GGEOMETRIC PROGRESSIONS IN SYNDETIC SETS
BHUWANESH RAO PATIL
Abstract.
In order to investigate multiplicative structures in additively largesets, Beiglb¨ock et al. raised a significant open question as to whether or notevery subset of the natural numbers with bounded gaps (syndetic set) containsarbitrarily long geometric progressions. A result of Erd˝os implies that syndeticsets contain a 2-term geometric progression with integer common ratio, butwe still do not know if they contain such a progression with common ratiobeing perfect square. In this article, we prove that for each k ∈ N , a syndeticset contains 2-term geometric progressions with common ratios of the form n k r and p k r , where p ∈ P (the set of primes), n is a composite number, r ≡ n ), r ≡ p ) and r , r ∈ N . We also show that 2-syndeticsets (sets with bounded gap two) contain infinitely many 2-term geometricprogressions with their respective common ratios being perfect squares. Introduction
Previous research (e.g. [1], [2], [3] and [7]) establishes that sets which are large inany of several multiplicative senses must have substantial additive structure. Forexample, a multiplicatively piecewise syndetic set in N must contain arbitrarilylong arithmetic progressions [1, Theorem 1.3]. However, very little appears to beknown as the existence of multplicative structures in “additively large” sets. Ifwe define “additive largeness” as having positive upper asymptotic density , thenone can observe that there are additively large sets that do not contain three termgeometric progressions. For instance, the set of square-free numbers is additivelylarge, because it has positive upper asymptotic density, but does not contain aconfiguration of the form { x, xr } . One may still ask if the property of containingmultiplicative structures holds for interesting classes of sets that are additivelylarge. This brings us to the following definition: Definition 1.1 (Syndetic set) . If l ∈ N , then A ⊂ N is called l -syndetic set if A has a non-empty intersection with every set of l consecutive natural numbers. Asubset of the natural numbers which is l -syndetic for some l ∈ N , is known as asyndetic set. An infinite arithmetic progression is the simplest example of a syndetic set.Plainly, syndetic sets have positive upper asymptotic density and thus are additivelylarge. Beiglb¨ock et al. [1] recognized the significance of looking for geometricprogressions in syndetic sets in order to study multiplicative structures in additivelylarge sets. They asked the following question.
Question 1. If A is syndetic, do there exist x, y ∈ N such that { x, xy, xy } ⊂ A ? In recent work[5], Daniel Glasscock et al. gave some evidence towards an affir-mative answer to this question by showing that many syndetic sets of dynamicalorigin contain arbitrarily long geometric progressions. The fact that syndetic setscontain 2-term geometric progression with integer common ratio is a consequence For definition of multiplicatively piecewise syndetic set, see [1]. Upper asymptotic density of a set A ⊂ N is defined by d ( A ) := lim sup | A ∩ [1 ,n ] | n a r X i v : . [ m a t h . N T ] A p r BHUWANESH RAO PATIL of the following propositions, namely Proposition 1.2 for dense sets and Proposition1.4 for additively piecewise syndetic sets.
Proposition 1.2 (Erd˝os [4]) . Suppose that A is a subset of natural numbers suchthat lower asymptotic density d ( A ) := lim inf | A ∩ [1 ,n ] | n > . Then A contains a -term geometric progression with integer common ratio. Definition 1.3 (Additively piecewise syndetic set) . Let A ⊂ N . Then A is calledadditively piecewise syndetic set if there exists l ∈ N such that for every n ∈ N ,there is a sequence ( x i ) ni =1 in A satisfying < x i +1 − x i ≤ l ∀ i ∈ [1 , n − . Forexample, every syndetic set is additively piecewise syndetic set. Proposition 1.4. [1, Corollary 2.17] If A is an additively piecewise syndetic set,then there exists a sequence ( y n ) ∞ n =1 in N \{ } such that for each n ∈ N , y n +1 ≡ (cid:81) ni =1 y i ) and (cid:81) ni =1 y i ∈ A . In other words, an additively piecewise syndetic set contains configurations of thetype { x, xy } for some x, y ∈ N . But there exists some additively piecewise syndeticset which does not contain configurations of the type { x, xy, xy } with x, y ∈ N .One can get this type of set inside the collection of thick sets where a thick set is asubset of the natural numbers containing arbitrarily long intervals in N . Using thefact that every thick set is additively piecewise syndetic, Proposition 1.5 guaranteesthe existence of a piecewise syndetic set not containing configurations of the type { x, xy, xy } with x, y ∈ N . Proposition 1.5. [1, Theorem 3.5]
There is a thick subset A of N such that theredo not exist a ∈ A and r ∈ Q \ { } such that ar ∈ A and ar ∈ A . The following weaker version of Question 1 is still an open question.
Question 2. If A ⊂ N is syndetic, do there exist x, y ∈ N such that { x, xy } ⊂ A ? Our first result relates to the above question and gives more information about2-term geometric progressions with integer common ratios in syndetic sets.
Theorem 1.6.
Let k ∈ N and H ∈ { P , N \ P } . Then any syndetic set contains a -term geometric progressions with common ratio n k r for some n ∈ H \ { } and r ∈ N with r ≡ n ) . The next result of this paper confirms an affirmative answer to Question 2 inthe case of 2-syndetic sets.
Theorem 1.7. A -syndetic set contains infinitely many -term geometric progres-sions whose common ratios are perfect squares. The proof of Theorem 1.6 uses Chinese remainder theorem extensively. In Sec-tion 2, we show that generating pairwise prime sets [see Definition 2.1] in a syn-detic set is enough for finding configurations as required in Theorem 1.6. Section3 describes about Triveni triplets [see Definition 3.1] to understand pairwise primesubsets of a syndetic set. Zorn’s lemma guarantees the existence of Triveni tripletsof order one with respect to a given syndetic set and then repeated use of Chineseremainder theorem at various stages generates Triveni triplets of higher order. Us-ing these observations, Section 3.3 explains the proof of Theorem 1.6. Section 4describes the proof of Theorem 1.7 by producing infinitely many explicit geometricprogressions.
Notation.
Let Q , N , P and Z denote, respectively, the set of rational numbers,the set of positive integers, the set of prime numbers and the set of integers. For k ∈ N and H ⊂ N , S k,l,H denotes the collection of l -syndetic sets that contain a EOMETRIC PROGRESSIONS IN SYNDETIC SETS 3 configuration of the form { x, xn k r } , where n ∈ H , r ∈ N satisfying r ≡ n ).Let [ a, b ] := { x ∈ Z : a ≤ x ≤ b } . Let (cid:40) denote the “proper subset of”. For A ⊂ N and x ∈ N , xA = Ax := { nx : n ∈ A } .2. Chinese remainder theorem and syndetic sets
Definition 2.1 (Pairwise prime set) . Let B ⊂ N . Then B is called a pairwiseprime set if gcd( a, b ) = 1 ∀ a, b ∈ B with a (cid:54) = b . Using the Chinese Remainder Theorem[8], the next lemma helps us to prove The-orem 1.6 in the case of syndetic sets containing pairwise prime sets with arbitrarilylarge cardinalities.
Lemma 2.2.
Let h ∈ N , let m , m , ..., m h be pairwise co-prime integers in N and let t , t , ..., t h be arbitrary elements in N ∪ { } . Then ∃ u ∈ N such thatif u t = u + t (cid:16)(cid:81) hi =1 m i (cid:17) with t ∈ N , then there exists ( r i,t ) hi =1 in N satisfying u t + t i = r i,t m i and r i,t ≡ m i ) ∀ i ∈ [1 , h ] .Proof. Consider the congruences x + t i ≡ m i (mod m i ) ∀ i ∈ [1 , h ] . Since { m i : i ∈ [1 , h ] } is a pairwise prime subset in N and { t i : i ∈ [1 , h ] } ⊂ N ∪ { } , the Chinese remainder theorem ensures the existence of u ∈ N such thatif u t = u + tw for w := (cid:81) hi =1 m i and t ∈ N , then u t + t i ≡ m i (mod m i ) ∀ i ∈ [1 , h ] . Therefore there exists a sequence ( r i,t ) hi =1 in N such that u t + t i = r i,t m i and r i,t ≡ m i ) ∀ i ∈ [1 , h ] . (cid:3) From the above lemma, we get the following important corollary which will beused recursively in the proof of Theorem 1.6.
Corollary 2.3.
Let h, k ∈ N , A ⊂ N be an infinite set and B ⊂ N be a pairwiseprime set such that | B | = h and uB ⊂ A for some u ∈ N . If H be a pairwise primesubset of N such that | H | = | B | and gcd( a, b ) = 1 ∀ a ∈ H, b ∈ B , then at least oneof the following is true.(1) A contains a configuration of the type { x, xn k r } for some x, r ∈ N , n ∈ H and r ≡ n ) .(2) There exists z (cid:48) ∈ N such that u [ z t , z t + h − ∩ A = ∅ ∀ t ∈ N where z t = z (cid:48) + t (cid:0)(cid:81) x ∈ B x (cid:1) (cid:0)(cid:81) x ∈ H x k (cid:1) . Proof.
Let B = { x , x , · · · , x h } . Since | B | = | H | , there exists a bijective map f : B → H . Define m i := x i f ( x i ) k ∀ i ∈ [1 , h ] . Since H and B are pairwise prime sets satisfying gcd( a, b ) = 1 ∀ a ∈ H, b ∈ B ,we get that the set { m i : i ∈ [1 , h ] } is a pairwise prime set. So, Lemma 2.2 gives z (cid:48) ∈ N which satisfies the property that if z t = z (cid:48) + t (cid:16)(cid:81) hi =1 m i (cid:17) for t ∈ N , thenthere exists a sequence ( r i,t ) hi =1 in N such that z t + i − r i,t m i and r i,t ≡ m i ) ∀ i ∈ [1 , h ] . (2.1)Since uB ⊂ A for some u ∈ N , we have y i := ux i ∈ A ∀ i ∈ [1 , h ]. Hence, applyingthe definitions of y i and m i in the equation (2.1), we have uz t + u ( i −
1) = y i f ( x i ) k r i,t and r i,t ≡ f ( x i )) ∀ i ∈ [1 , h ] . If u [ z t , z t + h − ∩ A (cid:54) = ∅ for some t ∈ N , then ∃ j ∈ [1 , h ] such that y j f ( x j ) k r j,t ∈ A and r j,t ≡ f ( x j )) where y j ∈ A and f ( x j ) ∈ H . This completes theproof. (cid:3) For any finite pairwise prime set B and an infinite pairwise prime subset of N (say H ), there exists pairwise prime set H ⊂ H such that | H | = | B | andgcd( a, b ) = 1 ∀ a ∈ H, b ∈ B . Then applying u = 1 in Corollary 2.3, we immediatelyget the following proposition. BHUWANESH RAO PATIL
Proposition 2.4.
Let k, l ∈ N , H be an infinite pairwise prime subset of N and A be an l -syndetic set. If B is a pairwise prime subset of A such that | B | ≥ l , then A ∈ S k,l,H . If H = P or an infinite pairwise prime subset of the set of composite numbers,we get Theorem 1.6 for those l -syndetic sets which contain a pairwise prime set B with | B | ≥ l . 3. Triveni triplets and syndetic sets
In the previous section, we observed that our proof of Theorem 1.6 depends onthe study of pairwise prime subsets of syndetic sets. For better understanding ofthese pairwise prime sets, we define Triveni triplet as follows. For l ∈ N and p ∈ P ,denote r ( p, l ) := max { t ∈ N ∪ { } : p t ≤ l + 1 } ,T ( l ) := (cid:110)(cid:81) p ∈ P ∩ [2 , l +1] p r p : 0 ≤ r p ≤ r ( p, l ) (cid:111) . Definition 3.1 (Triveni triplet) . Let l, h ∈ N , A ⊂ N and F ⊂ T ( l ) \ { } . Then ( F, h, l ) is called a Triveni triplet with respect to the set A if there exists a sequenceof pairwise prime sets ( B u ) u ∈ F such that(1) | B u | = h and uB u ⊂ A for each u ∈ F (2) For distinct u, v ∈ F , gcd( x, y ) = 1 ∀ x ∈ B u and y ∈ B v .Triveni triplets with respect to the set A are called A -Triveni triplets. | F | iscalled the order of the A -Triveni triplet ( F, h, l ) . Triveni triplets of order one with respect to syndetic sets.
One canproduce Triveni triplets of order one with respect to most of the syndetic sets usingZorn’s lemma[6].
Lemma 3.2.
Let r > and A ⊂ N . Let M r be the collection of pairwise primesubsets B of N satisfying rB = { rx : x ∈ B } ⊂ A . Then ∃ B r ∈ M r such that if C ⊃ B r and C ∈ M r , then C = B r . Here B r is a maximal element of M r .Proof. Let α be a chain in the partially ordered set ( M r , ⊂ ). Then the union ofevery elements of α belongs to the set M r . Hence, Zorn’s lemma guarantees theexistence of a maximal element. (cid:3) In the above lemma, B r may be an empty set. But certainly B r (cid:54) = ∅ for some r ∈ N . In particular, B (cid:54) = ∅ . The next proposition deals with the existence ofan infinite pairwise prime set B r for some r ∈ [1 , l ] with respect to those (2 l + 1)-syndetic sets which do not contain at least two elements of x N for each x ∈ N . Proposition 3.3.
Let A be a (2 l + 1) -syndetic set with | x N \ A | ≥ ∀ x ∈ N . Then ∃ d ∈ [1 , l ] and an infinite pairwise prime set B ⊂ N such that dB ⊂ A .Proof. Let B r be a maximal pairwise prime set satisfying rB r ⊂ A \ [1 , l ] for each r ∈ [1 , l ]. Lemma 3.2 assures the existence of such B r .By way of contradiction, assume that B r is finite for each r ∈ [1 , l ]. Define C := l (cid:91) r =1 rB r = { u , u , ..., u m } ⊂ A and t := m (cid:89) i =1 u i . Clearly, C (cid:54) = ∅ and t > l as B (cid:54) = ∅ . Since | t N \ A | ≥ ∃ s ∈ N \ { } such that st / ∈ A . Then the (2 l +1)-syndeticity of A ensures that z j ∈ A for some j ∈ [ − l, l ] \ z i := st + i ∀ i ∈ [ − l, l ]. Note that z j > t as j ≥ − l , s > t > l . Define d := max { r ∈ [1 , l ] : B r (cid:54) = ∅ and r | j } . EOMETRIC PROGRESSIONS IN SYNDETIC SETS 5
Clearly, d is well define as B (cid:54) = ∅ . Since dx divides t for all x ∈ B d and z j > t , wehave z j > dx ∀ x ∈ B d .Let x ∈ B d and d (cid:48) = gcd( dx, z j ). By the definition of t , dx | t . It follows that d (cid:48) = gcd( dx, j ). Using x ∈ B d and d (cid:48) | dx , we get that the pairwise prime set Z := { dxd (cid:48) } ⊂ N satisfies d (cid:48) Z ⊂ A \ [1 , l ] and so B d (cid:48) (cid:54) = ∅ . Since d | j and d (cid:48) | j , thedefinition of d gives d | d (cid:48) and d (cid:48) ≤ d . Hence d = gcd( dx, z j ).Therefore, we get z j ∈ A such that gcd( dx, z j ) = d and z j > dx for each x ∈ B d .It gives us pairwise prime set Y = B d ∪ { z j d } satisfying dY ⊂ A \ [1 , l ] and B d (cid:40) Y .This set Y contradicts the maximality of B d . So, we get a contradiction to theassumption that B r is finite for each r ∈ [1 , l ]. Hence, there exists r ∈ [1 , l ] suchthat B r is an infinite pairwise prime set. (cid:3) Therefore, Triveni triplets of order one can be found as a corollary of the aboveproposition in the following way.
Corollary 3.4.
Let k, l ∈ N , H be an infinite pairwise prime subset of N and A be a (2 l + 1) -syndetic set such that A / ∈ S k, l +1 ,H . Then there exists d ∈ [2 , l ] suchthat ( { d } , h, l ) is an A -Triveni triplet for each h ∈ N .Proof. Suppose for each d ∈ [2 , l ], there exists h d ∈ N such that ( { d } , h d , l ) is not an A -Triveni triplet. Applying Proposition 3.3, we get that either A ⊃ x N \ { xy } forsome x, y ∈ N or there exists an infinite pairwise prime subset B satisfying B ⊂ A .Therefore, Proposition 2.4 concludes the result. (cid:3) Triveni triplets of higher order with respect to syndetic sets.
Nowwe shall see a procedure for generating Triveni triplets of higher order with respectto syndetic sets. First we will prove some necessary results to demonstrate theprocedure. Due to the next lemma, one can construct sets with arbitrary largecardinalities in which the gcd of any two distinct elements belongs to the set T ( l ).Note that gcd of any two distinct elements in an interval with cardinality (2 l + 1)belongs to the set T ( l ). Lemma 3.5.
For l ∈ N , there exist an increasing function c : N −→ N and astrictly increasing sequence of positive integers ( x i ) ∞ i =1 such that x ≤ c ( h ) and gcd( x, y ) ∈ T ( l ) ∀ h ∈ N , x, y ∈ S h with x (cid:54) = y where S h = ∪ hi =1 [ x i , x i + 2 l ] . Proof.
Define x := 1 and c (1) := 2 l + 1. Then gcd( a, b ) ∈ T ( l ) for a, b ∈ S .For given h ∈ N , suppose there exist a sequence ( x i ) hi =1 in N and c ( h ) ∈ N suchthat a ≤ c ( h ) and gcd( a, b ) ∈ T ( l ) ∀ a, b ∈ S h with a (cid:54) = b . To complete the proof byinduction, we need to generate x h +1 and c ( h + 1) such that gcd( x, x h +1 + j ) ∈ T ( l ), x h < x h +1 ≤ c ( h + 1) − l and c ( h ) ≤ c ( h + 1) ∀ x ∈ S h and j ∈ [0 , l ]. For thispurpose, we define m := (cid:89) x ∈ S h x and Y := { p r : p ∈ P , p r | m and p r +1 (cid:45) m } x h +1 := 1 + (cid:89) u ∈ Y u and c ( h + 1) := max { l + x h +1 , c ( h ) } . The definition of c ( h + 1) ensures x h +1 ≤ c ( h + 1) − l and c ( h ) ≤ c ( h + 1). Since x h divides m , the definition of x h +1 gives us x h < x h +1 .Let x ∈ S h and j ∈ [0 , l ]. To show that gcd( x, x h +1 + j ) ∈ T ( l ), let q be a primedivisor of x such that q r ( q,l )+1 | x . Since x ∈ S h , it follows that q r ( q,l )+1 | m which isfollowed by q r ( q,l )+1 | u for some u ∈ Y . Then x h +1 + j ≡ j +1 (mod q r ( q,l )+1 ) by thedefinition of x h +1 . Hence, q r ( q,l )+1 (cid:45) x h +1 + j because of the fact that j +1 ∈ [1 , l +1]but q r ( q,l )+1 > l + 1 by the definition of r ( q, l ). Hence q r ( q,l )+1 (cid:45) gcd( x, x h +1 + j ).Since r ( p, l ) = 0 ∀ p ∈ P ∪ [2 l + 2 , ∞ ), we have gcd( x, x h +1 + j ) ∈ T ( l ) by definitionof T ( l ). (cid:3) BHUWANESH RAO PATIL
In the above lemma, the definition of the element x h +1 is inspired by the applica-tion of the Chinese remainder theorem on the congruences x ≡ u ) ∀ u ∈ Y .One can choose any positive integer which satisfies these congruences and is greaterthan 1. Corollary 3.6.
There exists a map m : N × N → N such that for h, l, n ∈ N , aninterval [ n, n + m ( h, l )] contains a set S with the following properties.(i) S = h (cid:91) i =1 [ x i , x i + 2 l ] for some strictly increasing sequence ( x i ) hi =1 in N .(ii) gcd( a, b ) ∈ T ( l ) ∀ a, b ∈ S with a (cid:54) = b .Proof. For given l, h ∈ N , Lemma 3.5 gives a strictly increasing sequence ( y i ) hi =1 in N and a positive integer c ( h ) such that a ≤ c ( h ) and gcd( a, b ) ∈ T ( l ) ∀ a, b ∈ R with a (cid:54) = b where R = ∪ hi =1 [ y i , y i + 2 l ]. Hence, define a map m : N × N → N suchthat m ( h, l ) := L + c ( h ) where L = (cid:81) q ∈ P ∩ [2 ,c ( h )] q r ( q,l )+1 .Let n ∈ N and a ∈ [ n + 1 , n + L ] be an integer divisible by L . Then we shallshow that the set S := { a + u : u ∈ R } satisfies S ⊂ [ n, n + m ( h, l )] along with theproperties (i) and (ii) given in the statement of Corollary (i). By construction, onecan note that property (i) is obvious. Using the fact that a ≤ c ( h ) ∀ a ∈ R and a ∈ [ n + 1 , n + L ], we get that S ⊂ [ n, n + m ( h, l )].To prove property (ii), Let v , v ∈ S with v (cid:54) = v . Then for each i ∈ { , } , v i = a + u i for some u i ∈ R . By way of contradiction, assume that p be a primesuch that p r ( p,l )+1 | gcd( v , v ). Since r ( p, l ) ∈ N ∪ { } , It follows that p | u − u .This implies p ≤ c ( h ) because u , u ∈ [1 , c ( h )]. Then, p r ( p,l )+1 | a by applying thedefinitions of a and L . It follows that p r ( p,l )+1 | gcd( u , u ) by using assumptionthat p r ( p,l )+1 | gcd( v , v ). But, this contradicts the fact that gcd( u , u ) ∈ T ( l ).So we get a contradiction to the assumption that p r ( p,l )+1 | gcd( v , v ).Therefore, p r ( p,l )+1 (cid:45) gcd( v , v ) for each p ∈ P and hence gcd( v , v ) ∈ T ( l )because r ( p, l ) = 0 ∀ p ∈ [2 l + 2 , ∞ ). (cid:3) The next lemma solves a Diophantine problem using Chinese remainder theorem.Define C ( l ) := (2 l + 1) l +1 and note that u < C ( l ) ∀ u ∈ T ( l ). Lemma 3.7.
Let ( a i ) ni =1 and ( u i ) ni =1 be sequences in N ∪ { } and T ( l ) respectively.If X = { x , x , · · · x n } be a pairwise prime subset of N , then there exist z ∈ N , asequence ( r i ) ni =1 in [0 , C ( l )] and a sequence ( t i ) ni =1 in N such that z + r i = a i + t i x i u i ∀ i ∈ [1 , n ] . (3.1) Proof.
Let u = lcm( u , u , · · · , u n ). Then u ∈ T ( l ) and so u < C ( l ). For each i ∈ [1 , n ], choose non-negative integers r i and b i such that a i = b i u + r i and0 ≤ r i < u . Note that r i ∈ [0 , C ( l )] as u < C ( l ). Since X is a pairwise prime set,the Chinese remainder theorem gives b ∈ N and a sequence ( v i ) ni =1 in N such that b = b i + v i x i ∀ i ∈ [1 , n ]. Hence, t i := v i uu i and z := bu satisfy the required equationin (3.1). (cid:3) Using the above corollaries and lemmas, the next two propositions demonstratethe complete procedure to generate Triveni triplets of higher order with respect tosyndetic sets. Define the map Λ : N × N → N such thatΛ( h, l ) = m ( h, l ) + 2 C ( l ) ∀ h, l ∈ N where the map m : N × N → N is taken from Corollary 3.6. For l ∈ N and F ⊂ T ( l ),define D ( l ) := | T ( l ) | and M ul ( F ) := { v ∈ T ( l ) : ∃ u ∈ F such that u | v } . EOMETRIC PROGRESSIONS IN SYNDETIC SETS 7
Proposition 3.8.
Let k, l, h ∈ N , A be a (2 l + 1) -syndetic set and H be aninfinite pairwise prime subset of N . If ( F, Λ( D ( l ) h, l ) , l ) is an A -Triveni triplet and A / ∈ S k, l +1 ,H , then ∃ w ∈ T ( l ) \ M ul ( F ) and a pairwise prime set C w such that | C w | = h and wC w ⊂ A .Proof. Since ( F, Λ( D ( l ) h, l ) , l ) is an A -Triveni triplet, there exists a sequence ofpairwise prime sets ( B u ) u ∈ F satisfying | B u | = Λ( D ( l ) h, l ) and uB u ⊂ A for each u ∈ F such that for distinct u, v ∈ F ,gcd( x, y ) = 1 ∀ x ∈ B u and y ∈ B v . Let B = ∪ u ∈ F B u . Since H is an infinite pairwise prime set, there exists asequence of pairwise prime subsets of H (say ( H u ) u ∈ F ) satisfying | B u | = | H u |∀ u ∈ F such that for distinct u, v ∈ F ,gcd( x, y ) = gcd( x, b ) = 1 ∀ x ∈ H u , y ∈ H v and b ∈ B. Given that A is a (2 l +1)-syndetic set satisfying A / ∈ S k, l +1 ,H . Applying Corol-lary 2.3, ∃ a sequence ( z u ) u ∈ F in N such that if z u,t = z u + t (cid:0)(cid:81) x ∈ B u x (cid:1) (cid:0)(cid:81) x ∈ H u x k (cid:1) with t ∈ N , then u [ z u,t , z u,t + Λ( D ( l ) h, l ) − ∩ A = ∅ ∀ t ∈ N , u ∈ F. (3.2)By Lemma 3.7, there exist sequences ( r u ) u ∈ F in [0 , C ( l )], ( t u ) u ∈ F in N and z ∈ N such that z + r u = uz u,t u ∀ u ∈ F. Then equation (3.2) guarantees that u N ∩ A ∩ [ z + r u , z + r u + u Λ( D ( l ) h, l ) − u ] = ∅ ∀ u ∈ F. Since r u ∈ [0 , C ( l )] and 1 < u < C ( l ) ∀ u ∈ F , it follows that u N ∩ A ∩ I = ∅ ∀ u ∈ F where I = [ z + C ( l ) , z + Λ( D ( l ) h, l ) − C ( l )] = [ z + C ( l ) , z + C ( l ) + m ( D ( l ) h, l )] . Hence, Corollary 3.6 gives us a sequence of intervals in I (say ( S i ) D ( l ) hi =1 ) such that(a) | S i | = 2 l + 1 and A ∩ S i ∩ u N = ∅ ∀ i ∈ [1 , D ( l ) h ], u ∈ F ,(b) v ∈ S i , v ∈ S j and v (cid:54) = v for i, j ∈ [1 , D ( l ) h ] ⇒ gcd( v , v ) ∈ T ( l ).Since A is (2 l +1)-syndetic set, there exists a sequence ( s i ) D ( l ) hi =1 such that s i ∈ S i ∩ A and gcd( s i , s j ) ∈ T ( l ) \ M ul ( F ) for i (cid:54) = j . Define g i := max { u ∈ T ( l ) \ M ul ( F ) : u | s i } ∀ i ∈ [1 , D ( l ) h ] . Using | T ( l ) | = D ( l ), there exist J ⊂ [1 , D ( l ) h ] and w ∈ T ( l ) \ M ul ( F ) such that | J | ≥ h and g j = w ∀ j ∈ J. Set C w := (cid:8) s j w : j ∈ J (cid:9) . Since g j = w ∀ j ∈ J and gcd( s j , s j ) ∈ T ( l ) \ M ul ( F ) ∀ j , j ∈ J with j (cid:54) = j ,we have gcd( s j , s j ) = w ∀ j , j ∈ J with j (cid:54) = j . Therefore, C w is a pairwiseprime set and wC w = { s j : j ∈ J } ⊂ A. (cid:3) Proposition 3.9.
Suppose that k, l ∈ N , H be an infinite pairwise prime sub-set of N and A be a (2 l + 1) -syndetic set with A / ∈ S k, l +1 ,H . If ( F, h, l ) is an A -Triveni triplet for each h ∈ N , then ∃ F (cid:48) ⊂ T ( l ) \ { } with F (cid:40) F (cid:48) such that ( F (cid:48) , h, l ) is an A -Triveni triplet for each h ∈ N .Proof. Let k ∈ N \ [1 , l + 1]. Since ( F, k , l ) is an A -Triveni triplet, there existsa sequence of pairwise prime sets ( B u ) u ∈ F such that for every distinct u, u ∈ F , | B u | = k and gcd( x, y ) = 1 ∀ x ∈ B u , y ∈ B u . Let B := ∪ u ∈ F B u , α := { p ∈ P : p | x for some x ∈ B } and W be the cardinality of α .Since A is a (2 l +1)-syndetic set with A / ∈ S k, l +1 ,H and ( F, h, l ) is an A -Triveni tripletfor each h ∈ N , Proposition 3.8 guarantees the existence of an element v ∈ T ( l ) \ M ul ( F ) and a pairwise prime set C v such that | C v | = W + k and vC v ⊂ A . Also,Proposition 2.4 ensures v (cid:54) = 1. Using the fact that α ⊂ P with | α | = W and C v is apairwise prime set of cardinality W + k , we get a pairwise prime set B v ⊂ C v suchthat | B v | = k and gcd( p, b ) = 1 ∀ p ∈ α, b ∈ B v . Then gcd( a, b ) = 1 ∀ a ∈ B, b ∈ B v because elements of the pairwise prime set B are made from primes in α . Moreover BHUWANESH RAO PATIL vB v ⊂ A , because B v ⊂ C v and vC v ⊂ A . Since ( F, k , l ) is an A -Triveni triplet,therefore, ( F k , k , l ) is also an A -Triveni triplet with F k = F ∪ { v } ⊂ T ( l ) \ { } .Here we constructed a sequence ( F n ) n ∈ N such that F (cid:40) F n ⊂ T ( l ) \ { } and( F n , n, l ) is an A -Triveni triplet ∀ n ∈ N . Since | T ( l ) | < ∞ , there exist subsequence( F n t ) t ∈ N of ( F n ) n ∈ N and set F (cid:48) with F (cid:40) F (cid:48) ⊂ T ( l ) \ { } such that F n t = F (cid:48) ∀ t ∈ N . Hence, ( F (cid:48) , t, l ) is an A -Triveni triplet for each t ∈ N . (cid:3) The combination of Corollary 3.4 and Proposition 3.9 generates Triveni tripletsof various orders with respect to those syndetic sets which do not contain config-urations of the form { x, xn k r } where r ∈ N , n ∈ H with r ≡ n ). Usingthese observations, we will now see the proof of Theorem 1.6.3.3. Proof of Theorem 1.6.
Let H = P or H be an infinite pairwise primesubset of the set of composite numbers. Since A is a syndetic set, there exists l ∈ N such that A is a (2 l + 1)-syndetic set.For k ∈ N , if possible assume that A / ∈ S k, l +1 ,H . Then Corollary 3.4 gives theexistence of an integer d ∈ [2 , l ] such that ( { d } , h, l ) is an A -Triveni triplet for each h ∈ N . Therefore, by Proposition 3.9, there exists a sequence ( F i ) ∞ i =0 such that(a) for each i ∈ [0 , ∞ ) and h ∈ N , ( F i , h, l ) is an A -Triveni triplet,(b) { d } = F (cid:40) F (cid:40) F (cid:40) · · · (cid:40) F i (cid:40) · · · ⊂ T ( l ).Since | T ( l ) | < ∞ , property (b) of sequence ( F i ) ∞ i =0 is a contradiction. Therefore, A ∈ S k, l +1 ,H . In other words, the syndetic set A contains configurations of theform { x, xn k r } where r ∈ N , n ∈ H satisfying r ≡ n ).4. Proof of Theorem 1.7
Now we are going to prove Theorem 1.7 by generating the required configurationsusing two different algorithms.
Lemma 4.1. A -syndetic set S contains infinitely many configurations of the type { x, xr } or infinitely many odd perfect squares.Proof. Let m ∈ N be an odd integer. If m + 1 / ∈ S , then the 2-syndeticity of S ensures that the odd perfect square m ∈ S . On the other hand, if m + 1 ∈ S ,then the identity (2 m + 1) − m ( m + 1) guarantees that { x, xr } ⊂ S for x = m +1 and r = 2 m whenever (2 m +1) − ∈ S . For the case (2 m +1) − / ∈ S ,the odd perfect square (2 m + 1) ∈ S due to the 2-syndeticity of S . Hence,infinitude of the odd integers completes the proof. (cid:3) Lemma 4.2. If y ∈ N + 1 and S is a -syndetic set such that y ( y + 2 i ) ∈ S for each i ∈ { , − } , then S contains a configuration of the type { x, xr } with x, r ∈ N \ { } and x ≥ y − .Proof. If y + 2 i ∈ S for some i ∈ { , − } , then { x, xr } ⊂ S for x = y + 2 i and r = y . On the other hand, if y +2 i / ∈ S for each i ∈ { , − } , then { y − , y +3 } ⊂ S due to 2-syndeticity of the set S . Since y is an odd integer, we get consecutivenatural numbers a y and b y satisfying 4 a y = y − ∈ S and 4 b y = y + 3 ∈ S. Since S is a 2-syndetic set, so it follows that one of a y and b y lies inside S . Hence, S contains { a y , a y } or { b y , b y } . Therefore, S contains the configuration { x, xr } for x ∈ { a y , b y } and r = 2. (cid:3) Proof of Theorem 1.7 (first method).
Let S be a 2-syndetic set. By Lemma 4.1,it is enough to show that if m ∈ S for some m ∈ N + 1, then S contains aconfiguration of the type { x, xr } where x, r ∈ N \ { } and x ≥ m − Let m ∈ N \ { } be an odd integer such that m ∈ S . If m n ∈ S for some n ∈ N \ { } , then we have { x, xr } ⊂ S for x = m and r = n . If m n / ∈ S foreach n ∈ N \ { } , then the 2-syndeticity of S ensures that m n − ∈ S ∀ n ∈ N \ { } . (4.1)Choose z ∈ N + 1 satisfying z ≡ m ). Define u ( z, i ) := z + i for each i ∈ {− , } . Clearly u ( z, i ) ≡ m ) for each i ∈ {− , } . Then, for each i ∈ {− , } ,there exists k u ( z,i ) ∈ N \ { } mk u ( z,i ) + 1 = u ( z, i ) . (4.2)Putting n = k u ( z,i ) in expression (4.1), we get m k u ( z,i ) − ∈ S . Inserting the valueof k u ( z,i ) from equation (4.2) in this, we have m k u ( z,i ) − u ( z, i ) − u ( z, i ) ∈ S. Therefore, { x, xr } ⊂ S for x = u ( z, i ) − r = u ( z, i ) whenever u ( z, i ) − ∈ S . For the remaining case u ( z, i ) − / ∈ S ∀ i ∈ {− , } , the 2-syndeticity of S guarantees that u ( z, i ) − z ( z + 2 i ) ∈ S ∀ i ∈ { , − } . Therefore we complete the proof by taking y = z in Lemma 4.2. (cid:3) The above algorithm also guarantees a configuration of the form { x, xr } with r < x inside syndetic sets containing the fourth power of some odd integer. Forthis, take m to be an odd perfect square and choose z = √ m for some i ∈ { , − } in the algorithm in the first method. The next algorithm generates a configurationof the form { x, xr } inside syndetic sets with the condition r > x using the identityin the following lemma. Lemma 4.3.
For a ∈ N , a (4 a + 3) + 1 = ( a + 1)(4 a + 1) .Proof of Theorem 1.7 (second method). Let S be a 2-syndetic set. If { a, a + 1 } ⊂ S for infinitely many a ∈ N , then applying the identity in Lemma 4.3, we get that { x, xr } ⊂ S for ( x, r ) = ( a, (4 a + 3)) or ( a + 1 , (4 a + 1)) for those a (cid:48) s. On theother hand, S contains an infinite arithmetic progression whenever { a, a + 1 } ⊂ S for only finitely many a ∈ N . Therefore, we finish the proof by using the fact thatany infinite arithmetic progression contains an infinite geometric progression. (cid:3) References [1] Mathias Beiglb¨ock, Vitaly Bergelson, Neil Hindman, Dona Strauss,
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