the combinatorics of solving linear equations origins: (i. schur) does the congruence always have...

The combinatorics of solving linear equations

Origins: (I. Schur)

n n nx y z (mod p)Does the congruencealways have non-trivial solutions? 1875 -

1941

The combinatorics of solving linear equations

Origins: (I. Schur)

n n nx y z (mod p)Does the congruencealways have non-trivial solutions?

Yes !

Theorem (Schur – 1916)

If the positive integers are partitioned into finitely many classes then at least one of the classes contains solutions to the equation

x y z.

1875 - 1941

Some notation:

N:: {1,2,3,......}, [n] : {1,2,...,n}

χ N [: [r] - an r-coloring of Nχ N [:

χ induces a partition of N 1 2 rC C ... C

into color classes χ 1iC (i)

Objects in a single color class will be called monochromatic.

Schur’s theorem restated:

In any r-coloring of there is always a monochromatic solution

to the equation x y z.

N





N


B. L. van der Waerden(1903 – 1996)

Theorem (van der Waerden – 1927)

In any r-coloring of there is always a monochromatic

arithmetic progression of length k (= k-AP).

N

Theorem (van der Waerden – 1927)

In any r-coloring of there is always a monochromatic


N




N

Special case k = 3.



to the equation x y 2z.

N

B. L. van der Waerden(1903 – 1996)

What about the equation x y 3z?


Fact. There is a 4-coloring of with no monochromatic

solution to

N

x y 3z.


Fact. There is a 4-coloring of with no monochromatic

solution to

N

x y 3z.

Proof. Define χ N [: [4] by setting

χ t(5 (5k i)) i, 1 i 4.

Then 31 2

1 2 3

tt t5 (5k i) 5 (5k i) 3 5 (5k i)

32 dd(1 5 3 5 ) i 0 (mod 5)

which is impossible.

In general, call an equation

r-regular if every r-coloring of contains a monochromatic

solution to E.

1 2 n 1 1 2 2 n nE E(x ,x ,...,x ) ax a x .. a x 0

N

Also, call E regular if it is r-regular for every r.



solution to E.

1 2 n 1 1 2 2 n nE E(x ,x ,...,x ) ax a x .. a x 0

N


Which equations are regular ?

Theorem (R. Rado – 1933)

E is regular iff it can be solved with all ix 0 or 1,

and not all 0.



solution to E.

1 2 n 1 1 2 2 n nE E(x ,x ,...,x ) ax a x .. a x 0

N


Which equations are regular ?

Theorem (R. Rado – 1933)

E is regular iff it can be solved with all ix 0 or 1,

and not all 0.

x y z, x y 2z, x y 3z

1 0 1, 1 1 2 1, ???

Conjecture (Rado – 1933) For every k, there are equations

which are k-regular but are not (k+1)-regular.

Conjecture (Rado – 1933) For every k, there are equations

which are k-regular but are not (k+1)-regular.

Theorem (Alexeev-Tsimerman – 2009)

The equation

i i

i i

k k2 2

i 02 1 2 1i 1 i 1

x ( 1 )x

is k-regular but not (k+1)-regular.

More generally, for an m x n integer matrix A,

let denote a system of homogenous linear equations.

Ax 0

For example, for

1 2 1 0 0

A 0 1 2 1 0 ,

0 0 1 2 1

the corresponding system is:

1 2 3 2 3 4 3 4 5x 2x x 0, x 2x x 0, x 2x x 0,

which corresponds to an arithmetic progression of length 5 (5-AP).

A matrix A is said to satisfy the columns condition (CC)

if it possible to partition the columns of A into blocks1 2 sB ,B ,...,B

so that:

(i) j 1

jc B

c 0;

(ii)j i

jc B

f or i 1, c

can be expressed as a rational linear

combination of the columns of 1 2 i 1B ,B ,...,B .

A matrix A is said to satisfy the columns condition (CC)

if it possible to partition the columns of A into blocks1 2 sB ,B ,...,B

so that:

For example,

2 4 2 1

0 2 2 1

1 3 2 0has CC since

2 4 2 0 2 4 1

0 23

2 0 , 0 2 1

1 3 2 0 1 3 0

12 2

(i) j 1

jc B

c 0;

(ii)j i

jc B

f or i 1, c

can be expressed as a rational linear

combination of the columns of 1 2 i 1B ,B ,...,B .

Theorem (Rado – 1933)The system is regular iff A satifies the columns condition.

Ax 0

Paul Erdös and Richard Rado1913-1996 1906-1989

Rado also had many other conjectures concerning systems

of regular equations, one of which was this.

Call a subset large if for any r, any r-coloring of S always

has monochromatic solutions for every regular system.

NS





NS

Conjecture. For any partition of a large set 1 2 tS S S ... S

into finitely many parts, at least one of the parts is large.

iS





NS

Conjecture. For any partition of a large set 1 2 tS S S ... S

into finitely many parts, at least one of the parts is large.

iS

This was finally proved in 1973 by Deuber.

Walter Deuber – (1943 – 1999)

Even though and are both regular, there is a

fundamental difference between them.

x y z x y 2z

Recall van der Waerden’s theorem:In any r-coloring of there is always a monochromatic


N

Erdös and Turán (1936) ask: Which color class has the k-AP’s?

Even though and are both regular, there is a


x y z x y 2z

Recall van der Waerden’s theorem:In any r-coloring of there is always a monochromatic


N

Erdös and Turán (1936) ask: Which color class has the k-AP’s?They conjectured that the “densest” class should have them.More precisely, they conjectured that if satisfies NS

then S should contain k-AP’s for every k.

| S [n]|limsup 0

n n

Equivalently, define to be the size of the largest subset

of [n] which contains no k-AP.

kr (n)

Conjecture (Erdös -Turán – 1936)

kr (n) o(n).For all k,



kr (n)



It became clear that even was not going to be so simple

to determine because of:

3r (n)

Theorem (Behrend – 1946)

3r (n) n exp( c logn).

For a suitable constant c > 0,



kr (n)



It became clear that even was not going to be so simple

to determine because of:

3r (n)

Theorem (Behrend – 1946)

3r (n) n exp( c logn).


This shows that for every 1 ε3r (n) n ε 0 as n .

The first non-trivial upper bound was given by Roth:

Theorem (Roth – 1954)

n3 log log nr (n) O( ).




This was subsequently improved by Szemerédi and Heath-Brown to

cn

3 (log n)r (n) O( ) with the current record being held by

Bourgain (2008)

2

23

(loglog n)3 (log n)r (n) O n .




This was subsequently improved by Szemerédi and Heath-Brown to

cn

3 (log n)r (n) O( ) with the current record being held by

Bourgain (2008)

2

23

(loglog n)3 (log n)r (n) O n .

There has also been a very recent breakthrough improvement

of Behrend’s lower bound by Michael Elkin.

Behrend (1946)

M. Elkin (2009)

22 2 log n

3 2r (n) cn 14log n

22 2 log n

3 2r (n) cn 14log n

Behrend (1946)

M. Elkin (2009)

22 2 log n

3 2r (n) cn 14log n

22 2 log n

3 2r (n) cn 14log n

Is this close to the “truth” ?

What about ?

kr (n)

What about ?

kr (n)

4r (n) o(n) Szemerédi - 1969

What about ?

kr (n)


kr (n) o(n) Szemerédi - 1974

+ $1000Endre Szemerédi – (1940 - )

What about ?

kr (n)



+ $1000

kr (n) o(n) Furstenberg – 1979 (ergodic theory)

Endre Szemerédi – (1940 - )

What about ?

kr (n)



+ $1000


c4r (n) O(n/ (loglogn) ) Gowers - 1998


What about ?

kr (n)



+ $1000



kk

cr (n) O(n/ (loglogn) ) Gowers - 2001


What about ?

kr (n)



+ $1000



kk

cr (n) O(n/ (loglogn) ) Gowers - 2001

The last result can be used to obtain the best currently

available bound for the van der Waerden function W(k).


A consequence of van der Waerden’s theorem on arithmetic progressions

is that for every n, there is a least number W(n) so that in any

2-coloring of [W(n)], there is always formed a monochromatic n-AP.

A consequence of van der Waerden’s theorem on arithmetic progressions

is that for every n, there is a least number W(n) so that in any

2-coloring of [W(n)], there is always formed a monochromatic n-AP.

n2222For every n, W(n) 2

9

Theorem (Gowers 2001)

$1000

2222

22

2

2

……

…

2

n 2

’s

W(n) <

The best known lower bound is:

nW(n 1) n 2 , (Berlekamp - 1968)

Elwyn Berlekamp - (1940 - )



Elwyn Berlekamp - (1940 - )The only known values are:

nW(n)

2 64353

31785

9 1132



Elwyn Berlekamp - (1940 - )

(Brave) Conjecture.$1000

For all n,2nW(n) 2 .

The only known values are:

nW(n)

2 64353

31785

9 1132

A new proof of van der Waerden for 3-AP’s

N

N

“corner”

G(N) = N by N triangular grid

TheoremTheorem: (RLG + J. Solymosi - 2006)

There is a universal constant c such that

if the points of G(N) are colored with at

most c log log N colors, then there is always

a monochromatic corner (x,y), (x+d,y), (x,y+d).

A new proof of van der Waerden for 3-AP’s

N

N

Sketch of proof:

Suppose G(N) is r-colored.

L1

Let L denote the set of points on the diagonal.

1

N

N

Sketch of proof:


L1

Let be points with the “most popular” color c .

1 1S L1


1

N

N

Sketch of proof:


L1

………

………

………


1 1S L1


1

Let be the points in the product

below L .

1 1 1T S xS

1

N

N

Sketch of proof:


L1

………

………

………


1 1S L1


1


below L .

1 1 1T S xS

1 Thus,

21 2

11

2r

N| S |r

2 2| T | ~ N

N

N

Sketch of proof:


L1

………

………

………


1 1S L1


1


below L .

1 1 1T S xS

1 Thus,

21 2

11

2r

N| S |r

2 2| T | ~ N

Hence, some line L parallel to L

has at least points of T on it.

21

2rN

2 1

1

L2

N

N

Sketch of proof:


L1

………

………

………


1 1S L1


1


below L .

1 1 1T S xS

1 Thus,

21 2

11

2r

N| S |r

2 2| T | ~ N

Hence, some line L parallel to L

has at least points of T on it.

21

2rN

2 1

1

L2

Let be points with

the most popular color c

2 2S L

2

and let be the points in the product below L .

2 2 2T S xS2

62 23

2

N2r

2

18r

| S |

2| T | ~ N

Thus, , etc.

Continue for r+1 steps.Then r 1r 1 r2 1 2 1

1| S | N

2 r

62 23

2

N2r

2

18r

| S |

2| T | ~ N

Thus, , etc.

Continue for r+1 steps.

L i

L j

Lk

a

b

c

d

ef

g

Observe that if abc, bde and cfg are corners

then so is adg.

Then r 1r 1 r2 1 2 1

1| S | N

2 r

62 23

2

N2r

2

18r

| S |

2| T | ~ N

Thus, , etc.


L i

L j

Lk

a

b

c

d

ef

g


then so is adg.

Hence, if then we have a

monochromatic corner.

u vc c , u v,

Then r 1r 1 r2 1 2 1

1| S | N

2 r

62 23

2

N2r

2

18r

| S |

2| T | ~ N

Thus, , etc.


L i

L j

Lk

a

b

c

d

ef

g


then so is adg.

Hence, if then we have a

monochromatic corner.

u vc c , u v,

Then r 1r 1 r2 1 2 1

1| S | N

2 r

This works providedcr2N 2

for a suitable absolute constant c.

This proves the Theorem.

N

Notice that we can project a corner onto a 3-AP

on a copy of . N


on a copy of .

Hence, we have the

Corollary. Corollary. If {1, 2, …, N} is colored with at

most c log log N colors, then a monochromatic

3-AP must be formed.

N

N


on a copy of .

Hence, we have the

Corollary. Corollary. If {1, 2, …, N} is colored with at

most c log log N colors, then a monochromatic

3-AP must be formed.

The best result of this type follows from

a difficult result of Bourgain, which

gives the above conclusion with

23

2

logn

log lognc colors.

N

N

N

N

We can actually continue the preceding argument for (m-1)r+1

steps to get a monochromatic “equilateral binary tree” with m levels.

……

……

……

N

N



……

……

……

We can project these trees onto the real axis to get

monochromatic affine “m-cubes”. These are sets of

the formm

k k kk 1

{a x , 0 or 1}

N

N



……

……

……



the formm

k k kk 1

{a x , 0 or 1}

Similar arguments yield monochromatic

connected crossed diagonals of a

hypercube, etc.

N

N



……

……

……



the formm

k k kk 1

{a x , 0 or 1}

Similar arguments yield monochromatic

connected crossed diagonals of a

hypercube, etc.

The partition regularity of these structures was first established by

Hilbert in 1892 in connection with his work on irreducibility of polynomials.

Combinatorial Combinatorial lineslines

For a finite set denote the set of n-tuples from A.

n1 2 mA {a ,a ,...,a }, let A

A combinatorial line in A is a set of m n-tuples wheren1 2 mX ,X ,...,X

k k k k k(1) (2) (j) (n)X (X ,X ,...,X ,...,X )

either all X (j) are equal, or k k k 1 k m.X (j) a f or

and for each j,

Combinatorial Combinatorial lineslines

For a finite set denote the set of n-tuples from A.

n1 2 mA {a ,a ,...,a }, let A

A combinatorial line in A is a set of m n-tuples wheren1 2 mX ,X ,...,X

k k k k k(1) (2) (j) (n)X (X ,X ,...,X ,...,X )

either all X (j) are equal, or k k k 1 k m.X (j) a f or

and for each j,

Theorem (Hales – Jewett – 1963)For every r, there is an N = N(A, r) such that if is r-colored

then there must always be formed a monochromatic combinatorial line.

NA

Hales – Jewett implies van der Waerden by taking A = {1, 2,…, m} and

identifying the n-tuple with the integer1 2 n(x ,x ,...,x ) 1 2 nx x ... x .



...a... ...b... ..1 1 .c...

...a... ...b... ..2 2 .c...

...a... ...b... ..m m .c...

… …S 2t

S mt

… …

S 1t



Of course, Szemerédi’s theorem is a (stronger) density form of

van der Waerden’s theorem since in any r-coloring of , some color

class must have density at least 1/r.

...a... ...b... ..1 1 .c...

...a... ...b... ..2 2 .c...

...a... ...b... ..m m .c...

… …S 2t

S mt

… …

S 1t

van der Waerden

SzemerédiHales - Jewett

We have this picture:

van der Waerden


density Hales – Jewett theorem ?


Is there a

van der Waerden


density Hales – Jewett


Yes !

TheoremTheorem – Density Hales–Jewett (Furstenberg - Katznelson - 1991)For every ε > 0, there is an N = N(ε) such that if

N NR A , with| R| | A| ,

then R must contain a combinatorial line.




Big Problem:Big Problem: Obtain reasonable (for example, any) bounds on N(ε).

(Furstenberg-Katznelson proof uses ergodic theory tools).





Warm-up Big Problem: Try this for m = 3.






Warm-up Big Problem: Try this for m = 3.

Very recent breakthrough !

Tim Gowers and the Polymath Polymath project.


It now appears that the Polymath project has succeeded

in finding 3 combinatorial proofs of the density Hales – Jewett

theorem, and consequently, for the first time we have

explicit (but still quite large!) bounds on N(m,ε).

It now appears that the Polymath project has succeeded

in finding 3 combinatorial proofs of the density Hales – Jewett

theorem, and consequently, for the first time we have

explicit (but still quite large!) bounds on N(m,ε).Is this the wave of the future ??

Back to linear equationsEven though and are both regular, there is a


x y z x y 2z

x y 2z always has solutions in any set of positive densitywhereas may not. x y z



x y z x y 2z


For example, has no solutions in either x y z

n2

{x [n] : x odd} or {x [n] : x }.



x y z x y 2z


In fact, a system of homogenous linear equations given by

Ax 0 is density regular iff the all 1’s vector is a solution.

For example, has no solutions in either x y z

n2

{x [n] : x odd} or {x [n] : x }.

If [n] is r-colored, how manyhow many monochromatic 3-AP’s must be formed?

It follows from general results of Frankl, RLG, and Rödl that for any regular

system of homogeneous linear equations that a positive fraction of all the

solutions must be monochromatic.




solutions must be monochromatic.For the equation let f (n) denote the minimum

possible number of monochromatic solutions formed over

all possible 2-colorings of [n].

1 2 tE(x ,x ,...,x ), E







1 2 tE(x ,x ,...,x ), E

For example, for E: x + y = z, it is known that

2

En22(1 o(1))f (n)

Robertson/Zeilberger and Schoen - 1998







1 2 tE(x ,x ,...,x ), E

For example, for E: x + y = z, it is known that

2

En22(1 o(1))f (n)

Robertson/Zeilberger and Schoen - 1998

What is the situation for 3–AP’s, i.e., the equation x + y = 2z ??


[n] contains 3-AP’s. 2n

4(1 o(1))



4(1 o(1))

Hence, if [n] is randomlyrandomly 2-colored then there will be

essentially monochromatic 3-AP’s. 2n

16(1 o(1))



4(1 o(1))



16(1 o(1))

Is this best possible ?



4(1 o(1))



16(1 o(1))

Is this best possible ? No!



4(1 o(1))



16(1 o(1))


It has recently been shown that there is a 2-coloring of [n] which has only 2117

2192(1 o(1)) n monochromatic 3-AP’s.



4(1 o(1))



16(1 o(1))




117 12192 18.735...



4(1 o(1))



16(1 o(1))




117 12192 18.735...

28 6 28 37 59 2828 63759116 116



4(1 o(1))



16(1 o(1))




117 12192 18.735...

In the other direction, it is known that there must

always be at least monochromatic 3-AP’s.2167532768(1 o(1)) n

28 6 28 37 59 2828 63759116 116

1675 11732768 2192

0.05111... 0.05337...

We believe the upper bound is the “truth”, i.e., any

2-coloring of [n] must have at least 21172192

(1 o(1)) n

monochromatic 3-AP’s.



(1 o(1)) n


Sketch of our approach (S. Butler, K. Costello, RLG)

Let c :[n] {0,1} be a 2-coloring of [n] and define

jS {x :c(x) j, 1 x n}, j 0, 1,



(1 o(1)) n




jS {x :c(x) j, 1 x n}, j 0, 1,

0 1V(S ,S ) # of monochromatic 3-AP’s formed



(1 o(1)) n




jS {x :c(x) j, 1 x n}, j 0, 1,


j

2πisxj

s S

f (x) e .



(1 o(1)) n




jS {x :c(x) j, 1 x n}, j 0, 1,


j

2πisxj

s S

f (x) e .

Thus,

1 2 2

0 1 0 0 1 102V(S ,S ) (f (x)f (2x) f (x)f (2x))dx

2 20 0 1 1f (x)f (2x) f (x)f (2x)

20 1 0 1

0 1 1 0 0 1

0 1 0 1

(f (x) f (x)) (f (2x) f (2x))

(f (x)f (2x) f (x)f (2x))(f (x) f (x))

f (x)f (x)(f (2x) f (2x))

Expanding the integrand, we get

2 20 0 1 1f (x)f (2x) f (x)f (2x)

20 1 0 1

0 1 1 0 0 1

0 1 0 1

(f (x) f (x)) (f (2x) f (2x))

(f (x)f (2x) f (x)f (2x))(f (x) f (x))

f (x)f (x)(f (2x) f (2x))

Expanding the integrand, we get

We can interpret these three terms respectively as:

3

0 1 1 0

0 1

| {(a,b,c) [n] : a b 2c}|

| {(a,b) (S S ) (S S ) :2b a [n]}|

| {(a,b) S S : a b is even}| .

3

0 1 1 0

0 1

| {(a,b,c) [n] : a b 2c}|

| {(a,b) (S S ) (S S ) :2b a [n]}|


The first term is and the third term can be

shown to be bounded above by

2

2n(1 o(1)) ,

2 2nodd even odd even2

(0 0 ) (0 0 )

0 12V(S ,S )

where and denote the number of odd and even numbers,

respectively, in [n] which have color 0.

odd0even0

3

0 1 1 0

0 1

| {(a,b,c) [n] : a b 2c}|

| {(a,b) (S S ) (S S ) :2b a [n]}|


The first term is and the third term can be

shown to be bounded above by

2

2n(1 o(1)) ,

2 2nodd even odd even2

(0 0 ) (0 0 )

If we assume that then this term isodd even0 0 2

8n(1 o(1)) .

0 12V(S ,S )

where and denote the number of odd and even numbers,

respectively, in [n] which have color 0.

odd0even0

Thus, to minimize we should maximize the size of

0 1V(S ,S ),

0 1 1 0T {(a,b) (S S ) (S S ) :2b a [n] }.


0 1V(S ,S ),

0 1 1 0T {(a,b) (S S ) (S S ) :2b a [n] }.

Let us color the points (a,b) in [n] by2

red if 0 1 1 0(a,b) {(S S ) (S S )}

and

blue if 0 0 1 1(a,b) {(S S ) (S S ) }


0 1V(S ,S ),

0 1 1 0T {(a,b) (S S ) (S S ) :2b a [n] }.

Let us color the points (a,b) in [n] by2

red if 0 1 1 0(a,b) {(S S ) (S S )}

and

blue if 0 0 1 1(a,b) {(S S ) (S S ) }

If and consist of blocks then this coloring

is a red – blue “checkerboard” coloring of [n] .0S 1S

2

2[n]

A red/blue “checkerboard”

(a,b)

The requirement in T that means that we can

restrict our considerations to the parallelogram P determined

by the lines

2b a [n]

1 12 2

y x and y (1 x).

P

2[n]0S

1S

2[n ] P

The intersection of the checkerboard with the parallelogram P

“Incremental improvement” algorithms produced the following pictures.

x y 2z random start

x y 2z random start

x y 2z another random start

x y 2z all blue start

Converging to a common block structure !

(0,0) (1,0)

(0,1) (1,1)

The “checkerboard” picture for the approximate block sizes obtained by the incremental algorithm (normalized by dividing by n ).

2

(0,0) (1,0)

(0,1) (1,1)

(1, 1/2)

(0,1/2)

Restricting the checkerboard to the relevant parallelogram P.)

kx

kx………

……

Local optimization of the red/blue blocks

kx

kx………

……

Increasing the value of by kx ε

kx

kx

= lost

………

……

kx

kx

= lost

= gained

………

……

*kx

= lost

= gained

Δ = 0 at local minimum………

……

*kx

(0,0) (1,0)

(0,1) (1,1)

We apply the “local minimum” condition to this picture.

This leads to the following set of linear equations:


The correspond to the interval widths (normalized so that they sum to 1).

iα

The remaining equations arise from forcing equal amounts of red and blue

along each line.


The correspond to the interval widths (normalized so that they sum to 1).

iα

The remaining equations arise from forcing equal amounts of red and blue

along each line. This results in the values etc.,

6 281 2548 548

α ,α ,...,

which gives the current best solution for x + y = 2z, namely there are

2-colorings which have only 21172192(1 o(1)) n monochromatic 3-AP’s.

x 2y 3z all blue start

Another equation

x 2y 3z random start

Two runs of x + y = z

(random starts)

Two runs of x + y = z

(random starts)

411

611

111

Known optimal coloring for x + y = z.

We next show the best currently known patterns for minimizing monochromatic

solutions to certain equations Ax + By = Cz for 2-colorings of [n], where A + B = C.



These were found by first doing a local perturbation of a coloring until it went to

a locally minimal pattern, in the sense that changing no single block gives any

improvement to the overall structure.





improvement to the overall structure. We then took the resulting (approximate)

structure and optimized so that on each line in the pattern intersected equal

amounts of each color.





improvement to the overall structure. We then took the resulting (approximate)

structure and optimized so that on each line in the pattern intersected equal

amounts of each color.

In the following figures we indicate the coefficient of n corresponding to

random ( ), the coefficient of the current best known configuration ( )

and we show what fraction of random we are currently achieving ( ).

We also show the coloring of the current best known configuration.

αγαγ

2

We collect information on all these patterns in the plot below. On the bottom axis is A/C (which will run between 0 and 1, with 1 corresponding to x + y = 2z) and on the vertical axis is (which again runs from 0 to 1).

αγ

With this “normalization” we would expect that the curve formed from all points would be continuous, i.e., a small change in A/C would result in a small change in since the same pattern should be a good approximation for both.

αγ

A/C


αγ


αγ

However, there are some gaps!. The most interesting (as of now) are the ones around 2/5 and 3/5, corresponding to the equation 2x + 3y = 5z (in blue). Something seems to be happening in that region, e.g., we haven’t been able to identify any stable behavior for the equation 7x + 10y = 17z.

A/C


αγ


αγ

However, there are some gaps!. The most interesting (as of now) are the ones around 2/5 and 3/5, corresponding to the equation 2x + 3y = 5z (in blue). Something seems to be happening in that region, e.g., we haven’t been able to identify any stable behavior for the equation 7x + 10y = 17z.

What happens as A/C 0 ? Does curve stay below 0.95 or does it tend to 1 ?

A/C

We have primarily just examined equations of the form

Ax + By = Cz with A + B = C. These are all density regular

equations and, in particular, solutions are invariant under

translation.




translation.

In every case except one, the best colorings we found are composed

of a finite number of anti-symmetrically colored blocks, such as

28 6 28 37 59 2828 63759116 116

for x + y = 2z.




translation.

In every case except one, the best colorings we found are composed

of a finite number of anti-symmetrically colored blocks, such as

28 6 28 37 59 2828 63759116 116

for x + y = 2z.

The one exceptional equation we found is 2x + 3y = 5z.

Here is a picture of what the heuristic algorithm always produced.

2x 3y 5z all blue start

This picture also consists of “blocks”, but this time the blocks

are formed by alternating blue/red (and red/blue) strings.Performing a similar local optimizing procedure on these blocks

gives a value of = 0.923755…, which is unexpectedly low.

αγ




αγ

Presumably, this is the optimal coloring for 2x + 3y = 5z although

we can’t prove it.

blue odd, red even

red odd, blue even

………




αγ

Presumably, this is the optimal coloring for 2x + 3y = 5z although

we can’t prove it. What is different about this equation ?

blue odd, red even

red odd, blue even

………

For the equation E: ax + by = cz with a + b = c, random 2-colorings

are never optimal.

In fact, it can be shown that the optimal 2-coloring of [n]

has at most times as many monochromatic solutions

to E as a random one does.

21

2bc1

For example, for 3-AP’s, (i.e., E: x + y = 2z), this ratio is 78

0.875

while the truth appears to be 0.854… .

Are random colorings ever optimal ?

Yes !Yes !


Yes !Yes !

Let E denote the equation x + y = z + w.


Yes !Yes !


If we count permutations of solutions as different then

it easy to compute that E has solutions in [n]. 3 22n O(n )

3


Yes !Yes !




3

Thus, in a random 2-coloring of [n] we would expect

to find monochromatic solutions.

33 2 22n1 O(n ) n12

O(n )8 3


Yes !Yes !




3

Thus, in a random 2-coloring of [n] we would expect

to find monochromatic solutions.

33 2 22n1 O(n ) n12

O(n )8 3

In fact, anyany 2-coloring of [n] must produce

at least this many monochromatic solutions.

Sketch of proof:

Let {red, blue} be a 2-coloring of [n] = {1,2,…,n}.

c :[n]

Think of where N > 2n.N[n] Z

Define by f(x) = 1 if x is red, and 0 otherwisef :[n] {0,1}

and by g(x) = 1 if x is blue, and 0 otherwise.

g:[n] {0,1}

Sketch of proof:

Let {red, blue} be a 2-coloring of [n] = {1,2,…,n}.

c :[n]

Think of where N > 2n.N[n] Z

Define by f(x) = 1 if x is red, and 0 otherwisef :[n] {0,1}

and by g(x) = 1 if x is blue, and 0 otherwise.

g:[n] {0,1}

Thus, [n]f g 1

The number of monochromatic solutions to x + y = z + w

is given by the expression

N

2 2

x

C: ((ff )(x) (g g)(x) )

Z

where denotes convolution.

With2πijk 2πi(N 1)k

N N

N1 0 k N,ke (1,...,e ,...,e ),

as an orthonormal basis for our Fourier transform , we have

F G F G N ||F|| ||F|| andN

2πixξN1

Nx

F(ξ) F(x)e

Z


N N

N1 0 k N,ke (1,...,e ,...,e ),


F G F G N ||F|| ||F|| and

Thus,

2 2|| ff || || g g|| N

2 2

x

C: ((ff )(x) (g g)(x) )

Z

N

2πixξN1

Nx

F(ξ) F(x)e

Z

2 2|| ff || || g g|| 2 2 2 2( || f || || f || || g || || g|| ) N


N N

N1 0 k N,ke (1,...,e ,...,e ),


F G F G N ||F|| ||F|| and

Thus,

2 2|| ff || || g g|| N

2 2

x

C: ((ff )(x) (g g)(x) )

Z

2 2|| ff || || g g|| 2 2 2 2( || f || || f || || g || || g|| ) N

4 4( || f || || g || ) N

N

4 4

ξ

(| f (ξ)| | g(ξ)| ) N

Z

N

2πixξN1

Nx

F(ξ) F(x)e

Z

Continuing,

N

4 4

x

(| f (x)| | g(x)| ) N

Z

C

N

4

x

18

(| f (x)| | g(x)|) N

Z

(by C-S twice)

Continuing,

N

4 4

x

(| f (x)| | g(x)| ) N

Z

C

N

4

x

18

(| f (x)| | g(x)|) N

Z

(by C-S twice)

N

4

x

18

(| f (x) g(x)|) N

Z

N

4

x

18

(| (f g)(x)|) N

Z

Continuing,

N

4 4

x

(| f (x)| | g(x)| ) N

Z

C

N

4

x

18

(| f (x)| | g(x)|) N

Z

(by C-S twice)

N

4

x

18

(| f (x) g(x)|) N

Z

N

4

x

18

(| (f g)(x)|) N

Z

218

|| (f g) (f g)||

2[n] [n]

18

||1 1 ||

3 32 22n n18 3 12

O(n ) O(n )

and the claim is proved.

In fact, none of the colorings we have produced for any ofthe equations ax + by = cz considered have been proved to be optimal (except for x + y = z).

What happens for other equations ?




Computational evidence strongly suggests that for the equation

E: x + ay = z, with \{1}, we havea N\

2

2E

1 o(1)2a(a 2a 3)

f (n) n .



Computational evidence strongly suggests that for the equation

E: x + ay = z, with \{1}, we havea N\

2

2E

1 o(1)2a(a 2a 3)

f (n) n .

What are the corresponding results for 4-AP’s4-AP’s (or k-AP’s) ??

A good coloring for avoiding monochromatic 4-AP’s

In this case, the number of monochromatic 4-AP’s

is at most where

24c n O(n)



is at most where

24c n O(n)

c = 4



is at most where

24c n O(n)

c =

Is this the “truth”?

4

Sensational result (Ben Green and Terry Tao – 2003):

There are arbitrarily long arithmetic progressions of primes.

Finding k-AP’s in sparse sets.

Sensational result (Ben Green and Terry Tao – 2003):

There are arbitrarily long arithmetic progressions of primes.

Ben Green – Cambridge Univ.

Terry Tao - UCLA

Finding k-AP’s in sparse sets.

In fact, they show that there exists a

k-term arithmetic progression of primes

222

222

2100 k

which are all less than

In fact, they show that there exists a

k-term arithmetic progression of primes

222

222

2100 k

(This is probably not best possible !)

which are all less than

An even tougher challenge.

Show that there exist arbitrarily long AP’s of consecutive primes.

The current record is 10, given be the following numbers:

k 0, 1,...,9,P 210 k,




k 0, 1,...,9,P 210 k,

where

P = 100 99697 24697 14247 63778 66555 87969 84032 95093 24689 19004 18036 03417 75890 43417 03348 88215 90672

29719(found by M. Toplic in 1998)




k 0, 1,...,9,P 210 k,

where

P = 100 99697 24697 14247 63778 66555 87969 84032 95093 24689 19004 18036 03417 75890 43417 03348 88215 90672

29719(found by M. Toplic in 1998)

Will we ever find 11 ??

Schur’s theorem (finite version):

For all r, there is a least Sc(r) such that any r-coloring

of [Sc(r)] contains a monochromatic solution of x y z.

Some open Some open problemsproblems




It is known that 1 2rc π Sc(r) c r !






(Old) Problem. Is

1/ rSc(r)lim ?r r






(Old) Problem. Is

1/ rSc(r)lim ?r r

$100


Suppose we consider the “off-diagonal” van der Waerden number W(k, l)

defined as the least integer such that if [W(k, l)] is 2-colored red and blue,

then there either is a red k-AP or a blue l-AP.

k 3 4 5 6 7 8 9 10 11 12 13

W(k,3) 9 18 22 32 46 58 77 97 114 135 160




k 3 4 5 6 7 8 9 10 11 12 13

W(k,3) 9 18 22 32 46 58 77 97 114 135 160

k

W(k,3)




k 3 4 5 6 7 8 9 10 11 12 13

W(k,3) 9 18 22 32 46 58 77 97 114 135 160

k

W(k,3)




Any conjectures ?Any conjectures ?

k 3 4 5 6 7 8 9 10 11 12 13

W(k,3) 9 18 22 32 46 58 77 97 114 135 160

k

W(k,3) Any conjectures ?Any conjectures ?

Only known that c' logkck W(k,3) k




We can prove the following related result.Consider the modified off-diagonal van der Waerden number W(k, l)


then there is either a red consecutive block of length k or a blue l-AP.

We can prove the following related result.

Theorem (RLG – 2006)


Consider the modified off-diagonal van der Waerden number W(k, l)



clogkW(k,3) k .







clogkW(k,3) k .

Unfortunately, the best upper bound we have is32c'kW(k,3) k .







clogkW(k,3) k .

Which bound is closer to the truth?

Unfortunately, the best upper bound we have is32c'kW(k,3) k .

Restricted van der Waerden

It was shown by Spencer (1975) that the following restricted

version of van der Waerden’s theorem holds.

For every k, there exists a finite set S(k) with the properties:

(i) any 2-coloring always produces a monochromatic k-AP;

(ii) S(k) does not contain a (k + 1)-AP.







We have no idea of the sizes s(k) = |S(k)| of the smallest

sets S(k) which satisfy the above conditions.







We have no idea of the sizes s(k) = |S(k)| of the smallest

sets S(k) which satisfy the above conditions.

For example, is it true that s(k)W(k)

as k ?

Recall that an equation is r-regular if for any partition

of into r color classes, there is a monochromatic solution to this equation.

1 2 n) 0E(x ,x ,...,x

Conjecture: (Rado – 1933)

For each n, there is an M = M(n) so that if the linear

homogeneous equation is M-regular,

then in fact the equation is regular.

1 2 n) 0E(x ,x ,...,x



1 2 n) 0E(x ,x ,...,x





1 2 n) 0E(x ,x ,...,x

The conjecture holds for n = 1 and n = 2. (Rado – 1933)



1 2 n) 0E(x ,x ,...,x





1 2 n) 0E(x ,x ,...,x


The conjecture also holds for n = 3 with M(3) = 24. (Fox and Kleitman – 2005)



1 2 n) 0E(x ,x ,...,x





1 2 n) 0E(x ,x ,...,x


The conjecture also holds for n = 3 with M(3) = 24. (Fox and Kleitman – 2005)

ChallengeChallenge: Prove that M(n) always exists, and determine (or estimate) its value.

Let W*(k) denote the least integer such that there is a set S(k) of integers

of size W*(k) such that any 2-coloring always produces a monochromatic k-AP.

Thus, W * (k) W(k).



Thus, W * (k) W(k).

It is known that W*(3) = 9 = W(3) but that 27W * (4) 35 W(4).

The characteristic function of one such S(4) is:

1 0 0 1 0 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 1 0 0 1



Thus, W * (k) W(k).


Does W(k) W * (k) as k ?


1 0 0 1 0 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 1 0 0 1

Is this the correct value of W*(4)?



Thus, W * (k) W(k).




1 0 0 1 0 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 1 0 0 1

Is boundedW(k)W*(k)

as k ?




Thus, W * (k) W(k).




1 0 0 1 0 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 1 0 0 1

Is boundedW(k)W*(k)

as k ?

What happens with r colors?


What about the regularity of nonlinear equations ?

For example, it is known that the equation

k1 2

1 2 k

aa a

x x x0...

is regular if and only if the corresponding equation

1 1 2 2 k kax a x ... a x 0

is regular.



k1 2

1 2 k

aa a

x x x0...


1 1 2 2 k kax a x ... a x 0

is regular.

Beautiful old Beautiful old questionquestion

Is regular ? 2 2 2x y z



k1 2

1 2 k

aa a

x x x0...


1 1 2 2 k kax a x ... a x 0

is regular.

Beautiful old Beautiful old questionquestion

Is regular ? 2 2 2x y z$100

Consider the linear equation L:

x + y + z = 4w

It is easy to 4-color so that L has no monochromatic solution

(for example, define ). a(5 (5b c)) c

An unexpected twistAn unexpected twist


x + y + z = 4w



Must every 4-coloring of produce a monochromatic solution of L ?

R



x + y + z = 4w




R

Well, it depends !



x + y + z = 4w




R

Well, it depends ! The answer is no in ZFC.



x + y + z = 4w




R


(LM is the axiom that every set of reals is Lebesgue measurable)

However, the answer is yes in ZF + LM.

Radoičič, Fox, Alexeev, RLG - 2007



x + y + z = 4w




R



A theorem of Solovay asserts that ZFC and ZF + LM are equally consistent.





x + y + z = 4w




R



A theorem of Solovay asserts that ZFC and ZF + LM are equally consistent.



For which (sets of) equations do we have this distinction ?


A classic conjecture of Erdös

If then X contains k-AP’s for every k.N 1x

x X

X with

$3000



x X

X with

$3000

Warm-up problem. Prove this holds for k = 3.

If then X contains 4 vertices of a square.



x X

X with

$3000


(Two-dimensional analogue - RLG).

N N 2 21

x y(x,y) X

X with

If then X contains 4 vertices of a square.



x X

X with

$3000


(Two-dimensional analogue - RLG).

N N 2 21

x y(x,y) X

X with

$1000

Paul is surprised when finally gets to read The Book

“Very creative. Very imaginative. Logic……that’s what’s missing.”

Front Row: Mrs. van Lint, Robert Rankin, Mrs. Rankin, Gina Birch, J. H. van Lint, L. J. Mordell, Mrs. Mordell, Paul Erdös (with Cheryl (" Che' ") Graham - Ron Graham's daughter), Mrs. Andrews.

Second Row: Mrs. J.B. Muskat (hidden), Mrs. T.R. Parkin (hidden), George Andrews, Stefan Burr, Mrs. S. Burr, Mrs. Pat Nickson, Francis Coghlan, Gérard Ligozat, John Tate, Bryan Birch, A.O.L. Atkin, Helmut Hasse, Marshall Hall Jnr.

Third Row: Pierre Barrucand, Juliette Barrucand, Yvette Amice, Harvey Cohn, Bernice Cohn, Bob Churchhouse, Julia Churchhouse, John Hunter, Richard Guy, Mrs. Kuroda, Michael Keates.

Fourth Row: Nelson Stephens, Ian Angell, Andy Mc Gettrick, Mrs. Straus, Daniel Straus, E. G. Straus, Paul Bateman, Fenella Rankin, Mohinder Cheema, K. Kloos, S. Kuroda, Terence Jackson.

Fifth Row: Ian Anderson, Marie-Nicole Montouchet, M. G. Gras, A. S. Fraenkel, Olga Taussky-Todd, Synolda Butler, ø. Rødseth, Susan Cohn, Debra Baumert, Anthony Cohn, Jean Pierre Serre, John Leech.

Back Row: George Greaves, Lyliane Bouvier, Jean Fresnel, Marvin Wunderlich, Horst Zimmer, Christiane Martinet, Jacques Martinet, John Todd, Albrecht Pfister, S. Mossige, J. Hjelle, John Brillhart.

http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Rankin.html

http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Mordell.html

http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Mordell.html

http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Erdos.html

http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Erdos.html

http://www.math.psu.edu/andrews/

http://www1.cs.columbia.edu/~sanders/graphtheory/people/Burr.SA.html

http://www.maths.man.ac.uk/DeptWeb/Homepages/fbc/

http://www.maths.man.ac.uk/DeptWeb/Homepages/fbc/

http://www2.math.uic.edu/~jeremy/tate.html

http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Hasse.html

http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Hasse.html

http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Hall_Marshall.html

http://www.genealogy.math.ndsu.nodak.edu/html/id.phtml?id=1295

http://www.math.ucalgary.ca/general/directory.php3?personid=36&capacityid=22

http://www.lse.ac.uk/people/[email protected]/

http://www.math.uiuc.edu/ResearchAreas/numbertheory/facultyresearch.html

http://www-users.york.ac.uk/~thj1/

http://www.wisdom.weizmann.ac.il/~fraenkel/

http://www.wisdom.weizmann.ac.il/~fraenkel/

http://www.awm-math.org/olgacelebration.html



Front Row: Mrs. Andrews, Mrs. H. Fredricksen, Mrs. K. Kloos, Mrs. Cassels, Jane Pitman, Mrs. J. Merriman, Felice Bateman, Sally Bateman, Harriet Cantor.

Second Row: E. S. Selmer, J. C. P. Miller, Ron Graham , John Selfridge, John Merriman, H. Tverberg, David Cantor, Andrzej Schinzel, Helen Alderson, J. W. S. Cassels.

Third Row: Michael Keates, Peter Pleasants, Elwyn Berlekamp, Emma Lehmer, D. H. Lehmer, Joseph Muskat, C.E. Fröberg, Robert Spira.

Fourth Row: H.G. Apsimon, Peter Swinnerton-Dyer, Albrecht Fröhlich, A. M. Macbeath, O. Hermann, J.C. Herz, Georges Poitou, John Conway, John Chalk.

Fifth Row: V. Felsch, J. Larmouth, O. Kolberg, Bob Laxton, Peter Weinberger, Roger Cook, Jonathan Hall, T.R. Parkin, J.H.E. Cohn, John Cosgrave.

Back Row: H. J. Godwin, Fred Lunnon, T. Kløve, H. Fredricksen, Jack Howlett, A. L. Dulmage, Nathan Mendelsohn, Leonard Baumert, Hans Riesel, Harold Stark

Frank Plumpton Ramsey(1903-1930)

Arthur Michael Ramsey

100th Archbishop of Canterbury

(1904-1988)

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