Sane Bounds on Some Polynomial van der Waerden Numbers
by William Gasarch1 , Clyde P. Kruskal2 , Justin D. Kruskal3
1
Introduction
CLYDE- I MOVED THIS NOTATION TO THE BEGINNING
We use the following standard notation and definitions.
Def 1.1 Let Z be the set of integers, N be the set of non-negative integers, and N+ be the
set of positive integers. If W ∈ N then [W ] is the set {1, . . . , W }.
Recall van Der Waerden’s Theorem [14] (see also [7], [10]).
Theorem 1.2 For any k ∈ N, for any c ∈ N, there exists W = W (k, c), such that for any
c-coloring of [W ], there exists a, d ∈ N, d 6= 0, such that a, a + d, . . . , a + (k − 1)d are all the
same color.
The original proof by van der Waerden was purely combinatorial and yielded a function
W whose growth was INSANE (called EEEEEEEEEENORMOUS by [7]). In particular, the
proof used an ω 2 induction and W (k, c) was bounded by a function that is not primitive
recursive. Shelah [13] gave a purely combinatorial proof that yielded bounds that were
HUGE, though not INSANE. In particular they were primitive recursive. Gowers [6] gave
an advanced proof (it used Fourier analysis) that yielded bounds that were relatively SANE
in that they only involved several stacks of exponentials.
We will now discuss a known generalization of van der Waerden’s theorem. Note that
the functions of d used are
d, 2d, . . . , (k − 1)d.
If these are replaced by other functions of d then does the result still hold? This of course
depends on what functions of d are used. The Polynomial van Der Waerden Theorem
addresses a large class of functions of d:
Theorem 1.3 For any p1 (x), . . . , pk (x) ∈ Z[x] such that each polynomial pi (x) has constant
term 0, for any c ∈ N, there exists W = W (p1 (x), . . . , pk (x); c) such that, for any c-coloring
of [W ], there exists a, d ∈ N, d 6= 0, such that a, a + p1 (d), . . . , a + pk (d) are all the same
color.
1
U. of MD at College Park, Dept of CS and UMIACS, [email protected]
U. of MD at College Park, Dept of CS, [email protected]
3
Eleanor Roosevelt High School, [email protected]
2
1
For k = 1, this theorem was proven independently by Furstenberg [4] and Sárközy [12].
Bergelson and Leibman [3] proved the general result using ergodic methods. These proofs
yielded no upper bounds on W (p1 (x), . . . , W (pk (x); c). Walters [15] proved it using combinatorial techniques which did yield bounds. However, the bounds are INSANE. The proof
of the polynomial van der Waerden theorem is an ω ω induction. To get a sense of how large
the bounds can be see Appendix VIII.
In this paper we show that, for some p(x) ∈ Z[x] and c = 2, 3, one can obtain sane
bounds on W (p(x); c). In particular, we will show the following.
• Let p(x) ∈ Z[x] be a polynomial of degree n with constant term 0. Then
(1) W (p(x); 2) ≤ mini∈N+ {|p(|2p(i)|)| + 1}, and
(2) W (p(x); 2) < 2 max{|p(1)|, . . . , |p(n + 1)|}.
• Let p(x) ∈ N[x] be a quadratic polynomial with constant term 0. Let g = gcd(p(1), p(2)).
Then
p(1) + p(2) − g + 1 if either p(1)/g or p(2)/g is even
W (p(x); 2) =
p(3) + 1
otherwise
Note that in this case we get an exact value for W (p(x); 2). Furthermore, there are
exactly 2g−1 proper colorings of [W − 1] up to isomorphism.
• If p(x) = ax2 + bx then W (p(x); 3) ≤ O(a2 b5 ).
• If p(x) = ax and c ∈ N+ then W (p(x); c) = |ac| + 1.
In some sense it is arbitrary to start the forbidden distances at 1 giving forbidden distances: p(1), p(2), . . .. They could start at any value, say s, giving forbidden distances
p(s), p(s + 1), p(s + 2), . . .. We define these new type of polynomial van der Waerden numbers and call them W (p(x); s, c).
It is an easy corollary of Theorem 1.3 that these numbers are bounded.
Theorem 1.4 For any p1 (x), . . . , pk (x) ∈ Z[x] such that each polynomial pi (x) has constant
term 0, for any s, c ∈ N , there exists W = W (p1 (x), . . . , pk (x); s, c) such that, for any
c-coloring of [W ] there exists a, d ∈ N, d 6= 0, a ≥ s such that a, a + p1 (d), . . . , a + pk (d) are
all the same color.
CLYDE- you had W (p1 (x), . . . , pk (x), s; c), I made it W (p1 (x), . . . , pk (x); s, c).
It turns out that for one polynomial, two colors (c = 2), and s large, these generalized
polynomial van der Waerden numbers have very sane bounds and a very simple form:
2
• For any p(x) ∈ Z[x] with constant term 0, there exists s0 ∈ N, such that for all s ≥ s0 ,
W (p(x); s, 2) = 2|p(s)| + 1.
CLYDE- I commented out the conjecture. We state it later.
For three colors (c = 3), we can get bounds for particular types of quadratic polynomials.
For example:
Bill.
Is this
• If p(x) = ax2 +bx, b is an odd multiple of a, and m = max{|a|, |b|} then W (p(x); s, 3) ≤ proved
O(m6 s7 ). (Much better results can be achieved for particular quadratic polynomials.) in the
paper?
2
Definitions and a Useful Lemma
Def 2.1 Let c ∈ N+ and W ∈ N+ .
1. A c-coloring of [W ] is a mapping [W ] → [c].
2. Let p(x) ∈ Z[x], c ∈ N+ , and W ∈ N+ . A (p(x); c)-proper coloring of [W ] is a c-coloring
of [W ] such that, for all x, y ∈ [W ], if y − x = p(d) for some d ∈ N + , then x and y
have different colors. When the context is clear, we will often write proper c-coloring
or simply proper coloring.
Note that the polynomial van der Waerden number W = W (p(x); c) is the least number
such that there is no (p(x); c)-proper coloring of [W ]. (By Theorem 1.3 there will always be
such a number.)
Although we care about proper (p(x); 2)-colorings, we need a more general notion: We
look at general sets of forbidden distances, not just distances forbidden by a polynomial
p(x).
We define proper coloring and van der Waerden number for sets of forbidden distances,
analogously with proper coloring and van der Waerden number for polynomials:
Def 2.2 Let F ⊆ Z, c ∈ N+ , and W ∈ N+ . Let f : N → Z.
• An (F, c)-proper coloring of [W ] is a c-coloring of [W ] such that, for all x, y ∈ [W ] with
y − x ∈ F , x and y have different colors.
• W = W (F ; c) is the least number such that there is no (F, c)-proper coloring of [W ].
If no such number exists, we set W (F ; c) = ∞.
• An (f (x), c)-proper coloring of [W ] is a c-coloring of [W ] such that, for all x, y ∈ [W ]
with y − x ∈ {f (i) : i ∈ N} x and y have different colors.
3
• W = W (f (x); c) is the least number such that there is no (f (x), c)-proper coloring of
[W ]. If no such number exists, we set W (f (x); c) = ∞.
CLYDE- above I add definition of W(f(x);c) since you use it later.
Def 2.3 Two colorings are isomorphic if you can obtain one from the other by permuting
the colors.
It is easy to see that if there are γ proper c-colorings of [W ] then there are γ/c! proper
colorings up to isomorphism.
CLYDE- I commented out the think about getting total number of proper colorings easier
than up-to-isom. Really the same.
Before getting started we make a couple of useful comments, which hold for any number
of colors:
• If 0 is a forbidden distance then the van der Waerden number is 1, because, in any
coloring, 1 and 1 + 0 automatically have the same color.
• We can always take the absolute value of a forbidden distance: If f < 0 is a forbidden
distance then a and a + f must have different colors. This is the same as saying a + f
and a must have different colors, but the difference is now positive.
Lemma 2.4 Let W (f (x); c) = Y +1. Then W (af (x); c) = aY +1. Furthermore, if there are
exactly γ (f (x), c)-proper colorings of [Y ] then there are exactly γ a (af (x), c)-proper colorings
of [aY ]; hence there are γ a /c! (af (x); c)-proper colorings of [aY ] up to isomorphism.
CLYDE- I changed the lemma now that we have the right notation for it. I also redid
the proof
Proof:
Consider the function af (x). For fixed 0 ≤ i < a, the equivalence class i mod a
can be treated as an isolated instance of f (x). Hence we can color it with any coloring of
[Y ]. These different colorings do not interfere with each other.
We proceed formally. Let COLi , 1 ≤ i ≤ γ, be all of the (f (x), c)-proper colorings of [Y ].
For each sequence (COLi0 , . . . , COLia−1 ) we define an (af (x), c)-proper coloring of [aY ].
The color we assign to x = ay + i where 0 ≤ i ≤ a − 1 is COLi (y). We leave it to the
reader to verify that this is a (af (x), c)-proper coloring. This shows that the number of
(af (x), c)-proper colorings of [aY ] is at least γ a .
Let COL be an (f (x), c)-proper coloring of [aY ]. For each 0 ≤ i ≤ a − 1 we define COLi
to be the coloring restricted to all y ≡ i (mod a). We leave it to the reader to show that
COLi is equivalent to an (f (x), c)-proper coloring of [Y ]. Hence we have mapped COL to
4
a sequence of a (f (x), c))-proper colorings of [Y ]. Hence the number of (af (x), c)-proper
colorings of [aY ] is at most γ a .
Combining the last two paragraphs we obtain that the number of (af (x), c)-proper colorings of [aY ] is γ a .
3
Linear polynomials
For completeness we cover linear polynomials, for which we obtain a complete solution.
Theorem 3.1 Let p(x) ∈ Z[x] be a linear polynomial with constant term 0 and c ∈ N+ .
Then
W (p(x); c) = |cp(1)| + 1 .
There are exactly (c!)|p(1)|−1 proper colorings of [W − 1] up to isomorphism.
Proof:
Let p(x) = x. Setting x = 1, 2, . . . , c gives forbidden distances 1, 2, . . . , c. So 1, 2, . . . , c + 1
must all be different colors, but there are only c colors. We can properly c-color [c] by giving
each value a different color. So, W (x; c) = c + 1 = cp(1) + 1, and there are c! proper colorings
of [c].
By Lemma 2.4, for any linear polynomial p(x) = ax, W (p(x); c) = |ac| + 1 = |cp(1)| + 1,
and there are c!|a|−1 = c!|p(1)|−1 proper colorings of [c] up to isomorphism.
4
4.1
2-Colorings
General polynomials
The following is our main lemma.
Lemma 4.1 Let s, t ∈ N+ . Let g = gcd(s, t). Then
(
s + t − g + 1 if either s/g or t/g is even
W ({s, t}; 2) =
∞
otherwise.
In both cases there are 2g−1 proper ({s, t}, 2)-colorings of [W ({s, t}; 2)−1] up to isomorphism.
5
Proof:
Temporarily assume s and t are relatively prime, so g = 1. Let z = s + t. We present a
proper ({s, t}, 2)-coloring of [z − 1], and show that it is unique up to isomorphism. Consider
the list
s mod z, 2s mod z, 3s mod z, . . . , (z − 1)s mod z.
Alternate coloring the elements in the list so that s mod z is RED, 2s mod z is BLUE,
3s mod z is RED, 4s mod z is BLUE, etc. Note that the coloring is forced after choosing
the first element to be RED: If (ks mod z) + s < z then ks mod z and (k + 1)s mod z must
have different colors because they are s apart. If (ks mod z) + s ≥ z then
(k + 1)s mod z = (k + 1)s − z mod z = (k + 1)s − (s + t) mod z = ks − t mod z
so ks mod z and (k + 1)s mod z must have different colors because they are t apart. So, the
coloring is unique up to isomorphism.
Furthermore, since s is relatively prime to t, it is also relatively prime to z. Hence
{s mod z, 2s mod z, . . . , (z − 1)s mod z} = [z − 1],
so this is a coloring of [z − 1].
We show that this is indeed a proper coloring: If x, x + s ∈ [z − 1] then x, x + s have
different colors by the construction. If x, x + t ∈ [z − 1] then x, x + t must have different
colors because x + t ≡ x − s mod z.
Whether this proper coloring can be extended beyond [z − 1] depends on the parity of z:
CASE (1): Assume that either s or t is even. (The other must be odd because we have
assumed that g = 1.)
Then z − 1 = s + t − 1 must be even, so that the first number in the above alternating
list of colors, s mod z, and the last number, (z − 1)s mod z, must have different colors. But
(z − 1)s ≡ zs − s ≡ −s ≡ t mod z.
So z = s + t cannot be RED or BLUE, implying that the coloring cannot be extended to z.
CASE (2): Assume s and t are both odd.
The above alternating list of colors, makes the odd numbers all have the same color, say
RED, and the even numbers BLUE (because each addition changes the parity of the number
being colored). Any number at or above s + t can be colored, but its color is forced by
subtracting s (or equivalently t). So the coloring can be uniquely extended to ∞.
6
Finally, assume that g > 1. Then there is no interaction of numbers x, y where x 6=
y mod g. So we can consider each equivalence class 1, 2, . . . , g mod g separately, and color
as above.
In CASE (1) above, each equivalence class has a finite number of elements:
1, g + 1, 2g + 1, . . . , (z/g − 2)g + 1 = ((s + t)/g − 2)g + 1 = s + t − 2g + 1;
2, g + 2, 2g + 2, . . . , (z/g − 2)g + 2 = ((s + t)/g − 2)g + 2 = s + t − 2g + 2;
..
.
g, 2g, 3g, . . . , (z/g − 2)g + g = ((s + t)/g − 2)g + g = s + t − 2g + g = s + t − g .
So, W ({s, t}; 2) = s + t − g + 1. Each equivalence class has 2 different colorings, depending
on whether it starts with RED or BLUE, so there are 2g proper colorings of [s + t − g].
In CASE (2) above, each equivalence class has an infinite number of elements:
1, g + 1, 2g + 1, . . . ;
2, g + 2, 2g + 2, . . . ;
..
.
g, 2g, 3g, . . . .
So, W ({s, t}; 2) = ∞, and there are 2g proper colorings of N+ .
Lemma 4.2 Let p(x) ∈ Z[x] be a polynomial with constant term 0. Let i, j ∈ N, such that
p(i), p(j) 6= 0, and g = gcd(p(i), p(j)). If either p(i)/g or p(j)/g is even then
W (p(x); 2) ≤ |p(i)| + |p(j)| − g + 1 .
Proof:
Follows from Lemma 4.1.
Note 4.3 The condition on p(i) and p(j) in Lemma 4.2 can be rephrased as p(i) and p(j)
have a different number of factors of 2 in them.
Theorem 4.4 Let p(x) ∈ Z[x] be a polynomial with and constant term 0. Then
W (p(x); 2) ≤ min+ {|p(|2p(i)|)| + 1} .
i∈N
7
Proof:
Fix an i ∈ N+ . If either p(i) = 0 or p(|2p(i)|) = 0 then W (p(x); 2) = 1 ≤ |p(|2p(i)|)| + 1.
Otherwise, 2p(i) has more factors of 2 than p(i) (in fact exactly one more factor), so p(|2p(i)|)
also has more factors of 2 than p(i) (since x is a factor of p(x)). By Lemma 4.2
W (p(x); 2) ≤ |p(i)| + |p(|2p(i)|)| − gcd(p(i), p(|2p(i)|)) + 1
= |p(i)| + |p(|2p(i)|)| − |p(i)| + 1
(since x is a factor of p(x))
= |p(|2p(i)|)| + 1
Theorem 4.5 Let p(x) ∈ Z[x] be a polynomial of degree n with constant term 0. Then
W (p(x); 2) < 2 max{|p(1)|, . . . , |p(n + 1)|} .
Proof:
Consider the following identity, which is proved in Appendix I (Lemma 7.3).
p(s) −
k−1 n X
s+k−1 X k−1
k=1
k
i=0
i
(−1)i (p(s + i + 1) − p(s + i)) = 0 .
Since each forbidden distance is paired except the initial p(s), it is the sum of an odd number
of forbidden distances (i.e., copies of the values from {p(s), p(s + 1), . . . , p(s + n)}). Thus,
since the sum is 0, some two values p(i) and p(j) must have a different number of factors
of 2. By setting s = 1, and using Lemma 4.2, this pair gives the desired upper bound on
W (p(x); 2).
Note 4.6 Theorems 4.4 and 4.5 are incomparable to each other. For different polynomials
either one could yield a smaller upper bound.
4.2
Quadratic polynomials
Corollary 4.7 Let s, t, u ∈ N+ . Let g = gcd(s, t). If u ≥ s + t, s/g and t/g are both odd,
and u/g is even then
W ({s, t, u}; 2) = u + 1
There are exactly 2g−1 proper colorings of [W − 1] up to isomorphism.
8
Proof:
By Lemma 4.1, [u] has 2g proper colorings. By the proof of Lemma 4.1,
1, g + 1, 2g + 1, . . . , u − g + 1
must alternate colors. Since u/g is even, 1 and u − g + 1 must have different colors. So,
in any proper coloring of [u + 1], u + 1 = (u − g + 1) + g must have the same color as 1,
but u + 1 must have a different color than 1 (because u is a forbidden distance). Thus, no
coloring can be extended to u + 1.
Lemma 4.8 Let p(x) ∈ N[x] be a quadratic polynomial with constant term 0. Let g =
gcd(p(1), p(2)). Then g|p(3).
Proof:
Let p(x) = ax2 + bx. Substituting x = 1, 2, 3 into p(x) gives
p(1) = a + b,
p(2) = 4a + 2b,
p(3) = 9a + 3b.
Subtracting p(1) from p(2) twice, and p(1) from p(3) thrice, we see that
g = gcd(p(1), p(2)) = gcd(a + b, 2a)
and gcd(p(1), p(3)) = gcd(a + b, 6a).
Thus, g|p(3).
Theorem 4.9 Let p(x) ∈ N[x] be a quadratic polynomial with constant term 0. Let g =
gcd(p(1), p(2)). Then
p(1) + p(2) − g + 1 if either p(1)/g or p(2)/g is even
W (p(x); 2) =
p(3) + 1
otherwise.
There are exactly 2g−1 proper colorings of [W − 1] up to isomorphism.
Proof:
CASE (1): Assume that either p(1)/g or p(2)/g is even.
The result follows from Lemma 4.1 (and the fact that p(1) + p(2) ≤ p(x), for all x ≥ 3).
CASE (2): Assume that p(1)/g and p(2)/g are both odd.
By Lemma 7.3, p(1)/g, p(2)/g, and p(3)/g cannot all be odd, so p(3)/g must be even. The
9
result follows from Lemma 4.8 and Corollary 4.7 (and the fact that not only is p(1) + p(2) ≤
p(x), for all x ≥ 3, but also p(3) < p(x), for all x ≥ 4).
Appendix VI gives a chart of some polynomial van der Waerden numbers for quadratic
polynomials and two colors.
5
3-Colorings
5.1
Method for obtaining upper bounds
Def 5.1
A coloring of [n] has repeat distance r if x and x + r have the same color, for all
1 ≤ x ≤ n − r.
(b)
(a) A coloring of [n] has repeat distance s under one-sided boundary condition b if x and
x + r have the same color, for all 1 ≤ x ≤ n − r − b.
(c) A coloring of [n] has repeat distance r under two-sided boundary condition b if x and
x + r have the same color, for all b < x ≤ n − r − b.
Lemma 5.2 In any 3-coloring of [n] with forbidden distances s, t, u, where 0 < s < t and
s + t = u:
(a) 2s + t is a repeat distance.
(b) t − s is a repeat distance under two-sided boundary condition s.
(c) 3s is a repeat distance under one-sided boundary condition t.
Proof:
(a) Consider a 3-coloring satisfying the conditions of the lemma. Let 1 ≤ x ≤ n − (2s + t).
Without loss of generality, we can assume that x is RED. Then x + s is not RED, say
BLUE, and x + u = (x + s) + t cannot be RED or BLUE so it must be GREEN. Then
(x + s) + u = (x + u) + s cannot be BLUE or GREEN so it must be RED. Since x and
x + u + s are both RED, (x + u + s) − x = u + s = 2s + t is a repeat distance,
10
(b) Consider a 3-coloring satisfying the conditions of the lemma. Let s < x ≤ n−(t−s)−s.
Without loss of generality, we can assume that x is RED. Then x − s is not RED, say
BLUE, and (x − s) + u = x + t cannot be RED or BLUE so it must be GREEN. Then
(x − s) + t = (x + t) − s cannot be BLUE or GREEN, so it must be RED. This process
requires that x − s > 0 and x + t ≤ n. So (x + t − s) − x = t − s is a repeat distance
under two-sided boundary condition s.
(c) Take 2s + t from part (a) and subtract t − s from part (b). The repeat distance is
(2s + t) − (t − s) = 3s. There is a one-sided boundary of size (t − s) + s = t from one
side of part (b).
Lemma 5.3 Assume [w] has a proper 3-coloring where s is a forbidden distance and r is
repeat distance under two-sided boundary condition b. If r|s then
w ≤ s + 2b + 1 .
Proof:
Assume w > s + 2b + 1. Assume, without loss of generality, that b + 1 is RED.
Then, by Lemma 5.2b, r + b + 1, 2r + b + 1, ..., s + b + 1 are also RED, since b + 1 > b and
(s + b + 1) + b = s + 2b + 1 ≤ n. But s is a forbidden distance so b + 1 and s + b + 1 cannot
both be RED. Contradiction.
5.2
Examples of 3-Colorings
Here are some examples of this method.
For linear polynomials we know by Theorem 3.1 that W (ax; 3) ≤ 3a + 1 (for a > 0). The
above method gives a tight upper bound:
Example 5.4 For a > 0, W (ax; 3) ≤ 3a + 1.
Proof:
Let p(x) = ax. Then
p(1) = a,
p(2) = 2a,
p(3) = 3a
are all forbidden distances. Since p(1)+p(2) = p(3), by Lemma 5.2b, p(2)−p(1) = 2a−a = a
is a repeat distance under two-sided boundary condition a. But p(1) = a is a forbidden
distance. Thus, by Lemma 5.3, W (ax; 3) ≤ a + 2a + 1 = 3a + 1.
11
For perfect squares the method gives a bound, but it is not tight.
Example 5.5 For a > 0, W (ax2 ; 3) ≤ 67a + 1.
Proof:
Let p(x) = ax2 . Then
p(3) = 9a,
p(4) = 16a,
p(5) = 25a .
Since p(3) + p(4) = p(5), by Lemma 5.2b, p(4) − p(3) = 16a − 9a = 7a is a repeat distance
under two-sided boundary condition 9a. But p(7) = 49a is a forbidden distance. Thus, by
Lemma 5.3, W (ax2 ; 3) ≤ 49a + 2 · 9a + 1 = 67a + 1.
In Appendix II we show that the exact bound is W (x2 ; 3) = 29, and that there are seven
proper colorings of 28 up to isomorphism. By Lemma 2.4 W (ax2 ; 3) = 28a + 1 and there are
42a /6 proper colorings of 28a up to isomorphism.
Example 5.6 For a, b > 0 and a|b, W (ax2 + bx; 3) ≤ 72b2 /a + 1.
Proof:
Let p(x) = ax2 + bx. Let
x = 5b/a,
y = 6b/a,
z = 8b/a .
Then
p(x) = 30b2 /a
p(y) = 42b2 /a
p(z) = 72b2 /a .
Since p(x) + p(y) = p(z), by Lemma 5.2b, p(y) − p(x) = 12b2 /a is a repeat distance under
two-sided boundary condition 30b2 /a. But p(3b/a) = 12b2 /a is a forbidden distance. Thus,
by Lemma 5.3, W (ax2 + bx; 3) ≤ 12b2 /a + 2 · 30b2 /a + 1 = 72b2 /a + 1.
In Appendix III we show that for W (x2 + x; 3) = 73, so the bound is tight for a = b.
By a computer program we have obtained that there are 54 proper 3-colorings of [72] up
to isomorphism. By Lemma 2.4 W (a(x2 + x); 3) = 72a + 1 and there are 324a /6 proper
colorings of 72a up to isomorphism.
5.3
Quadratics- General Upper Bounds
Theorem 5.7 Let p(x) = ax2 + bx. Then W (p(x); 3) ≤ O(|a5 b2 |).
12
Proof:
We prove the theorem for a, b ≥ 0. The other cases are similar.
CLYDE- ARE THE OTHER CASES REALLY SIMILAR
Let
x0 = (2a + 1)b,
y0 = (2a2 + 2a + 1)b,
z0 = (2a2 + 2a + 2)b .
Then
p(x0 ) = (4a3 + 4a2 + 3a + 1)b2
p(y0 ) = (4a5 + 8a4 + 8a3 + 6a2 + 3a + 1)b2
p(z0 ) = (4a5 + 8a4 + 12a3 + 10a2 + 6a + 2)b2
Thus p(x0 ) + p(y0 ) = p(z0 ). By Lemma 5.2b, 2p(x0 ) + p(y) is a repeat distance, and by
Lemma 5.2c, 3p(x0 ) is a repeat distance under one-sided boundary condition p(y). By a
Euclid’s algorithm-type argument (see Lemma 10.1 in Appendix IV) it is easy to check that
gcd(2p(x0 ) + p(y0 ), 3p(x0 )) = db2 for some constant d. Thus there exist integers j, k such
that j((2p(x0 ) + p(y0 )) + k(3p(x0 )) = db2 . By starting at 1 and adding repeat distance
2p(x0 ) + p(y0 ) j times and subtracting repeat distance 3p(x0 ) k times, we see that db2 is
also a repeat distance. Furthermore, by interspersing the adds and subtracts so that we
subtract whenever the sum is greater than 2p(x0 ) + p(y0 ), the one-sided boundary condition
is (2p(x0 )+p(y0 ))+p(y0 ) = 2(p(x0 )+p(y0 )). Thus for any integer α, αdb2 is a repeat distance
with the one-sided boundary condition 2(p(x0 )+p(y0 )) = O(a5 b2 ). But p(db) = ad2 b2 +b2 d =
(ad + 1)db2 is a forbidden distance. So, W (p(x); 3) ≤ p(db) + 2(p(x0 ) + p(y0 )) = O(a5 b2 ).
CLYDE- I DIDN’T”T LIKE HOW THIS LOOKED, THINKS LIKE
W (p(x); 3) ≤ p(db) + 2(p(x) + p(y))
HAVE p(x) APPEARING IN TWO DIFF WAYS. I MADE THE x, y, z you DEFINE IN
THE BEGINNING INTO x0 , y0 , z0 , BUT I MADE IT A MACRO SO EASY TO CHANGE
BACK.
CLYDE- you use O() HERE. IS THIS JUST FOR CONSTANTS OR DO YOU NEED
SOME QUANTITY TO BE LARGE? IF SO WE SHOULD DUMP THE O() AND GIVE
THE REAL CONSTANTS.
BILL- ONCE this is figured out I will do version for Carolyn Numbers.
Appendix VII gives a chart of some polynomial van der Waerden numbers for quadratic
polynomials and three colors.
13
5.4
Quadratics- Better Upper Bound in Some Cases
Assume that 1 and 2 are forbidden distances. This forces the colors to rotate
RBGRBGRBG... .
If some multiple of 3, say t, is also a forbidden distance then 1 and t + 1 could not be the
same color, so W (F ; 3) ≤ t + 1. This idea can be generalized:
Lemma 5.8 Assume that F is a set of forbidden distances with {f, 2f, tf } ⊆ F , where 3|t.
Then W (F ; 3) ≤ tf + 1.
We show several examples of using this lemma.
Example 5.9 Recall Example 5.6: For a, b > 0 and a|b, W (ax2 + bx; 3) ≤ 72b2 /a + 1.
Here is an alternative proof.
Let p(x) = ax2 + bx. Then
p(2b/a) = a(2b/a)2 + b(2b/a) = 6b2 /a
p(3b/a) = a(3b/a)2 + b(3b/a) = 12b2 /a
p(8b/a) = a(8b/a)2 + b(8b/a) = 72b2 /a .
By Lemma 5.8, W (ax2 + bx; 3) ≤ 72b2 /a + 1.
This method cannot be used to get a bound on W (p(x); 3) for all quadratic polynomials,
since p(x) = x2 has no solutions y, z ∈ N+ such that p(z) = 2p(y). However, this method
can give better results than the previous method for some special cases.
Example 5.10 Consider p(x) = x2 + 7x. Example 5.6 gives W (p(x); 3) ≤ 72 · 482 + 1 =
165889. But, by Lemma 5.8, p(3) = 30, p(5) = 60, and p(18) = 180, so W (p(x); 3) ≤ 181.
(Computer enumeration shows that the actual value is W (p(x); 3) = 80.)
Example 5.11 Consider p(x) = 4x2 +17x. Theorem 5.7, even with careful evaluation of the
constants, would yield a bound much worse than 295936 (which is a5 b2 ). But, p(15) = 1155,
p(22) = 2310, and p(70) = 20790, so W (p(x); 3) ≤ 20790. (Computer enumeration shows
that the actual value is W (p(x); 3) = 180.)
14
6
New starting place
Consider the polynomial van der Waerden numbers (W (p(x); s, c)), which have a new start
place, as described in the introduction. We will be particular interested in what happens
when s is large.
6.1
Linear Polynomials
Once again, for completeness, we cover linear polynomials, for which we obtain a complete
solution.
Theorem 6.1 Let p(x) ∈ Z[x] be a linear polynomial with constant term 0, and c ∈ N+ .
Then
W (p(x); s, c) = c|p(s)| + 1 .
There are exactly (c!)|p(s)|−1 proper colorings of [W − 1] up to isomorphism.
Proof:
Let p(x) = x. Setting x = s, 2s, . . . , cs gives forbidden distances s, 2s, . . . , cs. So 1, s+1, 2s+
1, . . . , cs + 1 must all be different colors, but there are only c colors. Thus W (p(x); s, c) ≤
|cp(s)| + 1.
We now show how to properly c-color [cp(s)]. For fixed 0 ≤ i < p(s), the equivalence
class i mod p(s) can be properly c-colored by giving each of the c values a different color.
So, W (p(x); s, c) = cp(s) + 1. Since there are c! ways to color each equivalence class, and
there are p(s) equivalence classes, there are c!p(s) proper colorings of [cp(s)]. Dividing by c!
shows that there c!p(s)−1 proper colorings up to isomorphism.
By Lemma 2.4, for any linear polynomial p(x) = ax, W (p(x); s, c) = c|p(s)| + 1, and
there are c!|p(s)|−1 proper colorings of [cp(s)] up to isomorphism.
6.2
c Colors, Any Polynomial
For nonlinear polynomials the situation is especially interesting when the starting place
becomes large:
Theorem 6.2 Let p(x) ∈ Z[x] be a polynomial with constant term 0. Then there exists an
s0 > 0 such that, for all s ≥ s0 , W (p(x); s, c) ≥ c|p(s)| + 1.
15
Proof:
Assume, WLOG, that the high order coefficient is positive. Let s0 be a value so
that, for all x ≥ s0 , p(x) is positive and monotonically increasing. It is clear that, for any
s ≥ s0 , [cp(s)] can be properly c-colored by coloring 1, . . . , p(s) color 1, p(s) + 1, . . . , 2p(s)
color 2, . . . , (c − 1)p(s) + 1, . . . , cp(s) color c. So W (p(x); s, c) ≥ cp(s) + 1.
We conjecture that the bound holds the other direction as well:
Conjecture 6.3 Let p(x) ∈ Z[x] be a polynomial with constant term 0. Then there exists
an s0 > 0 such that, for all s ≥ s0 , W (p(x); s, c) = c|p(s)| + 1.
As was just shown, the conjecture holds for linear polynomials and any number of colors (Theorem 6.1). In the next subsection we show that it holds for two colors and any
polynomial (Theorem 6.5).
6.3
Two colors
Lemma 6.4 Let T be a bag of not necessarily distinct numbers, whose sum is 0. Let N and
P be the magnitudes of the smallest (most negative) and largest (most positive) elements of
T , respectively. Then there is a sequence qi consisting all of the elements from T where every
partial sum si = q1 + · · · + qi satisfies 0 ≤ si < N + P . Furthermore, a positive number (qi )
occurs in the sequence only if the previous partial sum (si−1 ) is less than N .
Proof:
The proof is by induction. We construct the sequence qi by removing elements one at a
time from T . Assume that at step i − 1, 0 ≤ si−1 < N + P . At any step the current partial
sum plus all of the remaining elements in T must sum to 0. So, in particular, the final sum
is 0. There must always be some non-positive number in T .
This means that there are two cases:
Case (1): There are only non-positive numbers left in T : Then the remaining numbers can
be appended to q in any order (since they sum to 0 with the last partial sum).
Case (2): There are both positive and non-positive numbers left in T : There are two cases:
Case (a): The current partial sum si−1 < N : Then any positive number can be used as the
next number in the sequence (since the total will be si−1 + qi < N + P ).
Case (b): The current partial sum si−1 ≥ N : Then any non-positive negative number can
be used as the next number in the sequence (since the total will be si−1 + qi ≥ N − N = 0).
Theorem 6.5 Let p(x) ∈ Z[x] be a polynomial with constant term 0. Then there exists an
s0 > 0 such that, for all s ≥ s0 , W (p(x); s, 2) = 2|p(s)| + 1.
16
Proof:
Let p(x) be a polynomial of degree n > 1. (Theorem 6.1 covers the n = 1 case.)
By Theorem 6.2 there exists an s0 > 0 such that, for all s ≥ s0 , W (p(x); s, 2) ≥ 2p(s) + 1.
We now show that there is an s0 such that for all s ≥ s0 the bound holds the other direction
as well. Assume, WLOG, that the high order coefficient is positive. Let s0 be a value so
that, for all x ≥ s0 , p(x) is monotonically increasing. Other conditions below may increase
s0 .
Recall the identity from Appendix I (Lemma 7.3):
p(s) −
k−1 n X
s+k−1 X k−1
k
k=1
i
i=0
(−1)i (p(s + i + 1) − p(s + i)) = 0 .
Since each forbidden distance is paired except the initial p(s), it is the sum of an odd number
of forbidden distances that sum to 0, so it gives some upper bound on W (p(x); s, 2). We
now show that the values can be summed so that all of the partial sums are between 1 and
2p(s) + 1, which reduces the upper bound to W (p(x); s, 2) ≤ 2p(s) + 1.
Start with p(s) giving partial sum p(s) + 1. After that subtract pairs p(s + 1) − p(s)
until the partial sum is ≤ p(s + 1) − p(s). Subtract each pair by first adding p(s) and then
subtracting p(s + 1). This guarantees that the partial sum stays between 1 and 2p(s) + 1.
We need to make sure that there are enough such pairs to drive the final value down close
to 1. We can count such pairs by restricting i in the summation to be 0.
0 n X
s+k−1 X k−1
k=1
=
k
n X
s+k−1
k=1
n X
k
i=0
i
(−1)i (p(s + i + 1) − p(s + i))
(p(s + 1) − p(s))
!
s+k−1
=
− 1 (p(s + 1) − p(s))
k
k=0
s+n
=
− 1 (p(s + 1) − p(s))
by parallel summation
n
In this final value s+n
− 1 is a polynomial of degree n and p(s + 1) − p(s) is a polynomial
n
of degree n − 1. So their product is a polynomial of degree 2n − 1. Since p(s) has degree
only n, for s large enough there will be enough pairs (to drive the sum down).
For i even, the pairs p(s + i + 1) − p(s + i) contribute negative values to the sum, and
for i odd, the pairs p(s + i + 1) − p(s + i) contribute positive values. Negative values are
17
added by first adding p(s + i) and then subtracting p(s + i + 1). Positive values are added
by first adding p(s + i + 1) and then subtracting p(s + i). Let N be the magnitude of the
smallest negative pair, and P be the magnitude of the largest positive pair. By Lemma 6.4
the values can be summed down to 1 without going above N + P + p(s + n). Since n is fixed
and N and P are polynomials of degree n − 1, for s large enough, N + P + p(s + n) ≤ 2p(s).
In the particular case where p is a quadratic we obtain an estimate as to how large s0
has to be.
Theorem√ 6.6 Let p(x) = ax2 + bx with the coefficient a positive. Then, for s > −b/a and
2 −4a+b2
s ≥ 6a−b+ 56a
, W (p(x); s, 2) = 2p(s) + 1.
2a
Proof:
For x > −b/a, p(x) is positive and monotonically increasing. Setting n = 2 in the identity
from Appendix I (Lemma 7.3) gives
p(s) −
s(s + 1)
s(s + 3)
(p(s + 1) − p(s)) +
(p(s + 2) − p(s)) = 0 .
2
2
Now add forbidden distances as in the proof of Theorem 6.5 (the general polynomial result):
Start at 1 and add p(s). Subtract p(s + 1) − p(s) until the partial sum is close to 1. Then
add p(s + 2) − p(s + 1) or subtract p(s + 1) − p(s) as in Lemma 6.4 (until all of the forbidden
distances are used). The following
condition must hold: (p(s+1)−p(s))+p(s+2) ≤ 2p(s)+1.
√
6a−b+ 56a2 −4a+b2
This is equivalent to s ≥
.
2a
Corollary 6.7 For s ≥ 7, W (x2 ; s, 2) = 2p(s) + 1.
Appendix V gives exact values for W (x2 ; s, 2) for 1 ≤ s ≤ 6. It turns out that even for
s = 6, W (x2 ; s, 2) = 2p(s) + 1.
6.4
Three Colors
Theorem 6.8 Let s ∈ N . Let p(x) = ax2 + bx. Then W (p(x); s, 3) = O(a4 b2 s).
Proof:
We assume a, b ≥ 0. The other cases are similar.
18
We use a modification of the proof of Theorem 5.7. Let w be a number to be determined
later. Let
x0 = (2a + 1)bw,
y0 = (2a2 + 2a + 1)bw,
z0 = (2a2 + 2a + 2)bw .
By reasoning similar to that of Theorem 5.7 we get that (ad+1)db2 w is a repeat condition
j with
k
s
5 2
2
boundary condition O(a b w), and that (ad + 1)db w is a forbidden distance. If t = (2a+1)
then x0 , y0 , z0 ≥ s. Hence
W (p(x); s, 3) ≤ O(a5 b2 w) = O(a5 b2
s
) = O(a4 b2 s).
2a + 1
Lemma 5.8 can be used to obtain better upper bounds on W (p(x); s, c) for quadratic p
in some cases. The interested reader is invited to work out analogs of Examples 5.9, 5.10,
and 5.11.
19
7
Appendix I
CLYDE- CHECK HOW I HANDLE THE REFS HERE.
The following two Lemmas were known to Euler (see [5] equation 5.9) using what would
now be called the theory of finite differences. We provide a combinatorial proof of the first
Lemma that is based on a proof of a different theorem by Anonymous [1].
Lemma 7.1 Let m, n, s ∈ N with m < n.
n
X
n
(−1)
(s + i)m = 0.
i
i=0
i
Proof:
Consider the following problem:
How many ordered m-tuples of elements of {1, . . . , n+s} are there such that each element
of {1, . . . , n} appears at least once? This problem is as easy as it looks. The answer is 0.
However, we can also solve this problem a different way. We solve it by inclusionexclusion. How many ordered tuples are there with no constraints: (s + n)m ?
We subtract out those that do not use 1 or do not use 2 or · · · or do not use n. There
are n1 (s + n − 1)m of these. We then add back those that used two of {1, . . . , n}. There are
n
(s + n − 2)m of these. We keep doing this to obtain
2
n
1 n
m
2 n
m
n n
0 = (s + n) + (−1)
(s + n − 1) + (−1)
(s + n − 2) + · · · + (−1)
(s + n − n)m
1
2
n
If n is even this gives the result we seek. If n is odd then negate both sides and we obtain
the result we seek.
Theorem 7.2 Let p ∈ Z[x] be a polynomial of degree n − 1. Let s ∈ N, s ≥ 1. Then
n
X
n
(−1)
p(s + i) = 0
i
i=0
i
Proof:
Let
p(x) =
n−1
X
j=0
20
aj x j .
Then
n
X
i=0
n X
n−1
X
n
i
j n
p(s + i)
(−1) =
aj (s + i)
(−1)i
i
i
i=0 j=0
n−1
n
X
X
j n
=
aj
(s + i)
(−1)i .
i
j=0
i=0
By Lemma 7.1 all of the inner sums are 0. Hence the entire sum is 0.
We have been told that the following lemma can be derived from the method of finite
differences (see[2]); however, we have not been able to locate a self-contained proof or a
theorem to refer to to make the proof easier. Hence we provide our own proof. After
the proof we will note why we think it is unlikely there is an easier proof with the finite
differences.
Lemma 7.3 Let p(x) ∈ Z[x] be a polynomial of degree n with constant term 0. Then
p(s) −
k−1 n X
s+k−1 X k−1
k=1
k
i=0
i
(−1)i (p(s + i + 1) − p(s + i)) = 0 .
21
Proof:
(−1)i (p(s + i + 1) − p(s + i))
i
i=0
k=1
k n
X s+k−1 X
k
p(s) +
(−1)i p(s + i)
k
i
i=0
k=1
n
k
X s+k−1 X k
(−1)i p(s + i)
k
i
i=0
k=0
n X
n
X
s−1+j
j
i
(−1) p(s + i)
collecting the p(s + i) terms together, for fixed i
j
i
i=0 j=i
n n
X
X
s−1+j
j
i
(−1) p(s + i)
j
i
i=0
j=i
n
n X
X
s−1+j
s−1+i
i
(−1) p(s + i)
by a version of trinomial revision
s
−
1
+
i
i
i=0
j=i
n n
X
s−1+i X s−1+j
i
(−1) p(s + i)
i
s−1+i
j=i
i=0
n−i n
X
s−1+i X s−1+i+j
i
(−1) p(s + i)
i
s−1+i
j=0
i=0
n−i n
X
s−1+i X s−1+i+j
i
(−1) p(s + i)
i
j
i=0
j=0
n
X
s − 1 + i s − 1 + i + (n − i + 1)
i
(−1) p(s + i)
by parallel summation
i
n−i
i=0
p(s) −
=
=
=
=
=
=
=
=
=
k−1 n X
s+k−1 X k−1
k
22
=
=
=
=
=
=
n
X
s−1+i s+n
(−1) p(s + i)
s−1
n−i
i=0
n
X
s
s+i s+n
i
(−1) p(s + i)
s+i
s
n−i
i=0
n
X
s
s+i s+n
i
(−1) p(s + i)
s+i
s
s+i
i=0
n
X
n s+n
s
i
(−1) p(s + i)
s+i i
n
i=0
X
n
s+n
p(s + i) n
s
(−1)i
s+i
i
n
i=0
0
i
by extraction
by trinomial revision
by Lemma 7.3
The last equality holds because
p(s+i)
s+i
is a polynomial of degree n − 1.
CLYDE- What are TRINOMIAL REVISION AND PARALLEL SUMMATION.
Note 7.4 If f is a function then ∆f (s) = f (s + 1) − f (s) and ∆i+1 f (s) = ∆(∆i f (s)). A
basic theorem in the method of finite differences is that, for any function f ,
k
∆ f (s) =
k X
k
i=0
i
(−1)i p(s + i).
This looks very useful since this term came up on the third line of the proof of Lemma 7.3.
However, what is known about ∆k f (s) is not quite what we need.
1. (This is from [8] page 189.) If f is a polynomial of degree d and c0 , . . . , cd are such
that
x
x
x
+ cd−1
+ · · · + c0
f (x) = cd
0
d
d−1
then ∆m f (0) is cn if n ≤ d and is 0 if n > d. This is not useful to us since it is about
∆m f (0), not ∆m f (s).
2. (This is from [5] page 458.) If f is a polynomial of degree d with lead coefficient A.
23
Then ∆m f (0) is 0 if d < m and is Ad! if d = m. This is not useful since it says nothing
about the case of d > m.
8
Appendix II: POLYVDW({x2}, 3) = 29
In this section we will show that POLYVDW({x2 }, 3) = 29. We will show:
1. There is no proper 3-coloring of [29].
2. There are 10 proper 3-colorings of [28] up to isomorphism.
Theorem 8.1 POLYVDW({x2 }, 3) ≤ 29.
Proof:
Assume, by way of contradiction, that there exists χ, a proper 3-coloring of [29]. By
Lemma 5.2b we know that,
(∀x)[10 ≤ x ≤ 13 =⇒ χ(x) = χ(x + 7)].
We refer to this fact as FORCE.
We can assume, without loss of generality, that χ(10) = RED. Since 11 − 10 = 12 we
know that χ(11) 6= RED. We can, without loss of generality, assume that χ(11) = BLUE.
17: By FORCE χ(17) = χ(10) = RED
18: By FORCE χ(18) = χ(11) = BLUE.
10 11 12 13 14 15 16 17 18 19 20
R B
R B
19: Since χ(10) = RED and χ(18) = BLUE, χ(19) = GREEN.
12: By FORCE χ(12) = χ(19) = GREEN.
10 11 12 13 14 15 16 17 18 19 20
R B G
R B G
20: Since χ(11) = BLUE and χ(19) = GREEN, χ(20) = RED.
13: By FORCE χ(13) = χ(20) = RED.
10 11 12 13 14 15 16 17 18 19 20
R B G R
R B G R
Now we have that χ(17) = χ(13) = RED. But 17 − 14 = 22 . This is a contradiction.
24
8.1
POLYVDW({x2 }, 3) ≥ 29
We could just give a proper 3-coloring of [28] and obtain POLYVDW({x2 }, 3) = 29. However,
we will actually determine all of the non-isomorphic 3-colorings of [28]. Why? As the saying
goes, because we can.
Theorem 8.2 There are 7 non-isomorphic proper 3-colorings of [28].
Proof:
Let χ be an proper 3-coloring of [28]. By Lemma 5.2b we know that,
(∀x)[10 ≤ x ≤ 12 =⇒ χ(x) = χ(x + 7)].
We assume that χ(10) = RED and χ(11) = BLUE.
By reasoning similar to that of Theorem 8.1 the following is forced:
1 2 3 4 5 6 7 8 9 10 11 12 13 14
G R
B G R B G B G
15 16 17 18 19 20 21 22 23 24 25 26 27 28
R B R B G R B
G R
We are unable to force anything else. We will have two cases. Since χ(21) = BLUE
χ(22) ∈ {RED, GREEN}.
Our cases are based on χ(22).
Case 1: χ(22) = GREEN. This forces the following.
1 2 3 4 5 6 7 8 9 10 11 12 13 14
G R B G R B
B G R B G B G
15 16 17 18 19 20 21 22 23 24 25 26 27 28
R B R B G R B G
G R B G R
Nothing else is forced; however, we have constraints on how the coloring can be extended.
7: The numbers in [28] that are a square away from 7 are {3, 6, 8, 11, 16, 23}. All of these
are BLUE except for 23 which we have not colored yet. Hence
χ(7) ∈ {RED, GREEN} − {χ(23)}.
25
23: The numbers in [28] that are a square away from 23 are {7, 14, 19, 22, 24, 27}. All of
these are GREEN except for 7 which we have not colored yet. Hence
χ(23) ∈ {RED, BLUE} − {χ(7)}.
Hence there are 3 ways to extend this coloring to [28]:
• χ(7) = RED, χ(23) = BLUE.
• χ(7) = GREEN, χ(23) = RED.
• χ(7) = GREEN, χ(23) = BLUE.
Case 2: χ(22) = RED. The following is forced.
1 2 3 4 5 6 7 8 9 10 11 12 13 14
G R
B G R B G B G
15 16 17 18 19 20 21 22 23 24 25 26 27 28
R B R B G R B R B G R
Nothing else is forced; however, we have constraints on how this coloring can be extended.
27: The numbers that are a square away from 27 are {2, 11, 18, 23, 26, 28}. These numbers
are either colored BLUE or not colored at all (2,26,28). Hence
χ(27) ∈ {RED, GREEN} − {χ(2), χ(26), χ(28)}.
We take subcases based on this.
Subcase 2a: χ(27) = RED. The following is forced.
1 2 3 4 5 6 7 8 9 10 11 12 13 14
B G R G R B B B G R B G B G
15 16 17 18 19 20 21 22 23 24 25 26 27 28
R B R B G R B R B G R G R B
Subcase 2b: χ(27) = GREEN. The following is forced.
1 2 3 4 5 6 7 8 9 10 11 12 13 14
G R B G R
B G R B G B G
26
15 16 17 18 19 20 21 22 23 24 25 26 27 28
R B R B G R B
G R B G R
Nothing else is forced; however, we have constraints on how this coloring can be extended.
6: The numbers in [28] that are a square away from 6 are {2, 5, 7, 10, 15, 22}. All of these
are either RED or not colored (7,22). Hence
χ(6) ∈ {BLUE, GREEN} − {χ(7), χ(22)}.
7: The numbers in [28] that are a square away from 7 are {3, 6, 8, 11, 16, 23}. All of these
are either BLUE or not colored (6,23). Hence
χ(7) ∈ {RED, GREEN} − {χ(6), χ(23)}.
22: The numbers in [28] that are a square away from 22 are {6, 13, 18, 21, 23, 26}. All of
these are either BLUE or not colored (6,23). Hence
χ(22) ∈ {RED, GREEN} − {χ(6), χ(23)}.
23: The numbers in [28] that are a square away from 23 are {7, 14, 19, 22, 24, 27}. All of
these are either GREEN or not colored (7,22). Hence
χ(23) ∈ {RED, BLUE} − {χ(7), χ(22)}.
Using these constraints one can easily show that the following are the only proper 3colorings of [28] that extend the above coloring.
• χ(6) = BLUE, χ(7) = RED, χ(22) = RED, χ(23) = BLUE.
• χ(6) = BLUE, χ(7) = RED, χ(22) = GREEN, χ(23) = BLUE.
• χ(6) = BLUE, χ(7) = GREEN, χ(22) = RED, χ(23) = BLUE.
• χ(6) = BLUE, χ(7) = GREEN, χ(22) = GREEN, χ(23) = RED.
• χ(6) = BLUE, χ(7) = GREEN, χ(22) = GREEN, χ(23) = BLUE.
• χ(6) = GREEN, χ(7) = RED, χ(22) = RED, χ(23) = BLUE.
27
We have found ten proper 3-colorings of [28]. Nine of them begin GRB and are different
later, so they cannot be isomorphic. The remaining one has the same color at 6 as at 7,
which none of the other ones do. Hence it is not isomorphic to any of the others.
Appendix III: W (x2 + x; 3) = 73
9
Theorem 9.1 POLYVDW({x2 + x}, 3) = 73.
Proof:
We know that POLYVDW({x2 + x}, 3) ≤ 73 from Example 5.6 (for which we gave two
different proofs). Here is a 3-coloring of [73].
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
R R G G R R B B R R B B G G B B G G
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
R R G G R R B B R R B B G G B B G G
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
R R G G R R B B R R B B G G B B G G
55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72
R R G G R R B B R R B B G G B B G G
10
Appendix IV: GCD calculations
Lemma 10.1 Let
p(x) = (4a3 + 4a2 + 3a + 1)b2
and
p(y) = (4a5 + 8a4 + 8a3 + 6a2 + 3a + 1)b2
Then gcd(2p(x) + p(y), 3p(x)) = db2 for some d ≤ 34560.
28
Proof:
2p(x) + p(y) = (4a5 + 8a4 + 16a3 + 14a2 + 9a + 3)b2
and
3p(x) = 3(4a3 + 4a2 + 3a + 1)b2
They both have a common factor of b2 , so we can drop that term and put it back in at
the end. We will use a Euclid’s algorithm type argument along with the facts that
gcd(ma, b) ≤ m gcd(a, b)
gcd(a, b) ≤ gcd(ma, b)
and
Divide out 3 from 3p(x) this gives a starting point of
4a5 + 8a4 + 16a3 + 14a2 + 9a + 3
and
4a3 + 4a2 + 3a + 1
with 3b2 to be multiplied back in later.
Subtract a2 times the second from the first:
4a4 + 13a3 + 3a2 + 9a + 3
and
4a3 + 4a2 + 3a + 1
Subtract a times the second from the first:
9a3 + 10a2 + 8a + 3
and
4a3 + 4a2 + 3a + 1
Subtract 2 times the second from the first:
a3 + 2a2 + 2a + 1
4a3 + 4a2 + 3a + 1
and
Subtract 4 times the first from the second:
a3 + 2a2 + 2a + 1
and
− 4a2 − 5a − 3
Multiply the first by 4 and the second by -1:
4a3 + 8a2 + 8a + 4
and
4a2 + 5a + 3
Subtract a times the second from the first:
3a2 + 5a + 4
and
4a2 + 5a + 3
Subtract the first from the second:
3a2 + 5a + 4
and
29
a2 − 1
Subtract 3 times the second from the first:
5a + 7
and
a2 − 1
5a + 7
and
5a2 − 5
Multiply the second by 5:
Subtract a times the first from the second:
5a + 7
and
− 7a − 5
5a + 7
and
− 2a + 2
and
− 2a + 2
Add the first to the second:
Add two times the second to the first:
a + 11
Add two times the first to the second:
a + 11
and
24
Multiply the first by 24:
24a + 11 · 24
and
24
Subtract a times the second from the first:
11 · 24
and
24
These two numbers have gcd = 24. So the total is 24 · 3b2 = 72b2 .
Thus the gcd ≤ db2 for some natural number d ≤ 72.
30
11
Appendix V: W (x2; s, 2) for 1 ≤ s ≤ 6
By Corollary 6.7 we know W (x2 ; s, 2) = 2s2 + 1 for s ≥ 7. In this appendix we show what
happens for 1 ≤ s ≤ 6.
Theorem 11.1
1. W (x2 ; 1, 2) = 5.
2. W (x2 ; 2, 2) = 13.
3. W (x2 ; 3, 2) = 25.
4. W (x2 ; 4, 2) = 37.
5. W (x2 ; 5, 2) = 60.
6. W (x2 ; 6, 2) = 72.
Proof:
1) By Lemma 4.1 W ({1, 4}; 2) = 5. Since 5 ≤ 9, W (x2 ; 1, 2) = 5.
2) By Lemma 4.1 W ({4, 9}; 2) = 13. Since 13 ≤ 16, W (x2 ; 2, 2) = 13.
3) By Lemma 4.1 W ({9, 16}; 2) = 25. Since 2 ≤ 25, W (x2 ; 3, 2) = 25.
4a) W (x2 ; 4, 2) ≤ 37. Assume, by way of contradiction, that COL : [37] → [2] is a proper
(x2 , 4; 2)-coloring.
Let 1 ≤ x ≤ 12. Then x + 25 ∈ [37]. Note that
COL(x) 6= COL(x + 25) 6= COL(x + 25 − 16) = COL(x + 9).
Hence
1 ≤ x ≤ 9 =⇒ COL(x) = COL(x + 9).
Let 17 ≤ x ≤ 28. Then x − 16, x + 9 ∈ [37]. Note that
COL(x) 6= COL(x − 16) 6= COL(x − 16 + 25) = COL(x + 9).
Hence
17 ≤ x ≤ 28 =⇒ COL(x) = COL(x + 9).
31
By the above implications
COL(1) = COL(10) = COL(19) = COL(28) = COL(37).
But 37 − 1 = 36 = 62 . This is a contradiction.
4b) We show that W (x2 ; 4, 2) ≥ 37 by showing that there is a proper (x2 , 4; 2)-coloring of
[36].
By Lemma 4.1 W ({16, 25}; 2) = 41. Hence there is a 2-coloring of [40] with forbidden
distances 16 and 25. The restriction of this coloring to [36] is clearly a (x2 , 4; 2)-coloring of
[36].
5a) W (x2 ; 5, 2) ≤ 60. Assume, by way of contradiction, that COL : [60] → [2] is a proper
(x2 , 5; 2)-coloring.
Let 1 ≤ x ≤ 11. Then x + 49 ∈ [60]. Note that
COL(x) 6= COL(x + 49)
6= COL(x + 49 − 36) = COL(x + 13)
6= COL(x + 13 + 25) = COL(x + 38)
6= COL(x + 38 − 36) = COL(x + 2)
Hence
1 ≤ x ≤ 11 =⇒ COL(x) = COL(x + 2).
Let 1 ≤ x ≤ 9. Then x + 51 ∈ [37]. Note that
COL(x) 6= COL(x + 25)
6= COL(x + 25 + 25) = COL(x + 50)
6= COL(x + 50 − 49) = COL(x + 1)
Hence
1 ≤ x ≤ 9 =⇒ COL(x) 6= COL(x + 1).
By the above implications COL(1) 6= COL(12). Clearly COL(12) 6= COL(12 + 25) =
COL(37). Hence COL(1) = COL(37) which is a contradiction.
5b) W (x2 ; 5, 2) ≥ 60.
We define a coloring COL : [59] → [2] and then prove that it is a proper (x2 , 5; 2)-coloring.
1. x is RED if
32
(x ∈ ([1, 11] ∪ [23, 36] ∪ [48, 59]) and x is odd) or
(x ∈ ([12, 22] ∪ [37, 47]) and x is even.
2. x is BLUE if
(x ∈ ([1, 11] ∪ [23, 36] ∪ [48, 59]) and x is even) or
(x ∈ ([12, 22] ∪ [37, 47]) and x is odd.
We show that there cannot be two RED points that are 25, 36, or 49 apart. The case for
BLUE is similar.
Let x be RED.
Case 1: x ∈ [1, 11] and x is odd.
(a) x + 25 ∈ [26, 36] and x + 25 is even. Hence x + 25 is BLUE.
(b) x + 36 ∈ [37, 47] and x + 36 is odd. Hence x + 36 is BLUE.
(c) x + 49 ∈ [50, 60] and x + 49 is even. Hence x + 49 is BLUE or x + 49 is not in the
domain of COL.
In all later cases x + 49 is not in the domain of COL.
Case 2: x ∈ [12, 22] and x is even.
(a) x + 25 ∈ [37, 47] and x + 25 is odd. Hence x + 25 is BLUE.
(b) x + 36 ∈ [48, 58] and x + 36 is even. Hence x + 36 is BLUE.
Case 3: x ∈ [23, 36] and x is odd.
(a) x + 25 ∈ [48, 61] and x + 25 is even. Hence x + 25 is BLUE.
(b) x + 36 ∈ [59, 72] and x + 36 is odd. Hence x + 36 is BLUE or x + 36 is not in the
domain of COL.
Case 4: In all remaining cases x + 25 and x + 36 are not in the domain of COL.
6a) W (x2 ; 6, 2) ≤ 73. Assume, by way of contradiction, that COL : [73] → [2] is a proper
(x2 , 6; 2)-coloring.
Let 1 ≤ x ≤ 9. Then x + 64 ∈ [73]. Note that
COL(x) 6= COL(x + 64)
6= COL(x + 64 − 49) = COL(x + 15)
6= COL(x + 15 + 36) = COL(x + 51)
6= COL(x + 51 − 49) = COL(x + 2)
Hence
1 ≤ x ≤ 9 =⇒ COL(x) = COL(x + 2).
In particular COL(1) = COL(9).
33
Also note that
COL(1) 6= COL(1 + 36) = COL(37)
6= COL(37 + 36) = COL(73)
6= COL(73 − 64) = COL(9)
Hence COL(1) 6= COL(9).
This is a contradiction.
6b) W (x2 ; 6, 2) ≤ 73. We define a proper (x2 , 6; 2)-coloring of [72]. Color {1, . . . , 36} RED
and {37, . . . , 72} BLUE. We leave it as an easy exercise to show that this works.
34
12
Appendix VI: Polynomial van der Waerden numbers for a quadratic polynomial and two colors
Chart of W (p(x); 2) for p(x) = ax2 + bx for 0 ≤ a ≤ 10 and −10 ≤ b ≤ 10.
b
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
0
21
19
17
15
13
11
9
7
5
3
1
3
5
7
9
11
13
15
17
19
21
1
1
1
1
1
1
1
1
1
1
1
5
13
11
13
17
25
23
25
29
37
35
2
1
9
1
7
1
5
1
3
1
7
9
13
25
19
21
25
25
31
33
37
49
3
9
1
7
5
1
13
5
1
9
25
13
17
21
37
27
29
31
49
39
37
45
4
9
7
1
5
5
7
1
11
13
17
17
23
25
29
49
35
37
41
41
47
49
a
5
1
5
7
25
9
1
13
37
19
21
21
49
31
33
37
61
43
45
49
73
51
6
25
7
9
11
1
15
17
19
49
27
25
33
33
37
41
45
73
51
53
55
57
7
11
37
13
1
17
49
23
25
29
61
29
37
41
73
47
49
53
85
59
61
65
8
9 10
13 17
1
15
1 23
1 21 25
19 61 29
21 25 73
25 29 31
25 33 37
31 73 41
33 39 41
37 41 47
33 37 41
43 85 53
45 51 97
49 49 59
49 57 61
55 97 61
57 61 65
61 65 71
97 69 73
67 109 77
69 75 121
1. These numbers are far smaller then the bounds one can obtain from Walters [3] proof
of the polynomial van der Warden theorem. See Appendix VIII for a sense of how
large the bounds obtained from Walters proof are.
2. The numbers tend to increase with increasing a and |b|. One notable exception is that
if a = b then W (p(x); 2) jumps, hence when a = b + 1 or a = b + 1 the value drops.
CLYDE- DO WE HAVE MORE COMMENTARY?
35
13
Appendix VII: Polynomial van der Waerden numbers for a quadratic polynomial and three colors
Chart of W (p(x); 3) for p(x) = ax2 + bx for 0 ≤ a ≤ 5 and −5 ≤ b ≤ 5.
b
-5
-4
-3
-2
-1
0
1
2
3
4
5
0
16
13
10
7
4
1
4
7
10
13
16
a
1
2
3
4
5
1 64 61 217
1
1
1 91
1 289
1 10
1 135 171
1
1 68 97 171
1 49 105 190 183
29 57 85 113 141
73 76 65 156 253
64 145 123 151
?
37 95 217
?
?
65 127
? 289
?
55
? 109
? 361
1. These numbers are far smaller then the bounds one can obtain from Walters [3] proof
of the polynomial van der Warden theorem. See Appendix VIII for a sense of how
large the bounds obtained from Walters proof are.
CLYDE- ANY MORE COMMENTARY?
14
Appendix VIII: Bounds Obtained by Walters’ Proof
CLYDE- THIS APPENDIX IS NEW
The following bounds W (p(x); c) for quadratic p can be obtained from Walters’ proof of
the Polynomial Van Der Waerden Theorem [3].
Def 14.1 Let p(x) be a quadratic polynomial with zero constant term. For 1 ≤ r ≤ c we
define U (c, r) as follows.
1. U (c, 1) = min{p(x) : x ∈ N} + 1.
2. U (c, r) = (U (c, r − 1)W (2U (c, r − 1), cU (c,r−1) ))2 + U (c, r − 1)W (2U (c, r − 1), cU (c,r−1) )
36
Theorem 14.2 W (p(x); c) ≤ U (c, c).
We give some examples.
Example 14.3 W (x2 , 2). The largest value x2 obtains on the naturals is 1. Hence U (c, 1) =
2. Therefore
U (2, 2) = (U (2, 1)W (2U (2, 1), 2U (2,1) ))2 +U (2, 1)W (2U (2, 1), 2U (2,1) ) = 4W (4, 4))2 +2W (4, 4).
The best known bounds on W (4, 4) are
24
22
13
1048 ≤ W (4, 4) ≤ 2
.
The lower bound is from [11, 9], the upper bound follows from [6]. This leads to rather
large bounds for W (x2 ; 2). By contrast we have that W (x2 ; 2) = 5.
Example 14.4 W (x2 , 3). The largest value x2 obtains on the naturals is 1.
U (3, 2) = (U (3, 1)W (2U (3, 1), 3U (3,1) ))2 +U (3, 1)W (2U (3, 1), 3U (3,1) ) = (2W (4, 32 ))2 +2W (4, 32 ) = 4W (4, 9)2 +
The best known bounds on W (4, 9) are
24
393469 ≤ W (4, 9) ≤ 2
22
18
.
The lower bound is from [11, 9] (its actually the bound on W (4, 7)), the upper bound
follows from [6]. This leads to rather large bounds for W (x2 ; 3). By contrast we have that
W (x2 ; 3) = 29.
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39