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Formulas for two-color Rado numbers have been established for many families of linear equations. However, there are no explicit formulas for two-color Rado numbers for any nonlinear equations. In this paper, we will establish formulas for the two-color Rado numbers for three families of equations: x + y ^sup n^ = z, x + y ^sup 2^ + c = z, and x + y ^sup 2^ = az , where c and a are positive integers.[PUBLICATION ABSTRACT]
Graphs and Combinatorics (2013) 29:407415 DOI 10.1007/s00373-012-1140-8
ORIGINAL PAPER
An Excursion into Nonlinear Ramsey Theory
Adam Doss Dan Saracino Donald L. Vestal Jr.
Received: 13 May 2010 / Revised: 2 February 2012 / Published online: 23 Febuary 2012 Springer 2012
Abstract Formulas for two-color Rado numbers have been established for many families of linear equations. However, there are no explicit formulas for two-color Rado numbers for any nonlinear equations. In this paper, we will establish formulas for the two-color Rado numbers for three families of equations: x+yn = z, x+y2+c = z,
and x + y2 = az, where c and a are positive integers.
Keywords Ramsey theory Nonlinear Rado number
Mathematics Subject Classication (2000) 05D10
1 Introduction
Ramsey theory, loosely speaking, is the study of conditions which will guarantee that certain kinds of order will arise. There are several major theorems which fall in the realm of Ramsey theory: Ramseys theorem, Van der Waerdens theorem, the Hales-Jewett theorem, Schurs theorem, and Rados theorem. A good overview of these can be found in [7]. In this paper, we will look at results that are related to Schurs theorem and Rados theorem.
Schurs theorem was proved in 1916 (thus predating Ramseys theorem). He proved [20] that if the positive integers are colored using t colors, then there will be three
A. DossEpic Systems Corporation, Verona, WI, 53593, USA
D. SaracinoDepartment of Mathematics, Colgate University, Hamilton, NY, 13346, USA
D.L. Vestal Jr. (B)
Department of Mathematics and Statistics, South Dakota State University, Brookings, SD, 57007, USA e-mail: [email protected]
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(not necessarily distinct) integers x, y, and z, all colored with the same color, for which x + y = z. Such a solution (x, y, z) is referred to as a monochromatic solution.
For a given number of colors t, since a monochromatic solution must eventually occur, there will be a smallest integer S such that any coloring of the integers in {1, 2, . . . , S}
using t colors must contain a monochromatic solution. This smallest number is called the t-color Schur number and is denoted by S (t) . The only values which are known precisely are the following:
S (1) = 2, S (2) = 5, S (3) = 14, and S (4) = 45.
Richard Rado, a student of Schurs, generalized this result to include linear equations with more variables and also to systems of linear equations [5,1517]. Rado found a necessary and sufcient condition for a linear system to have a monochromatic solution. For more information on this, see [13].
2 Some Examples Involving Linear Equations
Rados Theorem was an existence result: under certain conditions, we can guarantee that a monochromatic solution exists. For example, if we color the positive integers using two colors, say green and blue, then we can conclude that there will be a monochromatic solution to the equation x1 + x2 + x3 + x4 = x5. A related question is, what
is the smallest integer R such that for any 2-coloring of {1, 2, . . . , R} , there must be a
monochromatic solution to this equation? This number, an analog of the Schur number, is referred to as the 2-color Rado number for the equation x1 + x2 + x3 + x4 = x5.
The two-color Rado number for this particular equation is 19. In 1982, Beutelspacher and Brestovansky [2] proved that the 2-color Rado number for the equation x1 + x2 + + xm1 = xm (where m 3) is m2 m 1.
Since then, there have been many similar results of 2-color Rado numbers for families of linear equations. We list some of the families here with references.
x1 + x2 + c = kx3 k > 0 [14]
x1 + x2 + + xm1 + c = xm c < 0 [12]
x1 + x2 + + xm1 = 2xm [19]
a1x1 + a2x2 + + am1xm1 = xm a1, a2, . . . , am1 > 0 [8, 9].
Of course, one thing that all of these equations have in common is that they are linear. In this paper, we explore nonlinear equations.
3 Some Examples Involving Nonlinear Equations
First, are there any Rado-type results involving nonlinear equations? The answer appears to be yes, although the results are more of the existence type rather than nding explicit Rado numbers. For example, Croot [4] proved the following conjecture of Erdos and Graham: there is a constant b > 0 such that for any t-coloring of the integers
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in [bracketleftbig]2, bt[bracketrightbig] , there exists a monochromatic set S such that [summationtext]
nS
= 1. Graham [6] reports that Rdl has proved that any t-coloring of the positive integers must yield a monochromatic solution to 1x + 1y = 1z. Similarly, Bergelson [1] proved that any t-coloring
of the positive integers must yield a monochromatic solution to x y = p (z) , where
p (z) is a polynomial with integer coefcients and constant term 0. In [11], Khalfalah and Szemerdi proved that if k is an integer and p(z) is a polynomial with integer coefcients such that p(a) is even for some integer a, then there exists an integer N(p, k) such that for all N N(p, k) and any k-coloring of the set {1, 2, 3, . . . , N},
there exists a monochromatic pair x < y N such that x + y = f (z) for some
integer z. For f (z) = z2, this proves a conjecture of Erdos, Roth, Srkzy, and Ss.
Finally (and perhaps of greatest interest) Graham [6] offers a monetary reward ($250) to anyone who can determine whether or not x2 + y2 = z2 has a monochromatic
solution for any t-coloring.
In the remainder of this paper, we will examine three families of nonlinear equations with the goal of nding an explicit formula for the 2-color Rado number.
4 The 2-Color Rado Number for x + yn = z
In this section, we prove our rst result.
Theorem 1 For n 2, the 2-color Rado number for the equation x + yn = z is
1 + 2n+1.
(Note that, in the case n = 1 this is the Schur equation, and we get 1+22 = 5 = S (2).)
Proof To prove this result, there are two things that have to be shown. First, the lower bound: there is a coloring of the set [braceleftbig]1, 2, . . . , 2n+1[bracerightbig] which avoids a monochromatic solution to our equation. And second, the upper bound: for any 2-coloring of the set 1, 2, . . . , 2n+1, 1 + 2n+1[bracerightbig] , there must be a monochromatic solution to our equation.
For the coloring of the set [braceleftbig]1, 2, . . . , 2n+1[bracerightbig] , we will let G, B denote the set of integers which are colored green, blue, respectively. We claim that the coloring below avoids monochromatic solutions to the equation x + yn = z:
G =
[braceleftBig]1, 3, . . . , 2n 1, 2n + 2, 2n + 4, . . . , 2n+1
[braceleftBig]2, 4, . . . , 2n, 2n + 1, 2n + 3, . . . , 2n+1 1
In other words, we color the odd integers in [bracketleftbig]1,2n[bracketrightbig] and the even integers in [bracketleftbig]2n+1,2n+1[bracketrightbig]
green, and then the even integers in [bracketleftbig]1, 2n[bracketrightbig] and the odd integers in [bracketleftbig]2n + 1, 2n+1[bracketrightbig] blue.
To show that this coloring avoids a monochromatic solution, we make the following observation: if x + yn = z, where x, y, z [braceleftbig]1, 2, . . . , 2n+1[bracerightbig] , then y must equal 1
or 2, since y 3 implies (recalling that n 2) that z = x + yn 1 + 3n > 2n+1.
Now, if x, y, and z are green, then y = 1, and therefore x and z are consecutive
1 n
[bracerightBig]
and
B =
[bracerightBig] .
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(green) integers. However, there are no consecutive integers in the set G. So that rules out a green solution. If x, y, and z are blue, then y = 2, and we must have
1 x 2n, 2n + 1 z 2n+1, and x and z both even or both odd. However, in the
set B, the only possible values for x would be even, while the possible values for z are odd. Thus, there is no blue solution.
An interesting thing to note regarding the coloring above is its uniqueness in the following sense: if 1 is colored green, then the coloring above is the only coloring of [braceleftbig]1, 2, . . . , 2n+1[bracerightbig] which avoids monochromatic solutions. Although this is not necessary for the lower bound, we will prove it for use in the upper bound argument.
Suppose that we have a coloring of [braceleftbig]1, 2, . . . , 2n+1[bracerightbig] , with 1 colored green, which avoids monochromatic solutions. From the solution x = 1, y = 1, z = 2, we see that
2 must be blue. From this, we can deduce two rules.
Rule 1: Any integer adjacent to a green integer must be colored blue.
Rule 2: Any integer that differs by 2n from a blue integer must be colored green.
The proof then follows from (nite) induction. Starting with an odd integer r < 2n which is colored green, we conclude (by Rule 1) that r + 1 is a blue, even integer;
by Rule 2, r + 1 + 2n is a green, even integer; by Rule 1, r + 2 + 2n is a blue, odd
integer; and then by Rule 2, r + 2 is an odd, green integer. In particular, starting with
r = 1 being green, we conclude that 2 is blue, 2 + 2n is green, 3 + 2n is blue, and 3 is
green, at which point, the process above can be repeated. As it is repeated, we nd that the odd integers in [bracketleftbig]1, 2n[bracketrightbig] are colored green, the even integers in [bracketleftbig]1, 2n[bracketrightbig] are colored blue, the even integers in [bracketleftbig]2n + 1, 2n+1[bracketrightbig] are colored green, and the odd integers in
2n + 1, 2n+1[bracketrightbig] are colored blue. This induction ends with r = 2n 1 colored green,
from which we conclude that 2n is blue, and 2n+1 is green (the next step would be to conclude that 1 + 2n+1 is blue, but this is outside of the range we are considering).
The only integer which does not get colored in the argument above is 1+2n, but since
2 + 2n is green, we must color 1 + 2n blue by Rule 1. Thus, the coloring is unique.
The uniqueness of the coloring above now comes in handy when we attempt to prove the upper bound. We must show that any coloring of the integers in the set 1, 2, . . . , 2n+1, 1 + 2n+1[bracerightbig] will yield a monochromatic solution to the equation
x + yn = z.
Without loss of generality, we may assume that 1 is colored green. If there is no monochromatic solution involving the integers in [braceleftbig]1, 2, . . . , 2n+1[bracerightbig] , then this must be due to those integers being colored as shown above. In particular, 1 and 2n+1 are colored green and 2 and 1 + 2n are colored blue. But then, depending on how we color
1 + 2n+1,
one of the solutions below will be monochromatic:
x = 2n+1, y = 1, z = 1 + 2n+1 or x = 2n + 1, y = 2, z = 1 + 2n+1.
This completes our proof.
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5 The 2-Color Rado Number for x + y2 + c = z
In this section, we establish the next theorem.
Theorem 2 For c 2, the 2-color Rado number for x + y2 + c = z is c2 + 7c + 7. Proof To show that c2 + 7c + 7 is a lower bound for the 2-color Rado number, we
give a coloring of {1, 2, . . . , c2 + 7c + 6} that admits no monochromatic solution of
x + y2 + c = z. Let
G =
[braceleftBig]1, 2, . . . , c + 1, c2 + 6c + 6, . . . , c2 + 7c + 6
[bracerightBig]
If we had a green solution, then y would have to satisfy 1 y c + 1, since if
y c2 + 6c + 6 then
z = x + y2 + c c4 + 12c3 + 48c2 + 73c + 37 > c2 + 7c + 6.
Also, z would have to satisfy c2 + 6c + 6 z c2 + 7c + 6, since z = x + y2 + c
1 + 12 + c > c + 1. But these bounds on y and z imply that x = z y2 c would
have to satisfy 3c + 5 x c2 + 6c + 5, and thus x would not be colored green. So
there is no green solution. There is no blue solution since x c + 2 and y c + 2
imply that z = x + y2 + c c2 + 6c + 6.
To show that c2+7c+7 is an upper bound for the 2-color Rado number, we suppose
(by way of contradiction) that c 2 and we have a 2-coloring of {1, 2, . . . , c2+7c+7}
that admits no monochromatic solution of x + y2 + c = z, and we obtain a contra
diction by establishing several facts about the coloring. Without loss of generality, we assume that 1 G.
Lemma 1 Both c2 + 6c + 6 and 2c + 3 are in G, and c + 2 B.
Proof Considering the solution (x, y, z) = (1, 1, c +2), we see that c +2 B. Then,
using the solution (c + 2, c + 2, c2 + 6c + 6), we see that c2 + 6c + 6 G. From the
solution (c2 + 6c + 6, 1, c2 + 7c + 7) we conclude that c2 + 7c + 7 B, and then
from the solution (2c + 3, c + 2, c2 + 7c + 7) we conclude that 2c + 3 G. Lemma 2 We have 2 G.
Proof Suppose that 2 B. We will show that then 2c + 1 must be colored both green
and blue.
First, from Lemma 1 we know that c + 2 B. Then, from the solution (2, c +
2, c2 + 5c + 6), we see that c2 + 5c + 6 G. From (c2 + 5c + 6, 1, c2 + 6c + 7) we
get c2 + 6c + 7 B, and then from (c + 3, c + 2, c2 + 6c + 7) we get c + 3 G.
Using (c + 3, 1, 2c + 4) we see that 2c + 4 B, and then using (c, 2, 2c + 4) we see
that c G. By considering (c, 1, 2c + 1) we conclude that 2c + 1 B.
and
B =
[braceleftBig]c + 2, c + 3, . . . , c2 + 6c + 5
[bracerightBig] .
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On the other hand, from (2, 2, c + 6) we get c + 6 G, so from (5, 1, c + 6) we
get 5 B. Since c + 2 B by Lemma 1, (c + 2, 2, 2c + 6) yields 2c + 6 G. From
the preceding paragraph, we know that c G, so from (2c + 6, c, c2 + 3c + 6) we
get c2 + 3c + 6 B. Since weve just seen that 5 B, from (5, c + 1, c2 + 3c + 6)
we obtain c + 1 G. Using Lemma 1 and (3c + 5, c + 1, c2 + 6c + 6) we see
that 3c + 5 B. Then from (2c + 1, 2, 3c + 5) we conclude that 2c + 1 G, a
contradiction.
Lemma 3 We have c 1 B.
Proof Note that c 1 1 since c 2. Using Lemmas 1 and 2 and the solution
(c 1, 2, 2c + 3), we see that c 1 B.
Lemma 4 We have 3 G.
Proof Suppose by way of contradiction that 3 B. We will show that then 2c + 8
must be in both B and G.From Lemma 1 and the solution (3, c+2, c2 +5c+7) we see that c2 +5c+7 G,
so from (c2 + 5c + 7, 1, c2 + 6c + 8) we obtain c2 + 6c + 8 B. Using Lemma 1 and
the solution (c + 4, c + 2, c2 + 6c + 8) we get c + 4 G, and then using Lemma 2
and the solution (c + 4, 2, 2c + 8) we get 2c + 8 B. But from Lemma 3 and the
solution (c 1, 3, 2c + 8) we get 2c + 8 G, a contradiction.
Lemma 5 If k is a positive integer and c k > 3 and c k B, then c k 3 B.
Proof Since c k B we conclude from Lemma 3 and the solution (c k, c
1, c2 k + 1) that c2 k + 1 G. Then from (c2 k + 1, 1, c2 + c k + 2) we get
c2 +ck +2 B, and from (2ck +1, c1, c2 +ck +2) we get 2ck +1 G.
Finally, from (c k 3, 2, 2c k + 1) we get c k 3 B.
We now derive the contradiction that proves Theorem 2, as follows. From Lemmas 2, 3, and 4 we conclude that c 5. Using k = 1 in Lemma 5, we con
clude from Lemma 3 that c 4 B, and since 1, 2, 3 G this forces c 8. Using
k = 4 in Lemma 5, we conclude that c 7 B, and since 1, 2, 3 G this forces
c 11. Using k = 7 in Lemma 5, we conclude that c 10 B, and since 1, 2, 3 G
this forces c 14. Continuing in this way, we conclude that c is greater than every
positive integer.
Our proof that c2 + 7c + 7 is an upper bound for the Rado number depends on the
assumption that c 2. But, when c = 1, similar arguments show that (again stipulat
ing that 1 G) the set {1, . . . , c2 +7c+7} has a unique 2-coloring that avoids a mono
chromatic solution: G = {1, 2, 5, 13, 14} and B = {3, 4, 6, 7, 8, 9, 10, 11, 12, 15}.
From the solutions (6, 3, 16) and (14, 1, 16) it then follows that the 2-color Rado number for the equation x + y2 + 1 = z is 16.
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6 The 2-Color Rado Number for x + y2 = az
In this section we prove our nal result.
Theorem 3 For a 2, the 2-color Rado number for x + y2 = az is a 1.
Proof Since (a 1, a 1, a 1) is a solution of x + y2 = az, it is clear that a 1 is
an upper bound for the 2-color Rado number. For a = 2, it is also obviously a lower
bound.We need to show that, for a 3, there is a coloring of the set {1, 2, . . . , a 2}
which contains no monochromatic solution to our equation. For a = 3 or 4, this set
contains no solutions at all. For a 5, there is always the solution (a 4, 2, 1). In
what follows we assume that a 5, so there is always a solution with z = 1.
Lemma 6 If (x1, y1, z1) and (x2, y2, z2) are solutions of x + y2 = az and y1 = y2,
then x1 = x2 and z1 = z2.
Proof By the stated assumptions, we have x1 x2 (mod a), so since 1 x1, x2
a 2 we have x1 = x2. It follows, by the stated assumptions, that z1 = z2.
Lemma 7 If (x, y, z) is a solution of x + y2 = az and 1 x, z a 2 then y > z.
Proof Suppose for a contradiction that y z, and write z = a k, where 2 k
a1. We have az = x + y2 x +z2 and az = (z+k)z = z2+kz, so z2+kz x +z2
and therefore kz x. Thus k(a k) x a 2, so (k 1)a k2 2. But
a = z + k 1 + k, so (k 1)a k2 1, a contradiction.
We now obtain a 2-coloring of {1, 2, . . . , a2} that admits no monochromatic solu
tion of x + y2 = az, as follows. Enumerate all the solutions with 1 x, y, z a 2
by using the antilexicographic order on the pairs (y, z). As noted above, the rst solution on the list has z = 1. Color 1 green, and color all ys that occur in solutions of
x + y2 = a 1 blue (all such ys are greater than 1, by Lemma 7).
Assume inductively that we have colored all the ys and zs of solutions with z < z0 so that in each of these solutions the y and the z are colored differently. Consider the solutions of x + y2 = az0. By Lemmas 6 and 7, the ys in these solutions are larger
than all the zs and different from all the ys in all solutions with z < z0, so none of the ys in solutions of x + y2 = az0 have been colored previously. If z0 has been
colored previously then retain this coloring of z0; otherwise color z0 arbitrarily. Then color all the ys in solutions of x + y2 = az0 the opposite color of z0.
By induction, then, we can unambiguously color all the ys and zs of all solutions so that in each solution the y and the z are colored differently. Finally, color any uncolored numbers arbitrarily, and we have a coloring that admits no monochromatic solution.
We required that a 2, but for completeness, we note that for a = 1, the equation
becomes x + y2 = z, which falls into our rst class of equations; thus the 2-color
Rado number is 1 + 23 = 9.
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7 Future Research
The search for the Rado numbers of various classes of equations has become a fruitful endeavor over the past two decades, largely because of the nice patterns that seem to occur. Similar nice patterns occur in the nonlinear equations we have above. But one thing that stands out here is the difference in the colorings used for the lower bounds. This raises a question of which type of coloring might be more typical. Although it is difcult to quantify this, there are many cases of block colorings that occur in linear families: typically three blocks (a string of numbers colored green, then a string colored blue, and a nal string colored green). In particular, in [3], Burr et al. prove that the 2-color Rado number for the family of linear equations x + y + c = z,
where c is a xed nonnegative integer, is 4c + 5. Our second family, x + y2 + c = z
is a nonlinear generalization of this, and it turns out to have a similar coloring for the lower bound. Based on this, we offer the following conjecture.
Conjecture 1 Let c and m be positive integers, with c > m. If m is odd or c is even, then the 2-color Rado number for the equation x1 + x2 + + xm + y2 + c = z is
t2 + mt + t + c 1, where t = c + m + 1.
Note that if m is even and c is odd in the situation above, then the 2-color Rado number will be innite: simply color the odd natural numbers green and the even natural numbers blue. As a specic case of the conjecture, the reader can verify that the 2-color Rado number for the equation x1+x2+y2+4 = z is in fact t2+mt+t+c1 = 73
and that the coloring of the integers from 1 to 72 which avoids a monochromatic solution turns out to be unique (in fact, it is the same three-block pattern that we saw with the second family of equations).
In the third family of equations, x + y2 = az, what happens if we add in another
linear term: x1 + x2 + y2 = az? Since (a 2, a 2, a 2, a 2) is a solution,
we clearly have a 2 as an upper bound for the 2-color Rado number. But based on
computer data, it looks like the Rado number is in fact consistently smaller than a 2
(for example, the 2-color Rado number for x1 + x2 + y2 = 30z is 19). If we allow for
a third color, however, we nd that the 3-color Rado number for x1 + x2 + y2 = 30z
is indeed 28.
Conjecture 2 Let a and m be positive integers, with a > m. For any t greater than m, the t-color Rado number for x1 + x2 + + xm + y2 = az is a m.
What other generalizations can be made? We could try systems of nonlinear equations. Or variations like off-diagonal or disjunctive systems (see [10] and [18] for linear examples of these). The possibilities are boundless.
References
1. Bergelson, V.: Ergodic theory of Zd -actions. In: Pollicott, M., Schmidt, K. (eds.) London Mathematical Society, Lecture Note Series, vol. 228, pp. 161 (1996)
2. Beutelspacher, A., Brestovansky, W.: Generalized Schur Numbers, Lecture Notes in Mathematics, vol. 969. pp. 3038. Springer, Berlin (1982)
3. Burr, S., Loo, S., Schaal, D.: On Rado numbers I (preprint)
123
Graphs and Combinatorics (2013) 29:407415 415
4. Croot, E. III.: On a coloring conjecture about unit fractions. Ann. Math. 157(2), 545556 (2003)5. Deuber, W.: Developments based on Rados dissertation Studien zur Kombinatorik. Survey in Combinatorics, pp. 5274. Cambridge University Press, Cambridge (1989)
6. Graham, R.: Some of my favorite problems in Ramsey theory. Comb. Number Theory 7(2), 229 236 (2007)
7. Graham, R., Rothschild, B., Spencer, J.: Ramsey Theory, 2nd edn. Wiley-Interscience Series in Discrete Mathematics. Wiley, New York
8. Guo, S., Sun, Z.: Determination of the two-color Rado number for a1x1 + a2x2 + + am1xm1 =
x0. J. Comb. Theory Ser. A 115(2), 345353 (2008)
9. Hopkins, B., Schaal, D.: On Rado numbers for
m1
[summationtext]
i=1
ai xi = xm. Adv. Appl. Math. 35(4), 433
441 (2005)
10. Johnson, B., Schaal, D.: Disjunctive Rado numbers. J. Comb. Theory Ser. A 112(2), 263276 (2005)11. Khalfalah, A., Szemerdi, E.: On the number of monochromatic solutions of x + y = z
2. Comb.
Probab. Comput. 15, 213227 (2006)
12. Kosek, W., Schaal, D.: Rado numbers for the equation
m1
[summationtext]
i=1
xi + c = xm. for negative values of c. Adv.
Appl. Math. 27(4), 805815 (2001)
13. Landman, B., Robertson, A.: Ramsey Theory on the Integers, American Mathematical Society. Providence, RI (2004)
14. Martinelli, B., Schaal, D.: On generalized Schur numbers for x1 + x2 + c = kx3. Ars Comb. 85, 33
42 (2007)15. Rado, R.: Verallgemeinerung eines Satzes von van der Waerden mit Anwendungen auf ein Problem der Zahlentheorie. Sonderausg. Sitzungsber. Preuss. Akad. Wiss. Phys. Math. Klasse 17, 110 (1933)
16. Rado, R.: Studien zur Kombinatorik. Math. Z. 36, 242280 (1933)17. Rado, R.: Note on combinatorial analysis. Proc. Lond. Math. Soc. 48, 122160 (1936)18. Robertson, A., Schaal, D.: Off-diagonal generalized Schur numbers. Adv. Appl. Math. 26(3),
252257 (2001)
19. Schaal, D., Vestal, D.: Rado numbers for x1 + x2 + + xm1 = 2xm. Congr. Numerantium 191,
105116 (2008)20. Schur, I.: ber die Kongruenz xm + y
m
z
m (mod p). Jahresber. Deutsch. Math. Verein. 25,
114117 (1916)
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