article12_5. 186KB Jun 04 2011 12:06:56 AM
Journal de Th´eorie des Nombres
de Bordeaux 18 (2006), 223–239
An arithmetic formula of Liouville
par Erin McAFEE et Kenneth S. WILLIAMS
´sume
´. Une preuve ´el´ementaire est donn´ee d’une formule arithRe
m´etique qui fut pr´esent´ee mais non pas prouv´ee par Liouville. Une
application de cette formule donne une formule pour le nombre de
repr´esentations d’un nombre entier positif comme ´etant la somme
de douze nombres triangulaires.
Abstract. An elementary proof is given of an arithmetic formula, which was stated but not proved by Liouville. An application of this formula yields a formula for the number of representations of a positive integer as the sum of twelve triangular
numbers.
This paper is dedicated to the memory of Joseph Liouville (1809-1882).
1. Introduction
In his famous series of eighteen articles, Liouville [3] gave without proof
numerous arithmetic formulae. These formulae are summarized in Dickson’s History of the Theory of Numbers [1, Volume 2, Chapter XI], where
references to proofs are given. The formula we are interested in involves a
sum over sextuples (a, b, c, x, y, z) of positive odd integers satisfying ax +
by + cz = n, where n is a fixed positive odd integer. It was stated without
proof by Liouville in [3, sixth article, p. 331] and reproduced by Dickson
[1, Volume 2, formula (M), p. 332].
Let n be a positive odd integer, and let F : Z → C be an odd function.
Then
X
(F (a + b + c) + F (a − b − c) − F (a + b − c) − F (a − b + c))
ax+by+cz=n
a,b,c,x,y,z odd
(1.1)
=
X
1X 2
σ(o(n − ax))F (a).
(d − 1)F (d) − 3
8
axy
x≡a,e≡b,y≡c
=
X
1=
X
k(x+y)+ey=n
x≡a+c, e≡b, y≡c
1=
X
1
kx+(k+e)y=n
x≡a+c, e≡b, y≡c
1 = L2 (a + c, k + b, c),
kx+ey=n
e>k
x≡a+c, e≡k+b, y≡c
as asserted.
For positive integers n and k we set
1, if k | n,
δ(n, k) =
0, if k ∤ n.
The next result is a simple consequence of Lemmas 2.1, 2.2 and 2.3.
Erin McAfee, Kenneth S. Williams
226
Lemma 2.5. If k and n are positive odd integers then
(n/k) − 1
δ(n, k).
L1 (1, 1, 0) − M1 (1, 1, 0) = L2 (1, 0, 0) − L2 (1, 1, 0) −
2
3. The equation kx + ey = n and the quantities R1 , R2 , R3 , S1 , S2 , S3
In this section we introduce the quantities R1 , R2 , R3 , S1 , S2 , S3 formed
from L1 , L2 , L3 , M1 , M2 , M3 respectively by taking the sums over e rather
than 1.
Definition 3.1.
R1 (a, b, c) =
X
e, R2 (a, b, c) =
X
e, S1 (a, b, c) =
X
e, S3 (a, b, c) =
kx+ey=n
ey
x≡a,e≡b,y≡c
e,
X
e,
X
e.
kx+ey=n
e>k
x≡a,e≡b,y≡c
kx+ey=n
xk
d≡k+b
n/d≡a
The following lemma gives a relationship between R2 (a, b, c) and
S2 (a, b, c). The proof is similar to that of Lemma 2.4.
An arithmetic formula of Liouville
227
Lemma 3.4.
S2 (a, b, c) = R2 (a + c, k + b, c) − kL2 (a + e, k + b, c).
The values of R3 (1, 1, 0) and S3 (1, 1, 0) when k and n are both odd follow
easily from Lemmas 3.2 and 3.3.
Lemma 3.5. If k and n are positive odd integers then
n−k
δ(n, k), S3 (1, 1, 0) = 0.
R3 (1, 1, 0) =
2
4. Solutions of kx + by + cz = n (k, n fixed odd integers)
Throughout this section k and n are restricted to be positive odd integers.
We are concerned with solutions (x, b, y, c, z) ∈ N5 of
kx + by + cz = n
with b, c, x odd and y, z even. For such solutions we have b+c 6= k, b−c 6= k
and x 6= z, so that
X
X
X
1
1+
1=
kx+by+cz=n
b+c>k
b,c,x odd, y,z even
kx+by+cz=n
b+ck, xk, y=z
b,c,x odd, y,z even
where we used the transformation (b, y, c, z) → (c, z, b, y) in the last step.
We now examine the above two sums. Firstly, we have
X
X
1=
1
kx+by+cz=n
b+c>k, y=z
b,c,x odd, y,z even
kx+(b+c)y=n
b+c>k
b,c,x odd, y even
=
=
X e
2
e even
1
2
X
1
kx+ey=n
e>k
x odd, y even
X
e
kx+ey=n
e>k
x≡1,e≡0,y≡0
1
= R2 (1, 0, 0),
2
by Definition 3.1, and secondly,
X
1=
kx+by+cz=n
b+c>k, y>z
b,c,x odd, y,z even
X
1
(y = z + y ′ )
X
1
(y = y ′ )
kx+b(z+y ′ )+cz=n
b+c>k
b,c,x odd, y ′ ,z even
=
kx+by+(b+c)z=n
b+c>k
b,c,x odd, y,z even
=
X
kx+by+ez=n
e=b+c,e>k
b,c,x odd, e,y,z even
1
Erin McAfee, Kenneth S. Williams
230
=
X
1
X
1
(e = k + c′ )
X
1
(c = c′ )
X
1
(x′ = x + z)
kx+by+ez=n
e>b,e>k
b,x odd, e,y,z even
=
kx+by+(k+c′ )z=n
k+c′ >b
b,c′ ,x odd, y,z even
=
k(x+z)+by+cz=n
b−c
de Bordeaux 18 (2006), 223–239
An arithmetic formula of Liouville
par Erin McAFEE et Kenneth S. WILLIAMS
´sume
´. Une preuve ´el´ementaire est donn´ee d’une formule arithRe
m´etique qui fut pr´esent´ee mais non pas prouv´ee par Liouville. Une
application de cette formule donne une formule pour le nombre de
repr´esentations d’un nombre entier positif comme ´etant la somme
de douze nombres triangulaires.
Abstract. An elementary proof is given of an arithmetic formula, which was stated but not proved by Liouville. An application of this formula yields a formula for the number of representations of a positive integer as the sum of twelve triangular
numbers.
This paper is dedicated to the memory of Joseph Liouville (1809-1882).
1. Introduction
In his famous series of eighteen articles, Liouville [3] gave without proof
numerous arithmetic formulae. These formulae are summarized in Dickson’s History of the Theory of Numbers [1, Volume 2, Chapter XI], where
references to proofs are given. The formula we are interested in involves a
sum over sextuples (a, b, c, x, y, z) of positive odd integers satisfying ax +
by + cz = n, where n is a fixed positive odd integer. It was stated without
proof by Liouville in [3, sixth article, p. 331] and reproduced by Dickson
[1, Volume 2, formula (M), p. 332].
Let n be a positive odd integer, and let F : Z → C be an odd function.
Then
X
(F (a + b + c) + F (a − b − c) − F (a + b − c) − F (a − b + c))
ax+by+cz=n
a,b,c,x,y,z odd
(1.1)
=
X
1X 2
σ(o(n − ax))F (a).
(d − 1)F (d) − 3
8
axy
x≡a,e≡b,y≡c
=
X
1=
X
k(x+y)+ey=n
x≡a+c, e≡b, y≡c
1=
X
1
kx+(k+e)y=n
x≡a+c, e≡b, y≡c
1 = L2 (a + c, k + b, c),
kx+ey=n
e>k
x≡a+c, e≡k+b, y≡c
as asserted.
For positive integers n and k we set
1, if k | n,
δ(n, k) =
0, if k ∤ n.
The next result is a simple consequence of Lemmas 2.1, 2.2 and 2.3.
Erin McAfee, Kenneth S. Williams
226
Lemma 2.5. If k and n are positive odd integers then
(n/k) − 1
δ(n, k).
L1 (1, 1, 0) − M1 (1, 1, 0) = L2 (1, 0, 0) − L2 (1, 1, 0) −
2
3. The equation kx + ey = n and the quantities R1 , R2 , R3 , S1 , S2 , S3
In this section we introduce the quantities R1 , R2 , R3 , S1 , S2 , S3 formed
from L1 , L2 , L3 , M1 , M2 , M3 respectively by taking the sums over e rather
than 1.
Definition 3.1.
R1 (a, b, c) =
X
e, R2 (a, b, c) =
X
e, S1 (a, b, c) =
X
e, S3 (a, b, c) =
kx+ey=n
ey
x≡a,e≡b,y≡c
e,
X
e,
X
e.
kx+ey=n
e>k
x≡a,e≡b,y≡c
kx+ey=n
xk
d≡k+b
n/d≡a
The following lemma gives a relationship between R2 (a, b, c) and
S2 (a, b, c). The proof is similar to that of Lemma 2.4.
An arithmetic formula of Liouville
227
Lemma 3.4.
S2 (a, b, c) = R2 (a + c, k + b, c) − kL2 (a + e, k + b, c).
The values of R3 (1, 1, 0) and S3 (1, 1, 0) when k and n are both odd follow
easily from Lemmas 3.2 and 3.3.
Lemma 3.5. If k and n are positive odd integers then
n−k
δ(n, k), S3 (1, 1, 0) = 0.
R3 (1, 1, 0) =
2
4. Solutions of kx + by + cz = n (k, n fixed odd integers)
Throughout this section k and n are restricted to be positive odd integers.
We are concerned with solutions (x, b, y, c, z) ∈ N5 of
kx + by + cz = n
with b, c, x odd and y, z even. For such solutions we have b+c 6= k, b−c 6= k
and x 6= z, so that
X
X
X
1
1+
1=
kx+by+cz=n
b+c>k
b,c,x odd, y,z even
kx+by+cz=n
b+ck, xk, y=z
b,c,x odd, y,z even
where we used the transformation (b, y, c, z) → (c, z, b, y) in the last step.
We now examine the above two sums. Firstly, we have
X
X
1=
1
kx+by+cz=n
b+c>k, y=z
b,c,x odd, y,z even
kx+(b+c)y=n
b+c>k
b,c,x odd, y even
=
=
X e
2
e even
1
2
X
1
kx+ey=n
e>k
x odd, y even
X
e
kx+ey=n
e>k
x≡1,e≡0,y≡0
1
= R2 (1, 0, 0),
2
by Definition 3.1, and secondly,
X
1=
kx+by+cz=n
b+c>k, y>z
b,c,x odd, y,z even
X
1
(y = z + y ′ )
X
1
(y = y ′ )
kx+b(z+y ′ )+cz=n
b+c>k
b,c,x odd, y ′ ,z even
=
kx+by+(b+c)z=n
b+c>k
b,c,x odd, y,z even
=
X
kx+by+ez=n
e=b+c,e>k
b,c,x odd, e,y,z even
1
Erin McAfee, Kenneth S. Williams
230
=
X
1
X
1
(e = k + c′ )
X
1
(c = c′ )
X
1
(x′ = x + z)
kx+by+ez=n
e>b,e>k
b,x odd, e,y,z even
=
kx+by+(k+c′ )z=n
k+c′ >b
b,c′ ,x odd, y,z even
=
k(x+z)+by+cz=n
b−c