RAMSEY TYPE RESULTS ON THE SOLVABILITY OF CERTAIN EQUATION IN ZM

Csikvári, Gyarmati and Sárközy asked whether there exist Ramsey type theorems for the equations a + b = cd and ab + 1 = cd in Zm for large enough m. In this paper it is proved that for any r-colouring of Zm the more general equation a1 + · · · + an = cd has a nontrivial monochromatic solution. Furthermore, an example is presented which shows that the corresponding statement does not hold for the equation ab + 1 = cd. We reformulate this problem with an additional natural condition, and give a partial positive answer.


Introduction
Sárközy [10], [11] proved that if A, B, C, D are "large enough" subsets of Z p , then the equations and can be solved with a ∈ A, b ∈ B, c ∈ C, d ∈ D. Gyarmati and Sárközy [5] generalized these results on the solvability of ( 1) and ( 2) to finite fields.Moreover, there are several papers written on the solvability of equations similar to (1) and ( 2) over a finite field, especially over Z p .(See for example, [3], [4].)It is natural to consider the solvability of these equations in Z m , as well ( [8]).However, in [1] and [5] the authors note that for composite m no density-type theorem can be proved for equations (1) and (2) in Z m , which shows that Z p and Z m behave differently.Furthermore, it is asked whether there exist Ramsey type results: Is it true that for every r-coloring of Z m equation (1) (or ( 2)) has a monochromatic solution, if r, the number of colors, is fixed and m > N (r)?
Problem 1. Are there Ramsey type results on the solvability of (1), resp.(2), in Z m ?
Hindman answered the analogue of this question over N positively ([6]).He showed that for every r-coloring of N the equation a In this paper we consider Problem 1 in Z m .First note that in the case of equation ( 1) trivial monochromatic solutions like 0 + 0 = 0 • 0 or 2 + 2 = 2 • 2 exist, naturally these have to be excluded.This kind of solution, where a = b = c = d is called trivial.In Section 2 we prove that a nontrivial monochromatic solution of (1) always exists.On the other hand in Section 3 a counterexample is presented in the case of equation ( 2), namely we show a coloring of Z m for infinitely many m such that (2) does not have a monochromatic solution.Therefore, instead of m > N (r) the condition p(m) > N (r) (where p(m) denotes the smallest prime divisor of m) has to be assumed, otherwise no Ramsey type result exists.Finally, we show that the answer is affirmative to this modified question in the special case when m is a squarefree number satisfying r To avoid confusion, throughout the paper the notion (a) m is going to be used for the modulo m residue class of a ∈ Z if more than one moduli are used.

The Equation
In this section the equation a + b = cd, and more generally, the equation a 1 + • • • + a n = cd will be studied.The case of prime moduli is well-known by the following theorem of Sárközy: In Theorem A the prime p cannot be replaced by an arbitrary m ∈ N.Moreover, there is no density theorem for equation (1) in Z m for arbitrary m, that is, there exists a constant c > 0 such that for infinitely many m there exists a set A ⊆ Z m having at least cm elements such that (1) does not have a solution in A. Now our aim is to prove that while there is no density theorem, a Ramsey type result exists for the equation a + b = cd over Z m .Note that in general there are many trivial solutions.First we have to determine all the trivial solutions, and to do this we have to solve the congruence a 2 ≡ 2a (mod m).Let m = r i=1 p αi i be the canonical form of the number m.By the Chinese Remainder Theorem, it is enough to determine the trivial solutions in Z p α i i .Let us denote the number of solutions of the congruence a 2 ≡ 2a (mod p α ) by s(p α ).The following cases have to be considered: • p > 2: the congruence a 2 ≡ 2a (mod p α ) has 2 solutions, namely a ≡ 0 and a ≡ 2, hence s(p α ) = 2.
• p = 2, α ≥ 3: there are four solutions: By the Chinese Remainder Theorem, the congruence a + b ≡ cd (mod m) has r i=1 s(p αi i ) trivial solutions.Naturally, our goal is to prove that there exists a nontrivial solution of (1), as well.To see this we will show that even the more general equation always has a monochromatic solution such that a 1 , . . ., a n , c, d ∈ Z m are pairwise distinct.These solutions, where a 1 , . . ., a n , c, d ∈ Z m are pairwise distinct, will be called primitive.The proof of this result is based on the following version of Rado's theorem ( [7], Theorem 9.4): c i y i = 0, then for every natural number r there exists some t such that for every r-coloring of the set {1, 2, . . ., t} the equation has a monochromatic solution b 1 , . . ., b v in {1, 2, . . ., t}, where the b i -s are distinct.

INTEGERS: 13 (2013)
For more details on Rado's theorems, see [2], [7] and [9].The following observation is also needed: . If m > N , then m has a prime power divisor greater than T .
Proof.For the sake of contradiction, suppose the contrary.Then each prime divisor of m is at most T , therefore m is the product of at most T prime powers.Since each prime power divisor is at most T , we have that m ≤ T T , which contradicts our assumption.
Theorem 4. For every n, r ∈ N there exists some N = N (n, r) such that for every N < m ∈ N and every r-coloring of Z m , equation (3) has a primitive monochromatic
Take an arbitrary r-coloring of Z m .By applying Lemma 3 with T = C 3 we obtain that if m > N = T T , then m has a prime power divisor greater than T .Now we prove that N = T T satisfies the condition of the theorem.In the proof we distinguish two cases: the prime power divisor guaranteed by Lemma 3 is itself a prime or it is not.
As the first case suppose that p > r 4 (n+2) 4 is a prime divisor of m such that p 2 m.Therefore, p and m/p are coprime, since p m/p.For 1 ≤ i ≤ p define the mod m residue class (x i ) m by the congruences x i ≡ i (mod p) and x i ≡ 0 (mod m/p).Now, we define an r-coloring of Z p depending on the given r-coloring of Z m in the following way: For 1 ≤ i ≤ p let the color of (i) p ∈ Z p be the color of (x i ) m .Note that Z p is colored by r colors, so we can choose (at least) p r elements having the same color.Let us denote the set of these (at least) p r elements by S. Now we partition S ⊆ Z p into n + 2 disjoint sets S 1 , . . ., S n+2 ⊆ S such that the size of any two of them differ by at most 1.Since p ≥ r 4 (n + 2) 4 ≥ 2r(n + 2), each of the sets S i has size at least p r(n+2) ≥ p 2r(n+2) .Now let A, B, C, D ⊆ Z p be defined in the following way:

4
> p 3 , so Theorem A can be applied, which yields that there exist a Let a 1 = a.Therefore, there exist a 1 , . . ., a n , c, d ∈ {1, 2, . . ., p} such that the corresponding mod p residue classes have the same color, and the congruence holds.The elements (x a1 ) m , . . ., (x an ) m , (x c ) m , (x d ) m ∈ Z m have the same color as (a 1 ) p , . . ., (a n ) p , (c) p , (d) p ∈ Z p ; moreover, they are distinct, since (a 1 ) p , . . ., (a n ) p , (c) p , (d) p ∈ Z p are distinct as well.Furthermore, (x a1 ) m , . . ., (x an ) m , (x c ) m , (x d ) m is a solution of (3), since x i ≡ 0 (mod m/p) for every i, and • (x a1 ) p . . .(x an ) p ≡ (x c ) p (x d ) p (mod p), as x i ≡ i (mod p) for every i and a 1 . . .a n ≡ cd (mod p).
Hence, (x a1 ) m , . . ., (x an ) m , (x c ) m , (x d ) m form a primitive monochromatic solution of (3) in Z m .As the second case, assume that for a prime power (but not prime) p t ≥ C 3 we have p t |m, where t ≥ 2 and t is the largest integer such that p t |m.Let t 0 = t/2.As t 0 ≥ t/3, we have p t0 ≥ C. We show that a monochromatic solution of the equation a 1 + • • • + a n ≡ cd (mod m) can be found among the residue classes of the form (y As the next step we define an r-coloring of N depending on the given r-coloring of Z m .Let the color of y ∈ N be the color of (y s theorem implies that there exist distinct integers α 1 , . . ., α n , γ, δ ∈ {1, 2, . . ., C} having the same color and satisfying give a solution of (3), moreover they are distinct, since α 1 , . . ., α n , γ, δ ∈ {1, 2, . . ., C} are distinct and p t0 ≥ C.
Finally, let us examine the case when n = 1.By Rado's theorem for every r ∈ N there exists some M = M (r) such that for every r-coloring of N the equation α = γ + δ has a primitive monochromatic solution in {1, . . ., M }.Suppose that 2 M < m and take an arbitrary r-coloring of Z m .Define a coloring of N in the following way: Let the color of a ∈ N be the color of (2 a ) m in Z m .Rado's theorem yields that there exist three distinct positive integers α, γ, δ ∈ {1, . . ., M } having the same color such that α = γ + δ.
Hence, we showed that if m > N = T T , then (3) has a nontrivial monochromatic solution in Z m .

The Equation ab + 1 = cd
In this section equation (2) will be studied.First, we will show that if m has a small prime divisor, then there is no Ramsey type theorem on the solvability of ab + 1 = cd in Z m in the classical sense: If we fix the number of colors r and m is large enough, then a monochromatic solution need not exist.
Example 5. Let p|m and the color of (x) m ∈ Z m be the mod p residue class containing x.If (a) m , (b) m , (c) m , (d) m ∈ Z m have the same color, then ab ≡ cd (mod p), so ab + 1 = cd in Z m .
In this example we colored Z m by p colors, where p|m, and there is no monochromatic solution of the equation ab + 1 = cd, which shows that the least prime divisor of m, denoted by p(m), has to be greater than the number of colors.To exclude counterexamples of this kind we reformulate the problem in the following way: Problem 6. Are there Ramsey type results on the solvability of ab + 1 = cd in Z m if r, the number of colors is fixed and p(m) is large enough in terms of r?
We give a partial positive answer to this question, namely we show that the answer is affirmative, if m is squarefree and To prove this result the following theorem of Sárközy is needed: Then there exist sets A 1 ⊆ Z m1 , A j (a 1 , . . ., a j−1 ) ⊆ Z mj (for every a 1 ∈ A 1 , a 2 ∈ A 2 (a 1 ), and so on, a j−1 ∈ A j−1 (a 1 , . . ., a j−2 )) satisfying the following conditions: • For every 2 ≤ j ≤ r, for every a 1 ∈ A 1 , for every a 2 ∈ A 2 (a 1 ), for every a 3 ∈ A 3 (a 1 , a 2 ) and so on, for every a j−1 ∈ A j−1 (a 1 , . . ., a j−2 ) the set A j (a 1 , . . ., a j−1 ) has at least α j m j elements.
So A j (a 1 , . . ., a j−1 ) ⊆ Z mj contains at least α j m j possible continuations of the vector (a 1 , . . ., a j−1 ) More precisely, we could add at least α j m j elements a j ∈ Z mj as the j-th coordinate to the vector (a 1 , . . ., a j−1 ) such that after the s-th step we have vectors belonging to We prove this assertion by induction on s.For s = 1 the statement holds trivially.
Thus the size of A 1 is at least α 1 m 1 , as needed.Applying this repeatedly we get that the statement is true for every s > 2 as well.
This implies that in Z m1 × • • • × Z ms at least α 1 m 1 first coordinates can be chosen, the set A 1 contains them.For every a 1 ∈ A 1 , α 2 m 2 second coordinates can be chosen, the set A 2 (a 1 ) contains them.And so on.Finally, α s m s s-th coordinates can be chosen in such a way that all of the elements (a 1 , a 2 , . . ., a s ) obtained in Z m1 × • • • × Z ms lie in A.
As the second step let m = p 1 . . .p s .As Z m = Z p1 × • • • × Z ps by the Chinese Remainder theorem, the modulo m residue class of a can be identified by an ordered s-tuple where the jth coordinate is the mod p j residue of the residue class of a: a ↔ (a 1 , . . ., a s ), where (a) pj = (a j ) pj for every 1 ≤ j ≤ s.Solving the equation ab + 1 = cd in A ⊆ Z m is equivalent to solve the system of equations a i b i + 1 = c i d i in Z pi (where 1 ≤ i ≤ s) in such a way that (a 1 , . . ., a s ), (b 1 , . . ., b s ), (c 1 , . . ., c s ), (d 1 , . . ., d s ) ∈ A. We have just proved that for every α 1 , . . ., α s ≥ 0 satisfying α 1 + • • • + α s ≤ α subsets A j (a 1 , . . ., a j−1 ) ⊆ Z pj can be chosen which satisfy the following conditions: 1 + • • • + a n = b 1 . . .b n has a solution where not only the numbers a 1 , . . ., a n , b 1 , . . ., b n , but also the sums i∈I a i (where ∅ = I ⊆ {1, . . ., n}) and products j∈J b j (where ∅ = J ⊆ {1, . . ., n}) are all distinct (except n i=1 a i and n j=1 b j ), and all of these sums and products have the same color.

Theorem 7 .
]).If p is a prime, A, B, C, D ⊆ Z p , |A||B||C||D| > 100p3  , then the equation ab+ 1 = cd has a solution in Z p satisfying a ∈ A, b ∈ B, c ∈ C, d ∈ D.Now we are ready to solve Problem 6 under a certain condition.Let m = p 1 . . .p s be the product of s different primes.Let A ⊆ Z m and α = |A| m exist a, b, c, d ∈ A satisfying the equation ab + 1 = cd.Proof.The main idea of the proof is to solve the congruence system ab + 1 ≡ cd (mod p i ) (for 1 ≤ i ≤ s) step by step.Our aim is to obtain a solution finally where (a) m , (b) m , (c) m , (d) m lie in A. As the first step we show that the following statement holds: Let m = m 1 m 2 . . .m s , where m 1 , m 2 , . . ., m s are pairwise coprime.Let A ⊆