On the Two-Color Disjunctive Rado Number for the Equations \(\sum_{i=1}^{m-2} x_i+ax_{m-1}-x_m=c_j, j=1,2\)

#A108 INTEGERS 25 (2025)
ON THE TWO-COLOR DISJUNCTIVE RADO NUMBER FOR THE
EQUATIONS Pm−2
i=1 xi+axm−1−xm=cj,j= 1,2
S ash i Dwi edi
Depa men o Pu e and Applied Ma hema ics, Alliance Uni e si y, Bengalu u,
India
[email p o ec ed]
Ami abha T ipa hi1
Depa men o Ma hema ics, Indian Ins i u e o Technology, New Delhi, India
[email p o ec ed]
Recei ed: 6/7/25, Re ised: 9/19/25, Accep ed: 11/10/25, Published: 11/25/25
Abs ac
Gi en a sys em o linea equa ions S, he disjunc i e Rado numbe o he sys em
Sis he leas posi i e in ege R=R(S), i i exis s, such ha e e y 2-colo ing o
he in ege s in [1, R] admi s a monoch oma ic solu ion o a leas one equa ion in S.
We de e mine R(S) when Sis he pai o equa ions Pm−2
i=1 xi+axm−1−xm=
c1,Pm−2
i=1 xi+axm−1−xm=c2 o some ange o alues o c1and c2.
1. In oduc ion
By an -colo ing o {1, . . . , N}we mean a mapping χ:{1, . . . , N}→{1, . . . , }. In
1916, Schu showed ha o e e y posi i e in ege , he e exis s a leas posi i e
in ege s=s( ) such ha o e e y -colo ing o he in ege s in he in e al [1, s],
he e exis s x, y, x +y∈[1, s] such ha χ(x)=χ(y) = χ(x+y). Schu ’s heo em
was gene alized in a se ies o esul s in he 1930’s by Rado leading o a comple e
esolu ion o he ollowing p oblem: cha ac e ize sys ems o linea homogeneous
equa ions wi h in eg al coe icien s Ssuch ha o a gi en posi i e in ege , he e
exis s a leas posi i e in ege n=R(S; ) such ha e e y -colo ing o he in ege s
in he in e al [1, n] yields a monoch oma ic solu ion o he sys em S. The e
has been a g owing in e es in he de e mina ion o he Rado numbe s R(S; ),
pa icula ly when Sis a single equa ion and = 2; o ins ance, see [1, 6, 7, 8, 9,
10, 12]. When = 2, we deno e his numbe simply by R(S).
DOI: 10.5281/zenodo.17711638
1Co esponding au ho
INTEGERS: 25 (2025) 2
The p oblem o disjunc i e Rado numbe s was in oduced by Johnson and Schaal
in [11]. The 2-colo disjunc i e Rado numbe o he se o equa ions E1,...,Ek
is he leas posi i e in ege Nsuch ha any 2-colo ing o {1, . . . , N}admi s a
monoch oma ic solu ion o a leas one o he equa ions E1,...,Ek; we deno e his
by R(E1,...,Ek). Johnson and Schaal ga e necessa y and su icien condi ions o
he exis ence o he 2-colo disjunc i e Rado numbe o he addi i e equa ions
x1−x2=aand x1−x2=b o all pai s o dis inc posi i e in ege s a, b, and also
de e mined exac alues when i exis s. They also de e mined exac alues o he
pai o mul iplica i e equa ions ax1=x2and bx1=x2whene e a, b a e dis inc
posi i e in ege s; o al e na e p oo s, see [2]. Dileep, Moond a and T ipa hi [3]
ex ended he esul s o Johnson and Schaal o he se o equa ions x1−x2=ai,
1≤i≤k, gi ing condi ions o he exis ence o he 2-colo disjunc i e Rado
numbe , exac alues in some cases, and uppe and lowe bounds in all cases. They
also in es iga ed and ob ained pa allel esul s o he se o mul iplica i e equa ions
y=aix, 1 ≤i≤k. Fu he , hey ga e a gene al sea ch-based algo i hm wi h a
un ime o O(kaklog ak) o he case o addi i e equa ions, which is exponen ially
be e han he b u e- o ce algo i hm o he p oblem. Lane-Ha a d and Schaal
[14] de e mined exac alues o 2-colo disjunc i e Rado numbe o he pai o
equa ions ax1+x2=x3and bx1+x2=x3 o all dis inc posi i e in ege s a, b. Sabo,
Schaal and Tokaz [15] de e mined exac alues o 2-colo disjunc i e Rado numbe
o x1+x2−x3=c1, and x1+x2−x3=c2, whene e c1, c2a e dis inc posi i e
in ege s. Kosek and Schaal [13] de e mined he exac alue o 2-colo disjunc i e
Rado numbe o he equa ions x1+···+xm−1=xm, and x1+···+xn−1=xn,
o all pai s o dis inc posi i e in ege s m, n.
Schaal and Zin e [16] s udied he 2-colo Rado numbe o he equa ion x1+
3x2+c=x3 o c≥3, gi ing a lowe bound in all cases and uppe bounds in some.
Dwi edi and T ipa hi [4] gene alized his o in es iga e he 2-colo Rado numbe
o he equa ion x1+ax2−x3=c o posi i e in ege s a, gi ing condi ions o
exis ence, uppe and lowe bounds in all cases, and exac esul s in a ew. The
same au ho s [5] u he gene alized his o in es iga e he 2-colo Rado numbe o
he equa ion Pm−2
i=1 xi+axm−1+xm=cwhen 4 ≤m≤a. They gi e a necessa y
and su icien condi ion o he Rado numbe o exis , gi e uppe and lowe bounds
in all cases, and exac alues in many cases. This pape in es iga es he disjunc i e
Rado numbe p oblem o he pai o equa ions Pm−2
i=1 xi+axm−1−xm=c1and
Pm−2
i=1 xi+axm−1−xm=c2. We ep oduce some pe inen esul s om [5] o
eady e e ence.
Theo em 1 ([5, Theo em 1]). Le a, c, m ∈Z, and 4 ≤m≤a. I a+m, and ca e
bo h odd, hen
R m−2
X
i=1
xi+axm−1−xm=c!
does no exis .
INTEGERS: 25 (2025) 3
P oposi ion 2 ([5, P oposi ion 1]). Fo a∈N, and 4 ≤m≤a,
R m−2
X
i=1
xi+axm−1−xm=a+m−3!= 1.
Theo em 3 ([4, Theo em 3], [5, Theo em 5]).
Le a, m be in ege s o he same pa i y, wi h a≥3, and m≥3. Le a′=a+m−3.
I ei he o
(i) m= 3, and c≤ −a(a−3)
2;
(ii) m≥4, and c<−(a′+ 3)(a−2)
is ue, hen
R m−2
X
i=1
xi+axm−1−xm=c!= (a′+ 3)(a′−c) + 1.
2. Resul s o Pm−2
i=1 xi+axm−1−xm=cj,j= 1,2
We s udy he disjunc i e Rado numbe o he pai o equa ions
m−2
X
i=1
xi+axm−1−xm=c1,(1a)
m−2
X
i=1
xi+axm−1−xm=c2,(1b)
whe e a≥3, m≥3, and c1,c2a e any in ege s. Th oughou his pape , we
deno e his 2-colo Rado numbe by Rad21,...,1
| {z }
m−2 imes
, a, −1; c1, c2, o mo e b ie ly
by R(c1, c2).
By assigning he colo o xiin he solu ion o Equa ion (1a) and Equa ion (1b)
o xi−1, we no e ha his is equi alen o de e mining he smalles posi i e in ege
R o which e e y 2-colo ing o [0, R −1] con ains a monoch oma ic solu ion o
m−2
X
i=1
xi+axm−1−xm=c′
1,(2a)
o
m−2
X
i=1
xi+axm−1−xm=c′
2,(2b)
whe e c′
j=cj−a′,j∈ {1,2}, and a′=a+m−3.
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P oposi ion 4. Le a, λ, n ∈Nsuch ha a≥3,n≥1, and λ≥a−1. Then o
each N∈ {0, . . . , λ(a+n)} he equa ion
n
X
i=1
xi+axn+1 =N(3)
admi s a solu ion wi h each xi∈ {0, . . . , λ}.
P oo . I N=λ(a+n), hen xi=λ o i∈ {1, . . . , n + 1}is a solu ion o Equa-
ion (3). I 0 ≤N < λ(a+n), we can w i e N=q(a+n)+ϵa + , whe e 0 ≤q < λ,
0≤ ≤n, and ϵ∈ {0,1}. Then, xi=q+ 1 o 1 ≤i≤ ,xi=q o + 1 ≤i≤n,
and xn+1 =q+ϵis a solu ion o Equa ion (3).
Theo em 5. Le 4≤m≤a, and cj=kj(a+m−3) wi h 1< kj≤a+m−2 o
j∈ {1,2}. Then,
R(c1, c2) = min{k1, k2}.
P oo . Le k= min{k1, k2}. The colo ing ∆ : [1, k−1] → {0,1}de ined by ∆(x)=0
is a alid colo ing, since
m−2
X
i=1
xi+axm−1−xm≤(a+m−2)(k−1) −1
=k(a+m−3) + (k−2) −(a+m−3)
< k(a+m−3).
Hence, R(c1, c2)≥k.
On he o he hand, since x1=· · · =xm=ksa is ies Equa ions (1a) o (1b) o
cj=k(a+m−3), j= 1,2, e e y colo ing χ: [1, k]→ {0,1}admi s a monoch oma ic
solu ion o Equa ions (1a) o (1b). Hence, R(c1, c2)≤k.
Theo em 6. Le a, m be in ege s o he same pa i y, wi h a≥3, and m≥4. Le
a′=a+m−3, and c′
j=cj−a′ o j∈ {1,2}. Then, o c1<−(a′+ 3)(a−2),
R(c1, c2) =











(a′+ 3)(a′−c1)+1 i c1−a′≤c2≤c1,
(a′+ 2)(a′−c1)+1 i (a′+ 2)c1−a′(a′+ 1) ≤c2< c1−a′,
(a′−c2)+1 i (a′+ 3)c1−a′(a′+ 2) < c2<(a′+ 2)c1−a′(a′+ 1),
(a′+ 3)(a′−c1)+1 i c2≤(a′+ 3)c1−a′(a′+ 2).
(4a)
(4b)
(4c)
(4d)
P oo . We no e ha a′=a+m−3, and ha
R(c1, c2)≤min{R(c1),R(c2)}= (a′+ 3)(a′−c1) + 1 = −(a′+ 3)c′
1+ 1
INTEGERS: 25 (2025) 5
by Theo em 1. We conside wo cases: (I) gi en by Equa ion (4a) and Equa-
ion (4d), and (II) gi en by Equa ion (4b) and Equa ion (4c). Thus, in Case I, i
su ices o p o e ha R(c1, c2)≥(a′+3)(a′−c1)+1 o comple e he p oo o Case I.
Case I. We exhibi a alid colo ing o [1,(a+m)(a+m−c1−3)] wi h espec o
Equa ion (1a) and (1b). Le ∆ : [1,(a+m)(a+m−c1−3)] → {0,1}be de ined by
∆(x) = 




0 i x∈[1, a +m−c1−3]
S[(a+m−1)(a+m−c1−3) + 1,(a+m)(a+m−c1−3)],
1 i x∈[a+m−c1−2,(a+m−1)(a+m−c1−3)].
Le A= [1, a +m−c1−3], B= [a+m−c1−2,(a+m−1)(a+m−c1−3)], and
C= [(a+m−1)(a+m−c1−3) + 1,(a+m−1)(a+m−c1−3)].
Suppose x1, . . . , xmis a solu ion o Equa ion (1a), wi h ∆(x1) = ··· = ∆(xm).
Suppose ∆(xi) = 0 o i∈ {1, . . . , m}. I x1, . . . , xm−1all belong o A, hen
a+m−c1−2≤xm=
m−2
X
i=1
xi+axm−1−c1
≤(a+m−2)(a+m−c1−3) −c1
≤(a+m−1)(a+m−c1−3).
Hence, xm∈B, and so χ(xm) = 1.
I a leas one o x1, . . . , xm−1belongs o Cand min Cdeno es he leas elemen
in C, hen,
xm= m−2
X
i=1
xi!+axm−1−c1≥(a+m−c1−3)+min C= (a+m)(a+m−c1−3)+1.
Hence, xmis ou side he domain o ∆. The e o e, ∆(xi) = 1 o i∈ {1, . . . , m},
and so
xm= m−2
X
i=1
xi!+axm−1−c1≥(a+m−2)·min B−c1≥(a+m−1)(a+m−c1−3)+1,
whe e min Bdeno es he leas elemen in B. Hence, xm∈C. This p o es ha ∆
is a alid colo ing o [1,(a+m)(a+m−c1−3)] wi h espec o Equa ion (1a).
Fo Equa ion (4a), he same a gumen applies wi h espec o Equa ion (1b). Fo
Equa ion (4d)
xm=
m−2
X
i=1
xi+axm−1−c2>(a+m)(a+m−c1−3).
Hence, xmis ou side he domain o ∆. Thus, ∆ is a alid colo ing o [1,(a+m)(a+
m−c1−3)] wi h espec o Equa ion (1b). This concludes he p oo o Case I.

INTEGERS: 25 (2025) 6
Case II. The colo ing ∆, wi h sui able modi ica ions, also p o ides a alid colo -
ing o Equa ion (4b) and (4c). Fo Equa ion (4b), we conside he unc ion ∆,
es ic ed o [1,(a+m−1)(a+m−c1−3)] = A∪B. A sub-a gumen used in
Equa ion (4a) shows ha his is a alid colo ing o Equa ion (1a) and (1b). Fo
Equa ion (4c), we conside he unc ion ∆, es ic ed o [1, a+m−c2−3]=A∪B∪C′,
whe e C′= [(a+m−1)(a+m−c1−3)+1, a+m−c2−3]. An a gumen simila o
he one o Equa ion (4a) shows ha his is a alid colo ing o Equa ion (1a) and
(1b). Since we ha e p o ided alid colo ings o Case II, i only emains o p o e
he uppe bounds in his case.
By assigning he colo o xiin he solu ion o Equa ion (1a) and (1b) o xi−
1, we equi alen ly conside monoch oma ic solu ions o Equa ion (2a) and (2b),
espec i ely, unde colo ings ha s a wi h x= 0. Le χ: [0,−(a′+ 3)c′
1]→ {0,1}
be any 2-colo ing o [0,−(a′+ 3)c′
1]. Wi hou loss o gene ali y, le χ(0) = 0. Each
s ep in he ollowing sequence o ces a colo on some numbe in he gi en ange in
o de o a oid a mononch oma ic solu ion o Equa ion (2a) o (2b):
•xi= 0 o 1 ≤i≤m−1 implies χ(−c′
1) = 1 and χ(−c′
2) = 1;
•xi=−c′
1 o 1 ≤i≤m−1 implies χ−(a′+ 2)c′
1= 0;
•xi= 0 o 2 ≤i≤m−1, and xm=−(a′+ 2)c′
1implies χ−(a′+ 1)c′
1= 1.
We cap u e his in o ma ion in Table 1.
0 1
0−c′
j
−(a′+ 2)c′
j−(a′+ 1)c′
j
Table 1
To comple e he p oo in Case II, we mus show ha :
•e e y 2-colo ing o χ: [0,−(a′+ 2)c′
1]→ {0,1}mus yield a monoch oma ic
solu ion o one o Equa ion (2a), (2b) o −(c′
1−a′)<−c′
2≤ −(a′+2)c′
1, and
•e e y 2-colo ing o χ: [0,−c′
2]→ {0,1}mus yield a monoch oma ic solu ion
o one o Equa ion (2a), (2b) o −(a′+ 2)c′
1<−c′
2<−(a′+ 3)c′
1.
We ha e assumed, wi hou loss o gene ali y, ha χ(0) = 0. The e a e wo
possibili ies o χ(1), o which he case χ(1) = 1 is common o Equa ion (4b) and
(4c). We i s assume χ(1) = 1. We claim ha
χ− c′
1−a′=(0 i is odd;
1 i is e en
INTEGERS: 25 (2025) 7
o ∈ {1, . . . , a′}. Wi h each xi= 1, 2 ≤i≤m−1, and xm=−(a′+ 1)c′
1in
Equa ion (2a), we ha e x1=−a′(c′
1+ 1), o cing χ−a′c′
1−a′= 0 in o de o
a oid a monoch oma ic colo ing. This p o es he claim o =a′.
Suppose ∈ {3, . . . , a′}, is odd, and ha χ− c′
1−a′= 0. We begin he
induc i e s ep a =a′. To comple e he claim, we show ha i χ− c′
1−a′= 0,
hen χ−( −1)c′
1−a′= 1 and χ−( −2)c′
1−a′= 0 o ∈ {3, . . . , a′}. Each
s ep in he ollowing sequence o ces a colo on some numbe in he gi en ange in
o de o a oid a mononch oma ic solu ion o Equa ion (2a):
•xi= 0 o 2 ≤i≤m−1, and xm=− c′
1−a′implies χ−( −1)c′
1−a′= 1;
•x1=−( −1)c′
1−a′, and xi= 1 o 2 ≤i≤m−1 implies χ− c′
1= 0;
•xi= 0 o 2 ≤i≤m−1, and xm=− c′
1implies χ−( −1)c′
1= 1;
•xi= 1 o 2 ≤i≤m−1, and xm=−( −1)c′
1implies χ−( −2)c′
1−a′= 0.
In pa icula , om he abo e claim, χ−c′
1−a′= 0. We no e ha χ(a+m−
1)c′
1= 0 om Table 1. Each s ep in he ollowing sequence o ces a colo on
some numbe in he gi en ange in o de o a oid a mononch oma ic solu ion o
Equa ion (2a) o (2b):
•x1=−c′
1−a′,xi=−c′
1+ 1 o 2 ≤i≤m−1, and xm=−(a+m−1)c′
1
implies χ−c′
1+ 1= 1;
•x1=−c′
1+ 1, and xi= 1 o 2 ≤i≤m−1 implies χ−2c′
1+ (a′+ 1)= 0;
•xi= 0 o 2 ≤i≤m−1, and xm=−2c′
1+(a′+1) implies χ−c′
1+(a′+1)= 1.
Now xi= 1 o 1 ≤i≤m−1, and xm=−c′
1+ (a′+ 1) o ms a monoch oma ic
solu ion o Equa ion (2a). This comple es he p oo when χ(1) = 1. Fo he
emainde o he p oo , we conside he case when χ(1) = 0. We claim ha
χ(n)=0 o 0≤n≤−2c′
1
a′=K. (5)
By way o con adic ion, assume χ(n) = 1 o some n≤K. We claim his implies
χ− c′
1=(0 i is e en;
1 i is odd
o ∈ {1, . . . , a′}. F om Table 1, we ha e χ(−c′
1) = 1. Le ∈ {1,...,a′−2}.
Assuming χ(− c′
1) = 1 when is odd, we show ha χ−( + 1)c′
1= 0 and
χ−( +2)c′
1= 1. Each s ep in he ollowing sequence o ces a colo on some numbe
in he gi en ange in o de o a oid a mononch oma ic solu ion o Equa ion (2a):
•x1=− c′
1, and xi=n o 2 ≤i≤m−1 implies χ−( + 1)c′
1+na′= 0;
INTEGERS: 25 (2025) 8
•x1=−( +1)c′
1+na′, and xi= 0 o 2 ≤i≤m−1 implies χ−( +2)c′
1+na′=
1;
•xi=n o 2 ≤i≤m−1, and xm=−( +2)c′
1+na′implies χ−( +1)c′
1= 0;
•x1=−( + 1)c′
1, and xi= 0 o 2 ≤i≤m−1 implies χ−( + 2)c′
1= 1.
In pa icula , we ha e χ−a′c′
1= 1. We no e ha he maximum allowable
alue o numbe s used is −a′c′
1+Ka′ om he second s ep, and his lies in he
domain o χ. In o de ha he numbe s lie in he domain o χ, we mus ha e
−a′c′
1+na′≤ −(a′+ 2)c′
1, in pa icula . This implies n≤ −2c′
1
a′. To comple e he
claim ha χ(n) = 0 o 0 ≤n≤K, we show ha χ−a′c′
1= 0 by using Table 1.
Each s ep in he ollowing sequence o ces a colo on some numbe in he gi en
ange in o de o a oid a mononch oma ic solu ion o Equa ion (2a) o (2b):
•xi=n o 2 ≤i≤m−1, and xm=−(a′+ 1)c′
1implies χ−a′c′
1−na′= 0;
•xi= 0 o 2 ≤i≤m−1, and xm=−a′c′
1−na′implies χ−(a′−1)c′
1−na′= 1;
•x1=−(a′−1)c′
1−na′, and xi=n o 2 ≤i≤m−1 implies χ−a′c′
1= 0.
This con adic ion comple es he p oo o he claim ha χ(n) = 0 o 0 ≤n≤K.
I can be shown ha a≤K, and so we ha e χ(n) = 0 o 0 ≤n≤a, in pa icula .
Fo he es o his p oo , we conside Equa ion (4b) and (4c) sepa a ely.
We i s conside Equa ion (4b). By assigning he colo o xiin he solu ion o
Equa ion (1a) and (1b) o xi−1, we no e ha he anges o c1and c2 ansla e o
−c′
1+a′+ 1 ≤ −c′
2≤ −(a′+ 2)c′
1.
We use χ(n) = 0 o 0 ≤n≤a, and p o e ha χ(n) = 0 o a+1 ≤n≤ −c′
1−1. Le
+ 1 = min{n:χ(n)=1}; we ha e shown ha + 1 > a. By way o con adic ion,
we may assume +1 ≤ −c′
1−1. By P oposi ion 4, he exp ession Pm−2
i=1 xi+axm−1
assumes e e y alue in he in e al [0,(a′+1) ] as each xi uns o e he se {0, . . . , }.
Unde he same ange o he xi’s, he exp ession
m−2
X
i=1
xi+axm−1−c′
2=xm
assumes e e y alue in he in e al I= [−c′
1+a′+ 1,(a′+ 1) −(a′+ 2)c′
1]. So in
o de o a oid a monoch oma ic solu ion o Equa ion (2b), we mus ha e χ(n) = 1
o each n∈I. Now choosing xi= + 1, 1 ≤i≤m−1 in Equa ion (2a) o ces
χ(a′+ 1)( + 1) −c′
1= 0 in o de o a oid a mononch oma ic solu ion. Bu
(a′+ 1)( + 1) −c′
1lies wi hin [−c′
1+a′+ 1,(a′+ 1) −(a′+ 2)c′
1], and his is a
con adic ion o he conclusion om he p e ious pa ag aph. The e o e, we ha e
he claim ha χ(n) = 0 o 0 ≤n≤ −c′
1−1.
INTEGERS: 25 (2025) 9
F om he abo e a gumen o =−c′
1−1, he exp ession
m−2
X
i=1
xi+axm−1−c′
2=xm
assumes e e y alue in he in e al J= [−c′
1+a′+ 1,−(a′+ 1)(c′
1+ 1) −(a′+ 2)c′
1].
Since −(a′+2)c′
1∈J, he e exis x1, . . . , xm−1, wi h each xi∈ {0,...,−c′
1−1}, such
ha Pm−2
i=1 xi+axm−1−c′
2=−(a′+ 2)c′
1. This gi es a monoch oma ic solu ion o
Equa ion (2b), since χ−(a′+ 2)c′
1= 0 by Table 1. This comple es he a gumen
o Equa ion (4b).
We now conside Equa ion (4c). By assigning he colo o xiin he solu ion o
Equa ion (1a) and (1b) o xi−1, we no e ha he anges o c1and c2 ansla e o
−(a′+ 2)c′
1<−c′
2≤ −(a′+ 3)c′
1−1.
Recall ha χ(n) = 0 o 0 ≤n≤−2c′
1
a′. Each s ep in he ollowing sequence o ces
a colo on some numbe in he gi en ange in o de o a oid a mononch oma ic
solu ion o Equa ion (2a):
•xi=−c′
1 o 2 ≤i≤m−1, and xm=−c′
2implies χ(a′+ 1)c′
1−c′
2= 0;
•x1= (a′+ 1)c′
1−c′
2, and xi= 0 o 2 ≤i≤m−1 implies χa′c′
1−c′
2= 1.
We cap u e his in o ma ion in Table 2.
0 1
0−c′
1
−(a′+ 2)c′
1−c′
2
(a′+ 1)c′
1−c′
2−(a′+ 1)c′
1
Table 2
A guing as in Equa ion (4b), wi h =K, he exp ession
m−2
X
i=1
xi+axm−1−c′
2=xm
assumes e e y alue in he in e al K= [−c′
2,(a′+1)K−c′
2]. Since (a′+1)c′
1−c′
2∈K,
he e exis x1, . . . , xm−1, wi h each xi∈ {0, . . . , K}, such ha Pm−2
i=1 xi+axm−1−
c′
2= (a′+ 1)c′
1−c′
2. This gi es a monoch oma ic solu ion o Equa ion (2b), since
χ(a′+ 1)c′
1−c′
2= 0 by Table 2. This comple es he a gumen in Equa ion (4c),
he eby comple ing he p oo o Theo em 6.