# A 64 INTEGERS 13 ( 2013 ) ON THE HIGHER-DIMENSIONAL GENERALIZATION OF A PROBLEM OF ROTH

Long ago Roth conjectured that for any k -coloring of the positive integers the equation x + x0 = n, x 6= x0 has a monochromatic solution in (x, x0) for more than cM integers n up to M (where c is an absolute constant independent of k). Later Erdős, Sárközy and T. Sós proved this conjecture with 2 " in place of c. In this paper we will prove a higher-dimensional generalization of this theorem by using a higher-dimensional extension of the well known Hilbert cube-lemma. We will also give bounds for the number of monochromatic solutions in higher dimension.

1. Introduction K. F. Roth conjectured (see [2] and [6]) that for an arbitrary k-coloring of the positive integers there are more than cM integers n  M such that the equation x + x 0 = n, x 6 = x 0 , has a monochromatic solution in (x, x 0 ).In [1] Erdős, Sárközy and T. Sós proved this conjecture in the following form: Theorem 1.For every k 2 there exists a positive integer M 0 (k) such that for any M M 0 (k) and an arbitrary k-coloring of the set N, the number of positive integers n  M for which there is a monochromatic solution of the equation x + x 0 = n, x 6 = x 0 , is greater than The proof of this theorem was based on the density version of Hilbert's cube lemma (for the original coloring version see [7] ).Szemerédi proved that if we consider a sequence of positive integers of positive density, then the sequence must contain a so-called Hilbert d-cube or a ne d-cube, i.e., for every d a set of the form u + P d i=1 " i v i , where " i =0 or 1 for every i.In [7] Hilbert used the coloring version of this lemma in studying irreducibility of polynomials with integer coe cients.Later, Szemerédi gave the density version of this lemma (see [3] and [10]).This density version of Hilbert's cube lemma is generally called Szemerédi's cube lemma.Erdős, Sárközy and T. Sós used the following quantitative version of this lemma: Lemma 1. (Szemerédi's cube lemma): If H is a subset of (1,M) for M large enough and H has at least 3M 1 2 d elements, then H contains a Hilbert-cube.
Our first goal is to give a higher-dimensional generalization of Lemma 1.With this generalization we will prove the following result: Theorem 2. For fixed positive integers r, s, k, there is a positive integer m 0 with the following property: for any positive integer m > m 0 and for any k-coloring of the elements of the set (1, m) r there are at least The special case r = 1 and s = 2 in Theorem 2 gives the result of Theorem 1.After the proof of Theorem 2 we will study the number of solutions of the equation r , where x and x0 are monochromatic.We will get the following result: Theorem 3.For every positive real number ↵ and with the property ↵ r + r  1 2 2r+1 k there is a positive integer m ↵ , such that for every m > m ↵ and for every k-coloring of N r the number of elements in (1, m) r having representations as a sum of two monochromatic distinct vectors in more than r 2 m r ways is more than ↵ r m r .
One can observe that the result of Theorem 2 is asymptotically independent of the number of the colors, if we fix r and s.In Theorem 3 we want to search for an arbitrary k-coloring a "large number" of vectors with a "large number of representations".Here we will see that these "large numbers" already depend on the number of the colors for fixed r and s.First we study only the case of two summands, later we show a way of studying the case of more summands.

The Generalization of Hilbert's Cube Lemma
Similarly to the original definition one can interprete d-cubes in the set of rdimensional vectors.
r and the set H has at least Proof.Our proof is a generalization of the proof given in [1].We will define sets We construct We give an estimate for L. We can majorize f (H j , h) by Thus we have L According to our assumption we have (for m large enough) the estimate So we have ) .This means that the vector h can play the role of ṽj+1 and we are able to define set H j+1 , too.Thus indeed we can define sets H 0 , H 1 , • • • , H d and r-dimensional vectors ṽ1 , ṽ2 , • • • , ṽd recursively.

The Proof of the Generalization of Roth's Problem
Here we give a proof of Theorem 2.
Proof.Assume to the contrary that there is an appropriate k-coloring for infinitely many positive integers m, such that the number of vectors x with the given property in (1, m) r is less than . We will get a contradiction via Lemma 2. Let S be the subset of (1, m) r , in which all the r coordinates of the elements are divisible by s.Let S 0 denote the set of those elements of S, which do not have a representation in the given form.According to our assumption we have 3 • 2 r 1 m r 1 2 ks+k 1 < |S 0 |.Now we can apply Lemma 2. In S 0 one can find an a ne (ks k+1)-cube, so that there are r-dimensional vectors ũ, ṽ1 , ṽ2 , • • • , ṽks k+1 in ( (m 1), (m 1))) r , such that all the sums ũ + P ks k+1 i=1 " i ṽi are in S 0 ,where " i =0 or 1 for every i.By the pigeonhole principle there are s vectors in the set

On the Number of Representations as the Sum of Two Monochromatic Distinct Vectors
In this section we study the number of representations, if the number of summands is s = 2. Our goal is to prove Theorem 3. The key will be the following lemma: Lemma 3. If for positive real numbers ↵ and there exist infinitely many positive integers m, such that there is a k-coloring of N r , for which at most ↵ r m r elements of (1, m) r have representations as a sum of two monochromatic distinct vectors of N r in more than , because one can order the vectors of this set (except at most one vector) into disjoint pairs such that each pair consists of two distinct vectors with sum x.For all vectors x of the set (1, m) r let g(x) denote the number of representations of x as the sum of two distinct monochromatic vectors of (1, m) r .Let m i be the number of vectors with the i-th color in 1, . We give an upper estimate for the sum 2 ways as the sum of two monochromatic distinct vectors, then Hence we have the following upper estimate: Using the Cauchy-Schwarz-inequality we get By dividing by m 2r we get If m tends to infinity, then the left-hand side is asymptotically 1 2 2r+1 k and the righthand side is asymptotically 1 2 r + 1 2 ↵ r .Clearly we have 1 2 r + 1 2 ↵ r < ↵ r + r , and hence ↵ r + r > 1 2 2r+1 k .
With Lemma 3 we can easily prove Theorem 3. The proof can be done in the following way: Proof.Assume, for a contradiction, that there are positive numbers ↵ and such that ↵ r + r  1 2 2r+1 k and for infinitely many positive integers m there is a k-coloring of N r such that the number of elements in (1, m) r having representations as a sum of two monochromatic distinct vectors in more than r 2 m r ways is at most ↵ r m r .This contradicts Lemma 3.
Remark.Let k 0 be the maximal odd integer such that k k 0 .We color the elements of N r with at most k colors in the following way: if the sum of the coordinates is 0 or 1 modulo k 0 , then we color this vector by the first color; the other vectors are colored by the other k 0 1 colors according to the other residue classes.In this case there are at most 3 r , for which the number of representations as a sum of two distinct monochromatic vectors is at least 2 • ⇥ m k 0 ⇤ m r 1 (to see this, one should only observe the vectors having the sum of coordinates equal to 0, 1 or 2 modulo k 0 ).Thus Theorem 2 cannot be improved significantly.

Further Remarks
We studied in the last section only the case s = 2 (i.e., the number of addends is two).We can do calculations in a similar way using the further asymptotic formulae (see [9]).The method we use will not be combinatorially new, one needs only the same technique, but with a little more work.Let us denote by p (n, s) the number of partititons of n into exactly s not necessarily distinct parts and q (n, s) the number of partititons of n into exactly s distinct parts.We formulate the relevant statements of [9] in Lemma 4.
m) r T (x), where T (x) is the number of ways vector x can be written as a sum of s monochromatic distinct vectors.A similar argument to that of Section 4 shows that we have to majorize H (x 1 , x 2 , • • • , x r ).
It is easy to verify that q (x 1 , s) By Lemma 4 we get the conclusion that H (x 1 , x 2 , • • • , x r ) is asymptotically equal to if s is fixed and x i 1.With further calculation one can achieve analogous results as formulated in Theorem 3.

r 2 m
r ways, then ↵ r + r > 1 2 2r+1 k .Proof.Let be m a positive integer with the given property.Let a be the minimal and b the maximal positive integer such that a m ↵ and b m  .In this case at most a r elements of (1, m) r have representations as a sum of two monochromatic distinct vectors in more than b r 2 ways.Let f (x) denote the number of representations of x = (x 1 , x 2 , • • • , x r , ) as the sum of two distinct vectors in (1, x 1 vectors ṽ1 , ṽ2 , • • • , ṽj have been defined.Let F be the set of the vectors in ( (m 1), (m 1)) r , whose first nonzero coordinate is positive.Denote by f (H j , h) the number of solutions of the equation b b0 = h, where b, b0 2 H j and h 2 F .
Let L be the maximum value of the numbers f (H j , h), where h 2 F and h 6 (1) where n >> 1 and s = O(1). g only a sketch of how our methods generally work.LetH (x 1 , x 2 , • • • , x r ) denote the number of representations of the vector x = (x 1 , x 2 , • • • , x r ) as a sum of s distinct vectors in (1, x 1 )⇥(1, x 2 )⇥• • •⇥(1, x r ).Let m i be the number of vectors with the i-th color in 1,