Lattice points and Number theory
July 26, 2023
Definition 1. A simplex is the convex combination of n + 1
points in the n−dimensional Euclidean space Rn , viz, all the points
of the form
n+1
P
θi Xi where
i=1
n+1
P
θi = 1 and θi ≥ 0.
i=1
Definition 2. A hyperplane is called integral if it can be
n
P
determined by the equation
ai xi = 0 with ai ∈ Z. Obviously,
i=1
there are infinitely many lattice points on an integral plane and
they form an (n − 1)−dimensional linear space on Z, denoted by S.
−−→
Definition 3. If a basis v1 , · · · , vn−1 with vi = OXi satisfies
the simplex △(X1 , · · · , Xn−1 ) contains only Xi ’s as lattice points,
it is called a primitive basis of S, and we call (n − 1)! times the
(n − 1)−dimensional volume of the simplex the character of the
basis.
Definition 4. Two non-empty and bounded sets U1 , U2 in Rn
are called similar if there is a shape-preserving transformation f
such that f (U1 ) = U2 .
Our first result is as follows:
Proposition 1. Assume that an n-dimensional integral plane P
has character c. B1 , B2 , · · · is a sequence of n-dimensional closed
balls on P with r(Bn ) → ∞, denote the number of lattice interior
to or on the boundary of Bn by xn , then
cxn
=1
n→∞ Vol(Bn )
lim
Here, the sequence of balls can be replaced by any relatively similar connected sets, whose n-volume exists and the (n − 1)-volume of
whose boundary also exists.
Proof. Let B = v1 , · · · , vn be a primitive basis of P . Let Sk
be the set of lattice points interior to or on the boundary of Bk .
Define Tk by {X − v|X ∈ Sk , v ∈ B}. For every point X, we
n
P
construct a hyper-parallelpiped Si to be {Xi +
θi vi |0 ≤ θi <
i=1
Note that, the proposition implies
that each primitive basis of S has the
same character.
lattice points and number theory
1}. It is trivial that Vol(Si ) = c. It is also trivial to verify that
S
S
⊆ Bn ⊆
.
X∈Sk ∩Tk
X∈Sk ∪Tk
Since the distance between any point in Sk − Tk and the
boundary of the ball is bounded by max |vi |, we have|Sk − Tk | =
O( ddr Vol(Bn )) = o(Vol(Bn )). So by the same token |Tk − Sk | =
k
o(Vol(Bn )). The conclusion follows.
■
We can now prove:
Proposition 2. Let B be a primitive basis of the lattice points
n
P
I on the integral plane P :
ai xi = 0 with ai ∈ Z and there is no
i=1
common
rdivisor of a1 , · · · , an greater thanr1. Then the character
of B is
n
P
a2i . Therefore we can define
i=1
n
P
a2i as the character
i=1
of the hyperplane.
Proof. We prove the proposition by induction on n.
When n = 2, (a, −b) is a primitive basis of the line, so the
√
character of the line is a2 + b2 .
When the conclusion holds for n − 1, let d be the greatest
common divisor of a1 , a2 , · · · , an−1 . Then by inductive
hypothesis,
s
n−1
n−1
P
P 2
ai xi = 0 is
the character of the n − 2-plane P ′ :
ai /d. That
i=1
i=1
is, there
sare Xi ’s (1 ≤ i ≤ n − 1) such that Vol(△(X1 , · · · , Xn−1 )) =
n−1
P 2
1
ai .
(n−2)!d
i=1
Bézout’s theorem guarantees the existence of mi ∈ Z such that
n−1
P
ai mi = −an d. Let B′ be a primitive basis of P ′ , denote the
−−→
point (m1 , m2 , · · · , mn−1 , d) by M . We first prove that B′ ∪ {OM }
i=1
is a primitive basis of P . If X is a lattice point in △(B, M ), since
it is on P , we shall have Xn = 0 or d because gcd(d, an ) = 1. For
the former, it reduces to the (n − 1)-plane P ′ . For the latter, we
must have θn = 1 so X = M .
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lattice points and number theory
Then the character of P is
v
un−1
uX
1
1
t
(n − 1)! ·
a2i ·
dist(X, P ′ )
(n − 2)!d i=1
n−1
v
un−1 v
X u
d2 a2
1u
u
a2i ud2 + n−1 n
= t
P 2
d i=1 t
ai
i=1
v
u n
uX
a2
=t
i
i=1
■
Proposition 3. Let a1 , · · · , ak be positive integers without a
common divisor greater than 1. Let An be the number of solutions
of the equation
a1 x1 + · · · + ak xk = n
where xi ’s are non-negative integers. Then
An
1
=
.
k−1
n→∞ n
a1 · · · ak (k − 1)!
lim
Proof 1. Let Sn =
n
P
Ai be the number of solutions of a1 x1 +
i=0
· · · + ak xk ≤ n. Denote the volume of the simplex a1 x1 + · · · +
1 n
ak xk ≤ n, xi ≥ 0 by Vn . It follows that Vn = k!
· · · ank . Since the
a1
character of character of Rk is 1, we can conclude that lim SVnn = 1.
n→∞
Noting that Vn is increasing and tends to infinity, we can also
n−1
= 1, which implies the proposition.
conclude that lim SVnn −S
−Vn−1
n→∞
Proof 2. The
s area of the (k − 1)-simplex a1 x1 + · · · + ak xk = n,
k
P
nk−1
xi ≥ 0 is (k−1)!
a2i . Since the character of the integral plane
i=1
s
k
P
a1 x1 + · · · + ak xk = n is
a2i , and An is the number of lattice
i=1
points on the plane, the conclusion follows.
■
Note also, the proposition can be proved using generating function.
♠
See G.Pólya’s book Problems and
theorems in Analysis I, page 5.
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