m1 (g) > 0, then m2 (g) > m1 (g).
2
Proof. Rewrite k(g) as follows:
g
1
1 1
1
g
1 b+c
1
k(g) = + g +
+
+
= + g+
+ .
a
a
b c
bc
a
a
bc
bc
Now b2 = b1 + δ and c2 = c1 − δ, so b2 + c2 = b1 + c1 and b2 c2 = b1 c1 + δ(c1 − b1 − δ) > b1 c1 ,
since c1 > b1 + δ. Hence k2 (g) ≤ k1 (g), and the conclusion follows as in Lemma 4.
.
Conclusion of proof of the Theorem. We show that for all admissible triples (α, β, γ)
compatible with a, b and c being pairwise coprime, we have m(2α2 ) > 0, so that g ∗ < 2α2 .
We consider cases determined by the values of α and S1 .
Case 1: α ≤ −2, S1 ≥ 2. By Lemma 5, it is enough to prove the statement for triples
of the form (α, α, γ), since it then follows for all (α, β, γ) with the same value of S1 . For
such a triple, write ρ = −α and S1 = S. Then γ = S + 2ρ and S2 = ρ2 − 2ργ = −2ρS − 3ρ2 .
By (3),
m(2ρ2 ) = 12Sρ4 − 2ρ2 (4Sρ + 6ρ2 + 9) + ρ2 (S + 2ρ) − S
= 12(S − 1)ρ4 − (8S − 2)ρ3 + (S − 18)ρ2 − S.
Since ρ ≥ 2 and S ≥ 2, we can insert ρ4 ≥ 2ρ3 and ρ3 ≥ 2ρ2 to obtain
m(2ρ2 ) ≥ (16S − 22)ρ3 + (S − 18)ρ2 − S ≥ (33S − 62)ρ2 − S > 0.
Case 2: α ≤ −2, S1 = 1. In this case, a + b + c is odd. Since a, b and c are pairwise
coprime, they are all odd, so α, β and γ are odd, and β ≥ α + 2. Again Lemma 5 ensures
that it is sufficient to prove the result for the case β = α + 2, so we take α = −ρ, β = 2 − ρ
and γ = 2ρ − 1. Then
S2 = ρ(ρ − 2) + (2 − 2ρ)(2ρ − 1) = −3ρ2 + 4ρ − 2,
αβγ = ρ(2ρ2 − 5ρ + 2),
so, since ρ ≥ 2,
m(2ρ2 ) = 12ρ4 + 2ρ2 (−6ρ2 + 8ρ − 13) + ρ(2ρ2 − 5ρ + 2) − 1
= 18ρ3 − 31ρ2 + 2ρ − 1
≥ 5ρ2 + 2ρ − 1 > 0.
Case 3: α = −1. First, suppose β = 0. Then γ ≥ 2: since a, b and c have to be
coprime and b = 3g, it follows that γ ≥ 5. Now S1 = γ − 1 and S2 = −γ, so
m(g) = 3(γ − 1)g 2 − (2γ + 9)g − γ + 1,
3
hence m(2) = 7γ − 29 > 0, so g ∗ < 2. Since we require g ≥ 2, this actually means that there
are no Carmichael numbers of this type.
Next, suppose that β = 1, so γ ≥ 2. Then S1 = γ and S2 = −1, so m(g) = 3γg 2 −
11g − 2γ, hence m(2) = 10γ − 22 and m(3) = 25γ − 33. If γ ≥ 3, this gives m(2) > 0, so
again no value for g. Lemma 4 now shows that the same applies whenever β ≥ 2. Finally,
if γ = 2, then m(3) > 0 and m(2) 6= 0, so again there is no value for g.
Note 1. In most cases, we have the stronger inequality a < 3g − g 1/2 . This will follow
if we can show that m(α2 ) > 0. Reasoning as in Case 3, it is easily seen that there are no
cases with α = −2. With α ≤ −3, by considering m(g) as above, one can see that any of
the following conditions implies that m(α2 ) > 0:
(1)
S1 ≥ 3 (as in Case 1, but using β ≥ α + 1);
(2)
S1 = 2 and β ≥ α + 2;
(3)
S1 = 1 and β > 0.
However, m(α2 ) < 0 in the cases S1 = 1, β = α + 2 and S1 = 2, β = α + 1. A specific
example in which there is an integer solution for g greater than α2 is (α, β, γ) = (−3, −1, 5),
which has the solution g = 14, leading to (a, b, c) = (39, 41, 47). However, this combination
fails to generate a Carmichael number, because 41 × 14 + 1 is the composite number 575 (in
Chick’s terminology, it generates a “K3 -number”).
Note 2. In practice, there are very few C3 -numbers with a > 2g. Computations
reported by [Ch, p. 15] have found that there are just eleven such numbers up to 1024 .
The first one is defined by (a, b, c, g) = (1049, 1841, 2304, 518), and the largest value of
a/g (≈ 2.683) is attained by (a, b, c, g) = (1497, 1601, 2002, 558). All these numbers satisfy
a < 3g − (λg)1/2 with λ > 10.
References
[Ch]
[Jam]
J.M. Chick, Carmichael number variable relations: three-prime Carmichael
numbers up to 1024 , arXiv:0711.2915v2[math.NT].
G.J.O. Jameson, Carmichael numbers with three prime factors, at
www.maths.lancs.ac.uk/~jameson.
G.J.O. Jameson
December 2011
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