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On a Decomposition of Integer Vectors, I
by Iskander ALIEV (Warszawa)
Presented by A. SCHINZEL on June 21, 2000
Summary. It is proved that for every non–zero vector n= (n1, n2, n3)∈Z3with
height h(n) = maxi∈{1,2,3}|ni|there exist linearly independent vectors p,q∈Z3, such
that n=up+vq,u, v ∈Zand
h(p)h(q)≤C(n)(h(n))1/2,
where C(n) is an explicitly given function and 1 < C(n)<2/√3.
Keywords: hexagon, height, integral vector.
2000 MS Classification: primary 11 H 06; secondary 11 D 04, 11 J 25.
1 Introduction
In this paper we shall consider integer vectors n= (n1, n2, n3) in Z3and write for such
vectors
h(n) = max
i∈{1,2,3}|ni|.
Chaladus and Schinzel [1] have proved that for every non–zero vector n∈Z3there
exist linearly independent vectors p,q∈Z3, such that n=up+vqwith u, v ∈Zand
h(p)h(q)<2
√3(h(n))1/2,
where the constant on the right hand side is the best possible.
We prove the following more precise result, which uses the estimates that depend
on the initial vector n.
1
Theorem. For every non–zero vector n= (n1, n2, n3)∈Z3(0 < n1< n2< n3)there
exist linearly independent vectors p,q∈Z3, such that n=up+vqwith u, v ∈Zand
h(p)h(q)≤2
pF(n1/n3, n2/n3)(h(n))1/2,
where
F(α, β) = 2 + 1
β+1
2β1
α2−1−(1 −α)(1 + α)(β−2α2+α2β)
4α3log 1 + α
1−α
−(1 −β)2(1 + β)
2β2log 1 + β
1−β,(1)
and 3< F (α, β)<4when 0< α < β < 1.
In the case when n1= 0 or two of the numbers n1,n2,n3are equal, one can trivially
find pand qwith h(p)h(q) = 1.
The proof of this theorem develops the ideas given by Schinzel in [2] and gives a
key for solving similar problems in higher dimensions.
2 On the area of hexagons
Let n= (n1, n2, n3) be a non–zero integer vector. We may assume without loss of
generality that 0 < n1< n2< n3and gcd(n1, n2, n3) = 1. Let m= (m1, m2, m3) be an
integer vector, such that mand nare linearly independent.
Define the hexagon
H:|miy−nix| ≤ 1, i ∈ {1,2,3},(2)
and introduce two quantities
δi=mi−m3
ni
n3
, i ∈ {1,2}.(3)
Then M1=n3δ1,M2=n3δ2,M3=n2δ1−n1δ2are the three minors of order two of
the matrix
m1m2m3
n1n2n3.
The numbers M1,M2,M3are not all equal to zero, since the vectors m,nare linearly
independent. Therefore, by the lemma 1 of [1], the area of the hexagon (2) equals
2|M1M2|+ 2|M1M3|+ 2|M2M3| − M2
1−M2
2−M2
3
|M1M2M3|,(4)
2
if each of the numbers |M1|,|M2|,|M3|is less than the sum of the two others, and
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max{|M1|,|M2|,|M3|} (5)
otherwise.
Let α,βbe real numbers, such that
0< α < β < 1.(6)
Denote by A(x, y) a two–variate function on R2which is formally defined by expressions
(4), (5) on substitution
M1=x, M2=y, M3=βx −αy ,
if (x, y)6= (0,0) and put A(0,0) = ∞.
Let ξ > 0 be a real number which will be defined later. Consider the inequality
A(x, y)≥4
ξ.(7)
Let Bξbe the solution set for (7), that is
Bξ=(x, y)∈R2:A(x, y)≥4
ξ.
We obtain the following formula for Bξ.
1. If xy ≥0, then Bξis defined by the inequalities
|x| ≤ ξ, if |y|<1−β
1−α|x|; (8)
−x2(1 −β)2+ 2|xy|(1 −α+β+αβ)−y2(1 + α)2
≥4
ξ(βx2|y| − α|x|y2),if 1−β
1−α|x|≤|y| ≤ 1 + β
1 + α|x|; (9)
|y| ≤ ξ, if 1 + β
1 + α|x|<|y|.(10)
3
2. If xy < 0, then Bξis defined by the inequalities
|x| ≤ ξ, if |y|<1−β
1 + α|x|; (11)
−x2(1 −β)2+ 2|xy|(1 + α+β−αβ)−y2(1 −α)2
≥4
ξ(βx2|y|+α|x|y2),if 1−β
1 + α|x| ≤ |y| ≤ 1 + β
1−α|x|; (12)
|y| ≤ ξ, if 1 + β
1−α|x|<|y|.(13)
Lemma 1. Bξis a centrally symmetric, convex set.
Proof. Since the inequalities (8)–(13) depend only on the absolute values of variables
x,ythe symmetry is obvious. Let us show that Bξis a convex set. The boundary of
Bξconsists of two horizontal segments
±Sh={±(ξt, ξ) : −(1 −α)/(1 + β)≤t≤(1 + α)/(1 + β)},
two vertical segments
±Sv={±(ξ, ξt) : −(1 −β)/(1 + α)≤t≤(1 −β)/(1 −α)},
and four curvilinear arcs, which correspond to (9), (12).
Let Lbe any such arc and A,Bbe the end points of L. Using a transformation
X=±x,Y=±ywith suitable signs, we may assume that A= (a, ξ), B= (ξ, b), where
0< a, b < ξ. If Lis the graph of a certain function y(x) on the interval I= [ a, ξ ],
then Bξis convex if for each such arc the following two conditions are valid
(i) y(x) has a horizontal tangent at the point x=aand a vertical tangent at the
point x=ξ,
(ii) y(x) is concave in the interval I.
Firstly consider the case xy ≥0 and assume without loss of generality that x≥0,
y≥0. Let Lbe an arc which corresponds to (9). Let us parametrize Lby the slope t
of radius–vector. We obtain
4
x(t) = ξ−t2(1 + α)2+ 2t(1 −α+β+αβ)−(1 −β)2
4t(β−αt),
y(t) = tx(t),
1−β
1−α≤t≤1 + β
1 + α
and
x((1 −β)/(1 −α)) = ξ , x((1 + β)/(1 + α)) = ξ(1 + α)/(1 + β).(14)
We have
dx
dt (t) = ξ((−2α+β+αβ)t+β(1 −β)) (−(1 −α)t+ 1 −β)
4t2(αt −β)2.
It is easy to check by using the condition (6) that dx/dt has not zeros in the inter-
val (1 −β)/(1 −α)<t<(1 + β)/(1 + α). Therefore by (14) one can interpret yas a
function of xon the interval I= [ ξ(1 + α)/(1 + β), ξ ]. We shall now find the first two
derivatives of this function. Calculation gives
dy
dx(t) = t2((1 + α)t−1−β)(α(1 + α)t+α−2β−αβ)
(−(1 −α)t+ 1 −β)((−2α+β+αβ)t+β(1 −β)) ,(15)
d2y
dx2(t) = 8t3(αt −β)3p(t)
(−(1 −α)t+ 1 −β)3((−2α+β+αβ)t+β(1 −β))3,(16)
where
p(t) = −(1 −α)(1 + α)2(−2α+β+αβ)t3−3α(1 −β)2(1 + α)2t2
+3β(1 −β)2(1 + α)2t−(1 + β)(1 −β)2(−α+ 2β+αβ).
By (14) and (15) y(x) has a horizontal tangent at the point x=ξ(1 + α)/(1 + β) and a
vertical tangent at the point x=ξ. Further, we have to check that y(x) is concave in
the interval I. By (14) and (16) the second derivative has the same sign at the points
x=ξ(1 + α)/(1 + β) and x=ξ. Therefore either y(x) is concave in Ior d2y/dx2
changes sign in this interval at least twice. The latter is possible only if d2y/dx2has at
least two roots in this interval. By (16) any such root must be a root of the polynomial
p(t). Let us find the discriminant of p(t). Straightforward calculation gives
Dp=−432(1 −β)4(1 + α)4(−α+β+αβ)(α−β)4<0.
Since deg p(t) = 3, the polynomial p(t) has exactly one real root and the concavity of
y(x) in the interval Iis proved.
5
Let us consider now the case xy < 0 and assume without loss of generality that
x > 0, y < 0. Let Mbe an arc which corresponds to (12) and Lbe the image of
Munder the transformation X=x,Y=−y. Similar to the previous case we can
parametrize Lby the slope tof radius–vector. We get
X(t) = ξ−t2(1 −α)2+ 2t(1 + α+β−αβ)−(1 −β)2
4t(β+αt),
Y(t) = tX(t),
1−β
1 + α≤t≤1 + β
1−α
and
X((1 −β)/(1 + α)) = ξ , X((1 + β)/(1 −α)) = ξ(1 −α)/(1 + β).(17)
We find
dX
dt (t) = ξ(−(α+ 1) t+ 1 −β) ((2α+β−αβ)t+β(1 −β))
4t2(αt +β)2.
It is easy to check by using the condition (6) that dX/dt has not zeros in the interval
(1 −β)/(1 + α)<t<(1 + β)/(1 −α). Hence by (17) we can consider Yas a function
of Xon the interval J= [ ξ(1 −α)/(1 + β), ξ ]. The first two derivatives of Y(X) are
dY
dX (t) = t2((−1 + α)t+1+β)(α(1 −α)t+α+ 2β−αβ)
(−(1 + α)t+ 1 −β)((2α+β−αβ)t+β(1 −β)) ,(18)
d2Y
dX2(t) = 8t3(αt +β)3q(t)
(−(1 + α)t+ 1 −β)3((2α+β−αβ)t+β(1 −β))3,(19)
where
q(t) = (1 + α)(1 −α)2(2α+β−αβ)t3−3α(1 −β)2(1 −α)2t2
−3β(1 −β)2(1 −α)2t+ (1 + β)(1 −β)2(α+ 2β−αβ).
The discriminant of the polynomial q(t) equals
Dq=−432(1 −β)4(1 −α)4(α+β−αβ)(α+β)4<0.
It is easy to see that all arguments given in the case xy ≥0 are valid with respect to
Y(X) and the interval J. Therefore, the conditions (i), (ii) are valid for all four arcs
and Bξis a convex set.
6
Denote by Aξ(α, β) the area of Bξ. We can obtain the formula for Aξ(α, β) by an
integration. Calculation gives
Aξ(α, β) = ξ2F(α, β).(20)
Define the square
Cξ={(x, y)∈R2:|x| ≤ ξ , |y| ≤ ξ},
and denote by Hξthe centrally symmetric hexagon with vertices
±(ξ , 0 ) ,±(ξ(1 + α)/(1 + β), ξ ),±(−ξ(1 −α)/(1 + β), ξ ).
Let “⊂” denote the strict inclusion. The following geometric observation gives us a
tool for an estimation of Aξ(α, β).
Lemma 2. Hξ⊂ Bξ⊂ Cξ.
Proof. All vertices of Hξbelong to the boundary of Bξ. By lemma 1, Bξis convex,
therefore Hξlies in Bξ. Moreover, the proof of lemma 1 immediately implies that Bξ
lies in Cξ. Since by (15), (18) and condition (6) Bξcan not coincide with a square or a
hexagon, the lemma is proved.
The area of the hexagon Hξis equal to 2ξ2(2 + β)/(1 + β). It is easily seen that this
expression is greater than 3ξ2. Therefore lemma 2 and (20) give us 3 < F (x, y)<4
when 0 < x < y < 1.
3 Proof of the theorem
Let us define the integer vector nas in section 2 and put
α=n1
n3
, β =n2
n3
, ξ =2
pn3F(α, β);
then by (20)
Aξ(α, β) = 4
n3
.(21)
Consider the three–dimensional cylinder Dwith base Bξand axis
(τ α, τβ , τ), τ ∈(−n3, n3).(22)
In view of lemma 1, Dis a centrally symmetric, convex set. By (21)
V(D) = 2n3Aξ(α, β) = 8 ,
7
where V(.) denotes the volume. In virtue of Minkowski’s first theorem, Dcontains a
non–zero integer vector m= (m1, m2, m3). By (22) we find |m3|< n3. Therefore in
view of condition gcd(n1, n2, n3) = 1, the vectors mand nare linearly independent.
Consider the plane z=m3with induced coordinates x,y. On this plane the point
(m1, m2) lies in the translation of Bξwith center at the point (m3α, m3β) and therefore
the point (δ1, δ2), where δ1,δ2are defined by (3), lies in Bξ. By definition of Bξthe
following inequality holds
A(δ1, δ2)≥4
ξ.(23)
Consider now the hexagon Hdefined by (2). Note that A(δ1, δ2)/n3gives us the
area of H. Hence, in virtue of Minkowski’s second theorem there exist two linearly
independent integer vectors (x1, y1) and (x2, y2), such that
|miyj−nixj| ≤ λj, i ∈ {1,2,3};j∈ {1,2},(24)
and
λ1λ2A(δ1, δ2)≤4n3.(25)
Put
p=y1m−x1n,q=y2m−x2n.
Then p,qare linearly independent and by theorem 2 of [2] we can assume without
loss of generality that
n=up+vq, u, v ∈Z.
Finally, by (23), (24) and (25) we obtain
h(p)h(q)≤λ1λ2≤4n3
A(δ1, δ2)≤n3ξ=2
pF(α, β)n1/2
3
=2
pF(α, β)(h(n))1/2.
Acknowledgment
It is a great pleasure for the author to thank Professor A. Schinzel for suggesting the
problem and help in the preparation of this paper.
8
INSTITUTE OF MATHEMATICS OF THE POLISH ACADEMY OF SCIENCES,
SNIADECKICH 8, 00–950 WARSZAWA, POLAND
(INSTYTUT MATEMATICZNY PAN)
e–mail: iskander@impan.gov.pl
References
[1] S. Chaladus, A. Schinzel, A Decomposition of Integer Vectors, II, PLISKA Studia
Mathematica Bulgarica, 11 (1991) 15–23.
[2] A. Schinzel, A Decomposition of Integer Vectors, I, Bull. Pol. Ac.: Math., 35
(1987) 155–159.
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