The 2-color Rado number of $x_1+x_2+...+x_{m-1}=ax_m$

0.12 MB·English

by  Dan Saracino

2
The 2-color Rado Number of
1
0
x + x + + x = ax
2 1 2 ··· m−1 m

l
u
J Dan Saracino

2 Colgate University

]
O
C Abstract
. In 1982, Beutelspacher and Brestovansky proved that for every
h
integer m≥3, the2-color Rado numberof theequation
t
a
m x1+x2+···+xm−1 =xm
[ is m2 −m−1. In 2008, Schaal and Vestal proved that, for every

m≥6, the2-color Radonumberof
1
v x1+x2+···+xm−1 =2xm
2
3 is ⌈m−1⌈m−1⌉⌉. Here we prove that, for every integer a ≥ 3 and
2 2
4 every m≥2a2(a−2), the 2-color Radonumberof
0
. x1+x2+···+xm−1=axm
7
0 is ⌈m−1⌈m−1⌉⌉.
a a
2
1
:
v 1. Introduction
i
X
r A special case of the work of Richard Rado [4] is that for every integer
a
m ≥ 3 and all positive integers a1,...,am there exists a smallest positive
integer n with the following property: for every coloring of the elements
of the set [n] = {1,...,n} with two colors, there exists a solution of the
equation
a1x1+a2x2+···+am−1xm−1 =amxm
usingelementsof[n]thatareallcoloredthesame. (Suchasolutioniscalled
monochromatic.) The integer n is called the 2-color Rado number of the
equation.
In 1982,BeutelspacherandBrestovansky[1]provedthat foreverym≥
3, the 2-color Rado number of
x1+x2+···+xm−1 =xm
1
is m2−m−1.Since then, Rado numbers fora number ofvariationsof this
equation have also been determined. For example, in 2008 Guo and Sun
[2] solved the problem for the equation
a1x1+a2x2+···+am−1xm−1 =xm,
forallpositiveintegersa1,...,am−1.Theyproved(confirmingaconjecture
of Hopkins and Schaal [3]) that the 2-color Rado number is aw2+w−a,
where a =min{a1,...,am−1} and w =a1+···+am−1. In the same year,
Schaal and Vestal [5] dealt with the equation
x1+x2+···+xm−1 =2xm.
They proved,in particular,that for every m≥6, the 2-colorRado number
is ⌈m−1⌈m−1⌉⌉. Ourpurpose inthe presentpaperis to obtainananalogue
2 2
of this result for all larger values of the coefficient on xm. We prove the
following result.
Theorem. For every integer a ≥ 3 and every m ≥2a2(a−2), the 2-color
Rado number of the equation
x1+x2+···+xm−1 =axm
is ⌈m−1⌈m−1⌉⌉.
a a
Notation. We will denote ⌈m−1⌈m−1⌉⌉ by C(m,a), and we will denote
a a
the equation indicated in the statement of the theorem by L(m,a).
To prove our theorem, we fix a ≥ 3 and m ≥ 2a2(a −2) and show
first,in Section2,thatC(m,a) is a lowerbound forthe Radonumber, i.e.,
there exists a 2-coloring of [C(m,a)−1] that admits no monochromatic
solutionof L(m,a). Then,in the remaining sections,weshow thatC(m,a)
is an upper bound for the Rado number, i.e., every 2-coloring of [C(m,a)]
admits a monochromatic solution of L(m,a).
Notation. In working with a fixed 2-coloring of a set, we will use the
colors red and blue, and we will denote by R and B, respectively, the sets
of elements colored red and blue.
2. Lower bounds, and someresults fordealingwith upperbounds
Proposition 1. For every a ≥ 3 and every m ≥ 2a2(a−2), the 2-color
Rado number of L(m,a) is at least C(m,a).
2
Proof. We have m ≥ 18, so the Rado number exists. We must show that
there exists a 2-coloring of [C(m,a) − 1] that yields no monochromatic
solution of L(m,a). We use the same coloring that Schaal and Vestal used
in [5] to establish their lower bounds, but in our less specific situation it
is easier to work directly from the meaning of C(m,a) than from algebraic
expressions for the ceiling function, as Schaal and Vestal did. We will use
the fact that, for every real number r,
r >⌈r⌉−1.
We 2-color[C(m,a)−1] by coloringall the elements of [⌈m−1⌉−1]red
a
and all remaining elements blue. For any red elements x1,x2,...,xm we
have
x1+···+xm−1 ≥ m−1 > m−1 −1≥xm,
a a (cid:24) a (cid:25)
sotherearenoredsolutionsofL(m,a).Foranyblueelementsx1,x2,...,xm
we have
x1+···+xm−1 ≥ m−1 m−1 >C(m,a)−1≥xm,
a a (cid:24) a (cid:25)
so there are no blue solutions either. (cid:3)
In considering upper bounds, we will often need to exhibit solutions of
L(m,a)in[C(m,a)]. To do this we willneedto knowthatcertainnumbers
are less than or equal to C(m,a).
Lemma 1. The numbers 2m−4,m−1, and a+1 are each less than or
equal to C(m,a). If a≥4, then 2m≤C(m,a).
Proof. To show that 2m−4≤C(m,a), it will suffice to show that
(m−1)2
≥2m−4.
a2
This is equivalent to
4 1
m≥ 2− a2+2− ,
(cid:18) m(cid:19) m
and is therefore clear when a≥4, because then m ≥2a2(a−2)≥2a2+2
(and we can replace 2m−4 by 2m). When a = 3 the displayed inequality
becomes m≥20− 37, and is true because m≥18.
m
Since 2m−4≥m−1whenm≥3,wehavem−1≤C(m,a). Itfollows
from our assumption on the size of m that m−1 ≥ a+1, and therefore
a+1≤C(m,a). (cid:3)
3
In their treatment of upper bounds in [5], Schaal and Vestal proceeded
by fixing the coloringofthe element1 andconsideringthe two possibilities
forthecoloringoftheelement2. Indealingwitha≥3,wefinditconvenient
tofix the coloringofthe elementa−2andconsiderthe twooptionsforthe
coloring of a−1.
Convention. In dealing with 2-coloringsof [C(m,a)] in the following sec-
tions, we will assume without loss of generality that a−2∈R.
3. Monochromatic solutions when a−1∈B
In this and the following sections , it will be convenient to have a com-
pact notation for indicating solutions of L(m,a).
Notation. If n1,...,nk are nonnegative integers whose sum is m, and
d1,...,dk are elements of [C(m,a)] such that we obtain a true equation
from L(m,a) by substituting d for the first n variables, d for the next
1 1 2
n variables, and so on, then we denote this true equation by
2
[n1 →d1; n2 →d2; ···; nk →dk].
For example, the true instance
a+a+···+a=a(m−1) (1)
of L(m,a) will be denoted by
[m−1→a; 1→m−1].
Suppose now that we have a 2-coloring of [C(m,a)] that yields no
monochromatic solution of L(m,a). We seek a contradiction. We will pro-
ceed by noting a number of solutions of L(m,a); all the numbers used in
these solutions will be in [C(m,a)] by Lemma 1.
Recall that we are assuming that a−2 ∈ R, and, in this section, that
a−1∈B.
Lemma 2. We have m−2∈R and m−3∈B.
Proof. Sincea−1∈B andtherearenomonochromaticsolutionsofL(m,a)
in [C(m,a)], the solution
[m−2→a−1; 2→m−2] (2)
4
tells us that m−2∈R. Likewise, since a−2∈R, the solution
[m−3→a−2; 3→m−3] (3)
tells us that m−3∈B. (cid:3)
Lemma 3. We have a∈R and m−1∈B.
Proof. Since a−1 and m−3 are in B, the solution
[2a→a−1; m−1−2a→a; 1→m−3]
tells us that a∈R, and then solution (1) above tells us that m−1∈B. (cid:3)
Now if m is even we obtain the desired contradictionby observing that
the solution
m−2 m−2
→a−2; →a; 2→m−2 (4)
(cid:20) 2 2 (cid:21)
isred. Ifmisoddweobtainacontradictionbyconsideringthe colorofthe
element a+1. If a+1∈B then the blue solution
m−1 m−1
→a−1; →a+1; 1→m−1 (5)
(cid:20) 2 2 (cid:21)
yields a contradiction. If a+1∈R then the red solution
m−1 m−5
→a−2; →a; 1→a+1; 2→m−2
(cid:20) 2 2 (cid:21)
yields a contradiction.
We have proved the following proposition.
Proposition 2. If a − 2 ∈ R then every 2-coloring of [C(m,a)] with
a−1∈B yields a monochromatic solution of L(m,a).
4. Consequences of assuming a−1 ∈ R and no monochromatic
solutions
Inthissectionweassumethata−1∈Randour2-coloringof[C(m,a)]
yieldsnomonochromaticsolutionsofL(m,a).Wederivesomeconsequences
that will be used in the next section.
Lemma 4. Each of m−1,m−2,m−3 is in B, and a∈R.
5
Proof. Since a−1 and a−2 are in R, it follows from solutions (2) and (3)
abovethatm−2andm−3areinB. To concludethe proof,it willsuffice,
by solution (1) above, to show that a ∈ R. But if a ∈ B, then by solution
(1) we have m−1∈R, and then the solution
[m−2a+1→a−2; 2a−4→a−1; 3→m−1] (6)
is red, a contradiction. (cid:3)
Lemma 5. We have 1∈R.
Proof. The solution
[m−a→1; a−2→m−1; 2→m−2]
would be blue if 1 were in B. (cid:3)
Lemma 6. We have 2∈R.
Proof. Ifa=3then2isa−1,whichis inR. Sowecanassumethata≥4,
and therefore, by Lemma 1, 2m and all smaller numbers are available in
[C(m,a)] for use in producing solutions of L(m,a).
From the solution
[m−a→1; a→m−a]
we conclude that m−a∈B, and from the solution
[m−2a+5→a; a−3→a−1; a−3→1;1→m−a+2]
we conclude that m−a+2∈B. If 2a∈B, then from the solution
[a−2→m−a; m−a→2; 1→2a; 1→m−a+2]
we infer that 2∈R.
For the remainder of the proof we assume that 2a ∈ R and 2 ∈ B and
seek a contradiction. First, by doubling all the entries in solution (1) we
conclude that 2(m−1)∈B. Then from the solution
[a−1→2m; m−a→2; 1→2(m−1)]
we deduce that 2m∈R. Using the solution
[m−2a→2(a−1); 2a−2→2a; 2→2m]
we see that 2(a−1) ∈ B, and then by doubling solution (6) we see that
2(a−2)∈R. From the solution
[m−a−1→2a; a→2(a+1); 1→2m]
6
we get 2(a+1)∈B.
If m is odd we now obtain a contradictionby doubling solution (5) and
noting that the resulting solution is blue. If m is even then by doubling
solution (4) we see that 2(m−2) ∈ B, and we obtain a contradiction by
noting that the solution
[m−a→2; a−2→2(m−1); 2→2(m−2)]
is blue. (cid:3)
Lemma 7. If a≥4 then the numbers m−3,m−2,m−1,...,2m−2 are
all in B.
Proof. By Lemma 4, we only need to prove this for m,m+1,...,2m−2.
Since a ≥ 4, we have 2m ∈ [C(m,a)] by Lemma 1. Since 2 ∈ R,
if 2a ∈ B we can repeat all the steps in the last two paragraphs of the
proof of Lemma 6, with all the colors reversed, to obtain a contradiction.
Therefore 2a∈R.
Now consider the number m+k, where 0≤k ≤m−2. The solution
[m−k−2→a; k+1→2a; 1→m+k]
shows that m+k ∈B. (cid:3)
Lemma 8. The numbers a−1,a,a+1,...,2a−2 are all in R.
Proof. We only need to prove this for a+1,a+2,...,2a−2.
Supposefirstthata=3,inwhichcaseweonlyneedtoshowthat4∈R.
Since 2m−6∈[C(m,a)] by Lemma 1 and
[m−3→2; 3→2m−6]
is a solution when a=3, we have 2m−6∈B. Since
[m−3→4; 2→m−3; 1→2m−6]
is also a solution when a=3, we conclude that 4∈R.
We nowassumethata≥4,sothatLemma7 applies. We wantto show
that 2a−2j ∈ R for all integers j such that 2 ≤ 2j ≤ a−1, and that
2a−(2j+1)∈R for all j such that 3≤2j+1≤a−1.
For 2a−2j we consider the solution
[m−(j+1)→2a−2j; j+1→2m−2(j+1)]
andneed to knowthat 2m−2(j+1)∈B. This will be true by Lemma 7 if
2m−2(j+1)≥m−3, i.e., if m≥2j−1. But this inequality holds, since
2j ≤a−1.
7
For 2a−(2j+1) we consider the solution
[m−(j+2)→2a−(2j+1); j →2m−(2j+5); 1→m−2; 1→2m−2(j+2)]
and need to know that 2m−(2j +5) and 2m−(2(j +2) are in B. This
will be true if 2m−(2j+5) ≥ m−3, i.e., if m ≥ 2j+2. This inequality
holds because (2j+1)≤a−1. (cid:3)
There is one more result that we will need in Section 5.
Lemma 9. If d is an integer such that a|d and m−1≤d≤2m−2, then
d
∈B.
a
Proof. Write d=m−1+k, with 0≤k ≤m−1. Then the solution
d
[m−1−k →1; k →2; 1→ ]
a
shows that d ∈B. (cid:3)
a
5. Monochromatic solutions when a−1∈R
In this section we suppose that a−1 ∈ R and there are no monochro-
matic solutions of L(m,a) in [C(m,a)], and we again seek a contradiction.
WewillusetheresultsofSection4,andwewillalsoneedtousealgebraic
expressions for C(m,a).
Lemma 10. Let m=ua2+va+c, where 0≤v,c≤a−1.
(i) If c=1 then C(m,a)= (m−1)2.
a2
(ii) If c=0 then C(m,a)= m2−m+va.
a2
(iii) If 2≤c≤a−1 then C(m,a)= m2+(a−c−1)m+c−ac−vac+va+ta2,
a2
where t= (c−1)(v+1) .
a
l m
Proof. If c = 1 then a|(m−1), so the claim is clear from the definition of
C(m,a). If c=0 then
m−1) m m2−m m2−m+va
C(m,a)= · = = ,
(cid:24) a a(cid:25) (cid:24) a2 (cid:25) a2
sincea2|m2 andvaisthesmallestnumberwecanaddtom2−mtoproduce
a multiple of a2.
8
If 2≤c≤a−1, then
c−1
C(m,a)= ua+v+ (ua+v+1)
(cid:24)(cid:18) a (cid:19) (cid:25)
so
C(m,a)=(ua+v)2+(ua+v)+(c−1)u+t.
m−c m−va−c
Replacing ua+v by and u by , and simplifying, we obtain
a a2
the final claim of the(cid:0)lemm(cid:1)a. (cid:3)
ThethreedescriptionsofC(m,a)inLemma10leadustoconsiderthree
cases.
Case 1: m≡1 (mod a)
In this case we have m−1 ∈ B, by Lemma 9. Since 1 ∈ R, we can use
a
an idea from [5] and let s be an integer such that s ∈ R, s+1 ∈ B, and
s+1≤ m−1. Then m−1(s+1)≤C(m,a) by Lemma 10, and the solution
a a
m−1
m−1→s+1; 1→ (s+1)
(cid:20) a (cid:21)
of L(m,a) shows that m−1(s+1)∈ R. We now obtain a contradiction by
a
noting that the solution
m−1 m−1 m−1
−a→a−1; a−1→a; (a−1) →s; 2→ (s+1)
(cid:20) a a a (cid:21)
is red.
Case 2: m≡0 (mod a)
m
In this case we have ∈ B by Lemma 9. We choose an s such that
a
m
s∈R, s+1∈B, and s+1≤ . Noting that
a
m−a m m2−m+va m2−am (a−1)m+va
C(m,a)− = − = ,
(cid:18) a (cid:19) a a2 a2 a2
we consider the element
m−a (a−1)m+va m−a m (a−1)m+va
α= (s+1)+ ≤ + ,
a a2 (cid:18) a (cid:19) a a2
m
so α ≤ C(m,a). Noting that +1 ∈ B by Lemma 9, we see that α ∈ R
a
by considering the solution
m m
m−a→s+1; v → +1; a−1−v → ; 1→α
a a
h i
9
of L(m,a).
WenowobtainaredsolutionofL(m,a)(andthereforeacontradiction)
by assigning the value α to xm−1 and xm and the value s to (a−1)(ma−a)
othervariables,andshowingthatwecanassignvaluesinRtotheremaining
m
+a−3 variables to complete the solution. In fact we will show that we
a
canaccomplish this by using only values in the set {2a−4,2a−3,2a−2}.
These values are all in R by Lemma 8 and the fact that 2a−4 ≥ a−1
when a≥3.
The values assigned to the remaining variables must add up to
a−1 a−1
(m−a)+ ((a−1)m+va).
a a2
If we can show that using only the value 2a−2 yields a sum that is at
leastthislarge,andusingonlythevalue2a−4yieldsasumthatisatmost
thislarge,thenthereisaunique solutionthatusesvaluesinoneofthesets
{2a−4,2a−3} or {2a−3,2a−2.}
Since v ≤a−1, we can achieve our first objective by showing that
m a−1 a−1
(2a−2) +a−3 ≥ (m−a)+ ((a−1)m+(a−1)a),
a a a2
(cid:16) (cid:17)
which simplifies to
1 1−a
2a2−8a+7− ≥m ,
a (cid:18) a2 (cid:19)
and this is easily seen to be true for a ≥ 3, since the right-hand side is
negative.
Since v ≥0, we can achieve our second objective by showing that
m a−1 a−1
(2a−4) +a−3 ≤ (m−a)+ ((a−1)m).
a a a2
(cid:16) (cid:17)
But this simplifies to 2a4−9a3+11a2 ≤m(a+1), which is easily verified
using our assumption that m≥2a2(a−2).
Case 3: m≡c (mod a), 2≤c≤a−1
In this case we have m+a−c ∈ B by Lemma 9. Choosing s such that
a
s∈R, s+1∈B, and s+1≤ m+a−c, we consider the element
a
m−c (c−1)m+c−c2−vac+va+ta2
β = (s+1)+ ,
(cid:18) a (cid:19) a2
10