Monochromatic sums and quotients in $\mathbb N$

Mono c hromatic sums and quotien ts in N Mauro Di Nasso ∗ 1 , Lorenzo Lup eri Baglini † 2 , Rosario Menn uni ‡ , Mariaclara Ragosta § 3 , and Alessandro V egn uti ¶ 2,4 1 Dip artimento di Matematic a, Università di Pisa, L ar go Bruno Ponte c orvo 5, 56127 Pisa, Italy 2 Dip artimento di Matematic a, Università di Milano, Via Saldini 50, 20133 Milano, Italy 3 Dep artment of Applie d Mathematics (KAM), Charles University, Malostr anské náměstí 25, Pr aha 1, Cze ch R epublic 4 Mathematisches Institut, Alb ert-Ludwigs-Universität F r eibur g, D-79104 F r eibur g, Germany 18th Marc h 2026 Abstract W e prov e partition regularit y of the configuration x, y , x + y , y/x in a strong infin- itary form that extends Hindman’s Theorem. W e study the related issue of partition regularit y of configurations in volving pro ducts of a degree one p olynomial in x with one in y , reducing the general problem to a handful of sp ecial cases. In tro duction Sums, pro ducts and colours. A m uch studied problem in the com binatorics of the natural num b ers is that of mono c hromaticity of arithmetic configurations. The archet ypal result in this area is Sch ur’s Theorem [17], sa ying that the pattern x, y , x + y is p artition r e gular : whenev er the natural n umbers 1 are finitely coloured, i.e. partitioned in finitely man y pieces, there are some x, y ∈ N such that x, y , x + y lie in the same colour. In this con text, a cornerstone result is Hindman’s infinitary extension of Sc hur’s The- orem [10]: in every finite colouring of N , one ma y find an infinite sequence suc h that all sums of finitely man y of its terms are mono chromatic. The same holds when the sum is replaced b y an arbitrary associative op eration [13, Theorem 5.8], and in particular the pattern x, y , xy is partition regular. Ho wev er, despite the thorough literature now av ailable on the problem of partition reg- ularit y of Diophantine equations 2 , whether x, y , x + y , xy is partition regular remains one of the most long-standing op en problems in the area. Ab out this pattern, what is kno wn at presen t essentially amounts to the following. • P artition regularity holds for colourings of Q , see [3]. 2020 Mathematics subje ct classification. Primary: 05D10. Secondary: 26E35, 03H15, 11U10, 54D80. Keywor ds: Arithmetic Ramsey theory , partition regularity , ultrafilter, nonstandard analysis. ∗ email: mauro.di.nasso@unipi.it † email: lorenzo.luperi@unimi.it ‡ email: R.Mennuni@posteo.net § email: mariaclara@kam.mff.cuni.cz ¶ email: alessandro.vegnuti@unimi.it 1 In order to av oid trivialities, throughout the paper we con vene that 0 / ∈ N . 2 See the introduction to [6] for a survey . 1 • Mono c hromatic x, y , x + y, xy ma y b e found in every colouring of N with 2 colours, see [11]. In fact, in suc h a colouring one may find mono c hromatic sums and pro ducts of arbitrarily large finite size, see [2]. • Ov er N , the pattern x, x + y , xy is partition regular, see [15]. • Ov er N , the Hindman version of partition regularity fails, already at the level of pairs. Namely , there is a finite colouring of N admitting no infinite sequence ( x i ) i ∈ N suc h that all x i + x j and all x i x j , for i  = j , lie in the same colour. See [12], or [7] for a compact pro of and a systematic study of similar problems. Ov er Q , the Hindman v ersion is still op en. Results. In this pap er we pro ve that sums and ratios are, instead, m uch more well-behav ed from this p oin t of view, as Hindman’s Theorem may b e extended to accommo date ratios of sums coming from a fixed sequence. Theorem A (Theorem 1.4) . F or every finite colouring N = C 1 ∪ . . . ∪ C r there are m ≤ r and an increasing sequence ( x i ) i ∈ N suc h that, for all k < ℓ ∈ N and all i 1 < . . . < i ℓ , the colour C m con tains all x i 1 + . . . + x i k and x i k +1 + . . . + x i ℓ x i 1 + . . . + x i k . In particular, this solves [7, Problem 6.3(3)], and gives that the pattern x, y , x + y , y /x is partition regular ov er N . In fact, so is every pattern x, y , x + y , q · y /x with q ∈ Q > 0 (Corol- lary 1.6) but, b esides this, Theorem A is sharp, in the sense made precise by Theorem 1.7. In fact, Theorem 1.4 tells us more. A simple c hange of v ariables ( a = x and b = y /x ) transforms x, y , x + y , y /x into a, b, ab, a ( b + 1) . Partition regularit y of the latter w as established by Goswami [9], answering a question of Sahasrabudhe [16], that in fact only ask ed ab out the pattern a, b, a ( b + 1) . Our Theorem 1.4 also provides a strong infinitary v ersion of Gosw ami’s result and cov ers, as a baby case, the configuration a, b, c, ab, bc, abc, a + ab, a + abc, b + bc, ab + abc, a + ab + abc. The partition regularit y of a, b, a ( b + 1) is somewhat unexp ected 3 , let alone that of a, b, ab, a ( b + 1) . One ma y therefore w onder what other partition regular configurations are out there inv olving pro ducts of t wo shifts. After proving the aforemen tioned results in Section 1, the remainder of the presen t w ork addresses a more general version of this problem. T o av oid trivialities, b elo w we strengthen the notion of partition regularit y b y requiring the existence of infinitely many mono c hromatic solutions, cf. Notation 1.1.1. In Section 3 w e consider configurations x, y , q ( ax + n )( by + m ) , for q ∈ Q > 0 , a, b ∈ N and n, m ∈ Z . The case n = m = 0 is known to be partition regular (Remark 3.1). When precisely one of n, m is zero, sa y n , the general problem remains op en but, for example, for every b, m the generalised Sahasrabudhe pattern x, y , 1 m x ( by + m ) is partition regular (Example 3.13). Our next main result is a necessary condition for partition regularit y in the case n  = 0  = m , inv olving consecutive squares and (the doubles of ) consecutive triangular n umbers. Theorem B (Theorem 3.5) . If nm  = 0 and the pattern x, y , q ( ax + n )( by + m ) is partition regular, then there is t ∈ Z such that 3 In fact, Sahasrabudhe conjectured its failure while stating his question in [16]. 2 (a) q am = t 2 and q bn = ( t + 1) 2 , or (b) q am = t ( t − 1) and q bn = t ( t + 1) . This shows that, for instance, the pattern x, y , ( x + 1)( y + 2) is not partition regular. This necessary condition is certainly not sufficient, e.g. the configuration x, y , ( x + 1)(4 y + 1) falls in case (a) but is not partition regular, as witnessed by colouring natural num b ers by their parity . This do es not happ en b y c hance, and is in fact a sp ecial case of the divisibility condition pro vided b y our next theorem, whic h also gives a canonical form for patterns satisfying the conclusion of Theorem B. Theorem C. Supp ose there is t ∈ Z satisfying the conclusion of Theorem B (p ossibly with nm = 0 ). The pattern x, y , q ( ax + n )( by + m ) is partition regular if and only if the pattern x, y , ( x + q bn )( y + q am ) is partition regular and either (a) q am = t 2 , q bn = ( t + 1) 2 and ( am + bn ) / 2 ab − 1 / 2 q ab ∈ Z , or (b) q am = t ( t − 1) , q bn = t ( t + 1) and ( am + bn ) / 2 ab ∈ Z . In Section 4 we study 4-piece configurations. As it turns out, very few of them hav e a c hance to b e partition regular. Theorem D. Let q i ∈ Q > 0 , a i , b i ∈ N and n i , m i ∈ Z . If the pattern x, y , q 1 ( a 1 x + n 1 )( b 1 y + m 1 ) , q 2 ( a 2 x + n 2 )( b 2 y + m 2 ) is partition regular, then 1. q 1 ( a 1 x + n 1 )( b 1 y + m 1 ) = q 2 ( a 2 x + n 2 )( b 2 y + m 2 ) , or 2. q 1 ( a 1 x + n 1 )( b 1 y + m 1 ) = q 2 ( b 2 x + m 2 )( a 2 y + n 2 ) , or 3. the pattern is x, y , q xy , q xy + x or the symmetric x, y , q xy, q xy + y , with q : = q 1 a 1 b 1 = q 2 a 2 b 2 . W e p oint out that, in case 3, partition regularity do es indeed hold (Prop osition 4.4). As for n -piece configurations with n ≥ 5 , observ e that every sub configuration of a partition regular one is itself partition regular. This, together with Theorem D, yields that, up to rescaling, there is only one 5-piece configuration with a chance to b e partition regular, and that there are no 6-piece partition regular configurations at all. Corollary E. Assume that the set { x, y , q i ( a i x + n i )( b i y + m i ) | i ≤ k } has cardinality k + 2 ≥ 5 and that the pattern is partition regular. Then k = 3 and the pattern has the form x, y , q xy , qxy + x, q xy + y . Moreo ver, the partition regularity of this pattern is equiv alent to that of x, y , xy , xy + x, xy + y . F urthermore, we ha ve the following 4-piece analogue of Theorem C, which holds without assuming a priori that we are in the conclusion of Theorem B. Theorem F. Let q ∈ Q > 0 , a, b ∈ N and m, n ∈ Z . Assume that q ( ax + n )( by + m )  = q ( bx + m )( ay + n ) . The pattern x, y , q ( ax + n )( by + m ) , q ( bx + m )( ay + n ) is partition regular if and only if the pattern x, y , ( x + q bn )( y + q am ) , ( x + q am )( y + q bn ) is partition regular and there is t ∈ Z such that p oint (a) or p oin t (b) of Theorem C holds. 3 W e lea ve op en the partition regularit y of patterns that are not excluded b y the theorems ab o v e, notable examples b eing x, y , x ( y + 1) , ( x + 1) y and x, y , x ( y + 2) , ( x + 2) y . W e collect these and other questions in Section 5. Metho dology . All results of this pap er were originally obtained b y nonstandard-analytic metho ds. The key observ ation is that, in mo del-theoretic parlance, partition regularity of a form ula φ ( x 1 , . . . , x k ) is equiv alent to the existence, in an elemen tary extension ∗ N of N , of a solution of the formula with all co ordinates of the same t yp e (F act 2.4). As well-kno wn, this is in turn equiv alent to the existence of an ultrafilter on N k con taining the set of solutions of φ ( x 1 , . . . , x k ) in N and pro jecting on every co ordinate to the same ultrafilter on N . The basics of these techniques are recalled in Section 2. With some effort, our pro ofs may b e translated in standard ultrafilter terms, making no use of nonstandard analysis nor of mo del theory . W e give suc h a presentation of the pro of of Theorem A, that makes use of a v ariant of the Milliken–T a ylor Theorem from [1] inv olving idemp oten t ultrafilters. Nev ertheless, when dealing with proofs of necessary conditions, the nonstandard ap- proac h yields more streamlined and—w e b elieve—conceptually clearer proofs. It will become eviden t in Sections 3 and 4 that there are obstructions of p -adic nature to the partition reg- ularit y of certain patterns. Said obstructions are particularly visible from the nonstandard viewp oin t; we exemplify this b y pro ving Theorem 1.7 b y these methods first, and then translating the pro of in standard terms in an app endix, Section 6. F unding The authors w ere supp orted b y the pro ject PRIN 2022 “Logical methods in com- binatorics”, 2022BXH4R5, Italian Ministry of Univ ersity and Research (MUR). M. Ragosta is supp orted by pro ject 25-15571S of the Czech Science F oundation (GAČR). This work has b een supported by Charles Universit y Research Cen tre programme No.UNCE/24/SSH/026. W e ackno wledge the MUR Excellence Department Pro ject aw arded to the Department of Mathematics, Universit y of Pisa, CUP I57G22000700001. M. Di Nasso is a member of the INdAM research group GNSAGA. 1 Quotien ts of sums This section is devoted to the pro of of Theorem A. W e assume familiarity with basic notions around ultrafilters, and refer the reader to [13] for an extensive treatment. Notation 1.1. W e adopt the following conv entions. 1. If w e sa y that the pattern f 1 ( x 1 , . . . , x n ) , . . . , f k ( x 1 , . . . , x n ) is p artition r e gular (or simply PR ) we mean that, for ev ery finite colouring of N , the set of mono chromatic k -tuples of the form ( f 1 ( a 1 , . . . , a n ) , . . . , f k ( a 1 , . . . , a n )) is infinite. E.g., in ev ery col- ouring of N , there is a monochromatic solution of x, y , 3 x − 2 y , 5 y /x obtained b y setting x = y = 5 , but as we will see in Theorem 1.7 this pattern is not PR in the sense men- tioned abov e. W e do this in order to av oid ha ving to handle separately trivial cases suc h as the one just mentioned. 2. By definition, saying that the pattern x, y , f ( x, y ) is PR is the same as sa ying that the equation z = f ( x, y ) is PR. W e use the tw o terminologies interc hangeably . 3. By D ( x, y ) we denote the function N 2 → N sending ( x, y ) to y /x if x | y , and to 1 otherwise. 4. As usual, FS( x i | i ∈ N ) denotes { x i 1 + . . . + x i k | k ∈ N , i 1 < . . . < i k } . 5. If F , G ∈ P fin ( N ) \ {∅} , by F < G we mean max F < min G . 4 6. If u ∈ β N k is an ultrafilter and f : N k → N is a function, w e denote b y f ( u ) the pushforw ard { A ⊆ N | f − 1 ( A ) ∈ u } of u along f . Remark 1.2. Let u ∈ β N b e an additive idemp oten t. 1. The ultrafilter u contains every n N . 2. The ultrafilter u is self-divisible , that is, { ( a, b ) | a divides b } ∈ u ⊗ u . 3. The ultrafilter D ( u ⊗ u ) contains ev ery n N . In particular, it is not the principal ultrafilter on 1 . Pr o of. 1. This is well-kno wn (and easy to prov e). 2. This is [8, Example 5.1(8)], but it can b e easily sho wn as follows. W e hav e { ( a, b ) | a divides b } ∈ u ⊗ u if and only if { n | n N ∈ u } ∈ u . It now suffices to apply p oin t 1. 3. W e ha ve n N ∈ D ( u ⊗ u ) if and only if { ( a, b ) | n divides b/a } ∈ u ⊗ u , if and only if { a | { b | b ∈ an N } ∈ u } ∈ u and we conclude by p oin t 1. F act 1.3. Let A ⊆ N 2 . There is a sequence ( x i ) i ∈ N suc h that ( X i ∈ F x i , X i ∈ G x i !      F , G ∈ P fin ( N ) \ {∅} , F < G ) ⊆ A if and only if there is an idemp oten t u ∈ β N such that A ∈ u ⊗ u . Pr o of. This is a sp ecial case of [1, Theorem 1.17]. Theorem 1.4. F or every finite colouring N = C 1 ∪ . . . ∪ C r there are m ≤ r and sequences ( x i ) i ∈ N and ( y j ) i ∈ N suc h that FS( x i | i ∈ N ) ∪ ( P i ∈ G x i P i ∈ F x i      F , G ∈ P fin ( N ) \ {∅} , F < G ) ⊆ C m (1) and ( X ℓ ∈ F ℓ Y i = k y i      F ∈ P fin ( N ) \ {∅} , k ≤ min F ) ⊆ C m . (2) Moreo ver, if u is any additively idempotent ultrafilter, w e may take as C m the colour be- longing to D ( u ⊗ u ) . Pr o of. Let u b e an additiv ely idempotent ultrafilter and let C m b e as in the “moreo ver” part. By Remark 1.2 the set { ( a, b ) | a divides b } b elongs to u ⊗ u . Apply F act 1.3 to the set D − 1 ( C m ) ∩ { ( a, b ) | a divides b } , obtaining a sequence ( z n ) z ∈ N suc h that, whenev er F , G ∈ P fin ( N ) \ {∅} are such that F < G , we hav e P i ∈ G z i P i ∈ F z i ∈ C m . T o conclude the pro of, w e simply set x n : = z n +1 /z 1 and y n : = z n +1 /z n . Then, for every F , G ∈ P fin ( N ) \ {∅} , we hav e X i ∈ F x i = P i ∈ F z i +1 z 1 and P i ∈ G x i P i ∈ F x i = 1 z 1 P i ∈ G z i +1 1 z 1 P i ∈ F z i +1 = P i ∈ G z i +1 P i ∈ F z i +1 . 5 It follows that FS( x i | i ∈ N ) ∪ ( P i ∈ G x i P i ∈ F x i      F , G ∈ P fin ( N ) \ {∅} , F < G ) ⊆ ( P i ∈ G z i P i ∈ F z i      F , G ∈ P fin ( N ) \ {∅} , F < G ) ⊆ C m . Moreo ver, if F ∈ P fin ( N ) \ {∅} and k ≤ min F , w e analogously ha ve X ℓ ∈ F ℓ Y i = k y i = X ℓ ∈ F ℓ Y i = k z i +1 z i = X ℓ ∈ F z ℓ +1 z k = P ℓ ∈ F z ℓ +1 z k ∈ C m . See [13, Section 17.3] for other results inv olving sums of pro ducts of elements from an infinite sequence. Note that the configurations there are things of the form e.g. x 1 + x 2 x 3 + x 4 + x 8 x 11 , while those in the previous theorem lo ok like x 1 + x 1 x 2 x 3 + x 1 x 2 x 3 x 4 . Corollary 1.5 (Gosw ami) . The pattern x, y , xy , x ( y + 1) is PR. Pr o of. By Theorem 1.4 the pattern a, b, a + b, b/a is PR. Apply the c hange of v ariables x : = a , y : = b/a . Corollary 1.6. F or every finite colouring N = C 1 ∪ . . . ∪ C r and q ∈ Q > 0 there are m ≤ r and a sequence ( x i ) i ∈ N suc h that FS( x i | i ∈ N ) ∪ ( q · P i ∈ G x i P i ∈ F x i      F , G ∈ P fin ( N ) \ {∅} , F < G ) ⊆ C m . Pr o of. If v is an ultrafilter witnessing that Theorem 1.4 holds, then q · v ∈ β Q con tains N and witnesses that the conclusion holds. If the reader prefers, they may directly run the pro of of Theorem 1.4 with D ( u ⊗ u ) replaced by q · D ( u ⊗ u ) . Corollary 1.6 ab o ve shows that one can add co efficien ts in fron t of the ratios. A natural question is whether the same holds for the linear term. The answer is negativ e. Theorem 1.7. Let q ∈ Q > 0 and c, d ∈ Z \ { 0 } . Then x, y , cx + dy , q · y /x is PR if and only if c = d = 1 . W e will give a nonstandard pro of of this in Section 2, and a standard one in Section 6. Remark 1.8. 1. As a sp ecial case of (1), w e see that sums, divisions and pro jections are Ramsey partition regular, in the sense of [7], in particular answering in the p ositiv e [7, Prob- lem 6.3(3)] (which only asked for sums and divisions). 2. A sp ecial case of Theorem 1.7 is that differences, divisions and pr oje ctions are not PR, let alone Ramsey PR. Ramsey partition regularit y of differences and pro ducts and of differences and divisions (the other p oin ts of [7, Problem 6.3]), as well as the long-standing problem of partition regularit y of sums, pro ducts and pro jections, remain op en. 6 2 A quic k review of nonstandard metho ds In what follo ws, we will use some metho ds and terminology coming from nonstandard ana- lysis that hav e pro ven very w ell-suited to study the partition regularit y of Diophan tine equations. In this section, we collect all the results w e need; w e refer to [4, 5, 14] for extended presentations of this approach. W e work in a nonstandard extension ∗ R of R (whic h we assume to b e sufficiently satur- ated). In particular, we are in terested in the substructure induced on ∗ N ⊆ ∗ R . In such a setting, it is p ossible to introduce the notion of u -equiv alence. Definition 2.1. Let α, β ∈ ∗ N . W e say that α, β are u -equiv alent, and write α ∼ β , if for all A ⊆ N α ∈ ∗ A ⇐ ⇒ β ∈ ∗ A. Giv en α ∈ ∗ N , the set { A ⊆ N | α ∈ ∗ A } is an ultrafilter, and every ultrafilter arises in this wa y b y saturation. F rom a mo del-theoretical p ersp ectiv e, b eing u -equiv alent amoun ts precisely to ha ving the same type o ver ∅ , in the language with, for eac h k , a predicate for ev ery subset A ⊆ N k . The main properties of u -equiv alence that we will use are listed here; the in terested reader can find a pro of in [4, Section 11.2]. F act 2.2. Let α , β ∈ ∗ N and f : N → N . 1. If α ∼ β then f ( α ) ∼ f ( β ) . 2. If α ∼ f ( α ) then α = f ( α ) ; in particular, if α ∼ β then α = β or | α − β | is infinite. 3. If β ∼ f ( α ) then there exists γ ∼ α such that β = f ( γ ) . 4. If α ∼ β and α ∈ N then α = β . Recall that tw o elements α , β ∈ ∗ R are in the same A r chime de an class if 1 n | α | < | β | < n | α | for some n ∈ N . This is equiv alent to saying that the quotient α/β ∈ ∗ R is in the Arc himedean class of 1. In this case, we will denote by st( α/β ) the unique real such that α/β − st( α /β ) is infinites imal. In our pro ofs, we will rep eatedly use the following fact, whose pro of can b e found in [7, Lemma 5.5.(1)], relating u -equiv alence and Arc himedean classes. See Lemma 6.3 for a standard version (and a standard pro of ) of the same result. F act 2.3. Let α ∼ β b e infinite and in the same Archimedean class. Then st ( α/β ) = 1 . It was prov ed in [14, Theorem 2.2.9] that the notion of u -equiv alence allows us to rephrase partition regularity in nonstandard terms. F or patterns, it reads as follo ws. F act 2.4. Given f 1 , . . . , f k : N n → N , the pattern f 1 ( x 1 , . . . , x n ) , . . . , f k ( x 1 , . . . , x n ) is PR if and only if there exist α 1 , . . . , α n ∈ ∗ N such that f 1 ( α 1 , . . . , α n ) ∼ · · · ∼ f k ( α 1 , . . . , α n ) / ∈ N . The requirement that f i ( α 1 , . . . , α n ) / ∈ N corresp onds to the non triviality requirement in Notation 1.1.1. 7 Remark 2.5. In what follo ws we will b e interested in configurations of the form α ∼ β ∼ f 1 ( α, β ) ∼ f 2 ( α, β ) (3) where f 1 , f 2 are certain p olynomials ov er Q , hence it will sometimes b e conv enient to work in ∗ Q instead of ∗ N . The analogue of F act 2.4 still holds (as it do es ov er any set), but the reader should keep in mind that finding α, β ∈ ∗ Q satisfying (3) only gives partition regularit y of the configuration ov er Q . Obtaining partition regularity o ver N requires to also pro ve that α ∈ ∗ N (in which case, all other terms in the pattern also lie in ∗ N , since they are equiv alent to α ). F acts 2.2 to 2.4 are essen tially all the nonstandard analysis that w e will use in our proofs. Moreo ver, we will use a small amount of p -adic metho ds, recalled b elo w. Remark 2.6. 1. In what follows, we will frequently fix a “sufficiently large” prime p . The “sufficiently large” is to be understo od with resp ect to the height 4 of the data of the problem at hand; in the case of Theorem 1.7, these would b e c, d, q . In particular, if a, b ∈ Z are small with resp ect to p , then a ≡ b (mo d p ) implies a = b . This kind of arguments will b e used rep eatedly in the pap er. 2. Giv en a p ositiv e rational q = k /h and a sufficiently large p , when we refer to the class of q mo dulo p we mean the class of k h − 1 , where the m ultiplicative inv erse of h is computed in F p . 3. Besides working in the finite field F p in the w ay just men tioned, some argument will use the ring of p -adic in tegers Z p , or its field of fractions Q p . In fact, w e could work in these structures also in the pro ofs using F p men tioned ab ov e, but we prefer to limit their use to a minim um, for the b enefit of the reader unfamiliar with Z p . Intuitiv ely , working in F p amoun ts to considering the last digit of the expansion in base p , and working in Z p to considering the last ω digits of said expansion; that is, if α = P i ∈ ∗ N ∪{ 0 } a i p i , the class of α in Z p ma y b e identified with the sequence ( a i ) i ∈ N ∪{ 0 } . 4. W e will mak e frequen t use of the p -adic v aluation v p and of the function smod p sending n to the class of n/p v p ( n ) in F × p , i.e., to its least significant nonzero digit in base p . Recall that this map is a multiplicativ e homomorphism. When p is clear from context, w e will simply write v and smo d resp ectiv ely . 5. In order not to ov erburden the notation, w e write e.g. h smo d( α ) = k smo d( β ) in place of h smo d( α ) ≡ k (smod β ) (mo d p ) . The following observ ations will b e crucial, and used throughout the pap er, sometimes without mention. Remark 2.7. Let p b e a prime. 1. Since smo d p has finite image, it follows from p oin ts 1 and 4 of F act 2.2 that if α ∼ β then smo d p ( α ) = smo d p ( β ) . 2. Similarly , there is natural map ∗ N → Z p , sending α to the sequence of its last ω digits mo dulo p (or to the sequence of its remainder classes mo dulo the standard p o wers of p , dep ending on the reader’s fav ourite men tal picture of Z p ). This map factors through the quotient by ∼ . In other words, if α , β ∈ ∗ N and α ∼ β , then α, β hav e the same class in Z p . 4 Recall that, if q = n/m with ( n, m ) = 1 , the height of q is max {| n | , | m |} . In particular, the height of an integer is its absolute v alue. 8 Lemma 2.8. Let h ∈ N , k , ℓ ∈ Z , η , ζ , ξ ∈ ∗ N , θ ∈ ∗ N ∪ { 0 } , and p > h + | k | + | ℓ | a prime. Assume that η ∼ ζ ∼ ξ , that hη ∼ θ + k ζ + ℓξ , and that v p ( θ ) > v p ( ζ ) , v p ( ξ ) . 1. If v p ( ζ ) < v p ( ξ ) , then either k = 0 or k = h . 2. If v p ( ζ ) = v p ( ξ ) then either k + ℓ = 0 or k + ℓ = h . Pr o of. Apply the map smod = smo d p to hη and θ + kζ + ℓξ , and observ e that smo d( η ) = smo d( ζ ) = smo d( ξ ) , call it s . If v p ( ζ ) < v p ( ξ ) , then either k = 0 or v ( θ + ℓξ ) > v ( k ζ ) , from which it follows that the least significan t nonzero digit in base p of θ + k ζ + ℓξ equals that of kζ , that is, smod( θ + k ζ + ℓξ ) = smo d( kζ ) = k s . This implies smod( hη ) = hs = k s , and as p is large enough w e obtain h = k . If v p ( ζ ) = v p ( ξ ) then either k + ℓ = 0 or smo d( θ + k ζ + ℓξ ) = smo d( k ζ + ℓξ ) = ( k + ℓ ) s , and we conclude similarly as ab ov e. As a first example of application of the metho ds, we pro ve b elow Theorem 1.7. Pr o of of The or em 1.7. Right to left is a w eakening of Corollary 1.6, so w e fo cus on left to righ t. By F act 2.4 there exist α, β ∈ ∗ N \ N such that α ∼ β ∼ cα + dβ ∼ q β /α . In particular q β /α is a p ositiv e infinite integer, so the whole Archimedean class of q β , hence of β , must b e larger than that of α . Therefore, the Archimedean class of cα + dβ equals that of β , and it follows from F act 2.3 and β ∼ cα + dβ that d = 1 . W ork mo dulo a sufficiently large prime p and recall Remark 2.7. Since q β /α ∈ ∗ Z and p is large, and in particular v ( q ) = 0 , we hav e v ( β ) ≥ v ( α ) . By Lemma 2.8 (applied with θ = 0 ) and the assumption c  = 0 we see that either c = 1 , and we are done, or v ( β ) = v ( α ) and c = − 1 . In this last case, b ecause v ( α ) = v ( β ) and smo d( α ) = smo d( β ) , we obtain the con tradiction 0 = v ( q ) + v ( β ) − v ( α ) = v ( qβ /α ) ∼ v ( β + cα ) = v ( β − α ) > 0 . As F acts 2.2 and 2.4 will b e used very often, we will no longer mention them explicitly . 3 Pro ducts of t w o linear p olynomials As a sp ecial case of (2), for | F | ≤ 2 we recov er the main result from [9] (whic h answers [16, Question 31]), namely , that the pattern x, y , xy , x + xy is PR. In fact, the case | F | ≤ 2 of (2) pro vides a Ramsey version of this. With | F | = 3 w e obtain (Ramsey) partition regularity of the pattern x, y , z , xy , yz , xy z , x + xy , x + xy z , y + y z , xy + xy z , x + xy + xy z . As x, y , xy , x ( y + 1) is PR, one may w onder whether other shifts of factors in this kind of configuration yield PR patterns. More precisely , we w ould like to determine for which q ∈ Q > 0 , a, b ∈ N and n, m ∈ Z the equation z = q ( ax + n )( by + m ) is PR. The case n = m = 0 follows from Corollary 1.6 and a change of v ariables, but can also b e settled directly by the following observ ation. Remark 3.1. The pattern x, y , q xy is PR. T o prov e it, it suffices to take any multiplicativ e idemp oten t u containing every n N and consider the ultrafilter q − 1 · u . When at least one of n, m is nonzero, the situation is more in volv ed. Let us write q = k /h and rephrase the problem, asking for whic h h, k , a, b ∈ N and n, m ∈ Z the configuration hx, hy , k ( ax + n )( by + m ) 9 is PR, equiv alently , there are α , β ∈ ∗ N \ N with hα ∼ hβ ∼ k ( aα + n )( bβ + m ) . W e first deal with the case where exactly one of n, m is nonzero, sa y m . By simple algebraic manipulations we may reduce to the c ase of patterns hx, hy , ax ( by + m ) where ( h, a ) = 1 = ( b, m ) . Prop osition 3.2. Assume that ( h, a ) = 1 = ( b, m ) . If the pattern hx, hy , ax ( by + m ) is PR then a = 1 . Pr o of. If not, let p b e a prime dividing a , and let v b e the p -adic v aluation. As ( h, a ) = 1 , w e hav e v ( h ) = 0 . Therefore, if α, β are witnesses of partition regularity , v ( α ) = v ( hα ) ∼ v ( aα ( bβ + m )) = v ( a ) + v ( α ) + v ( bβ + m ) . If v ( α ) ∈ N , then the ∼ ab o ve is an equality , and as v ( a ) > 0 it follows that v ( α ) > v ( α ) , a contradiction. If instead v ( α ) > N then v ( β ) > N as well, hence v ( bβ + m ) = v ( m ) . Therefore, v ( α ) ∼ v ( a ) + v ( α ) + v ( m ) , hence the function x 7→ x + v ( a ) + v ( m ) sends the infinite num b er v ( α ) to an equiv alent one, contradiction. W e now pro ceed to deal with the case n, m  = 0 and pro ve Theorem B. In its pro of, and in several arguments further down in the pap er, we will use the following easy observ ations. Lemma 3.3. Let p a prime, u, v ∈ Z and α ∈ ∗ N . 1. If the class r of α in F p satisfies ur = v and γ : = uα − v , then v p ( γ ) > 0 . 2. If the class ρ of α in Z p satisfies uρ = v and γ : = uα − v , then v p ( γ ) > N . Pr o of. By construction, the class of γ in F p [resp., Z p ] is 0 . This means precisely that γ is divisible by p [resp., every standard p ow er of p ]. Remark 3.4. Let u, v ∈ Z , f ∈ Q [ x, y ] , and α, β ∈ ∗ N satisfy α ∼ β ∼ f ( α, β ) . Then γ : = uα − v and δ : = uβ − v satisfy γ ∼ δ ∼ g ( γ , δ ) for a suitable g ∈ Q [ x, y ] . Supp ose that p is a prime sufficiently large with resp ect to u, v and the co efficients of g . If the class r of α, β in F p satisfies ur = v , it follows from Lemma 3.3 that the constant term of g is null. Hence, we do not need to calculate explicitly the constant term of g when applying this kind of transformation. Theorem 3.5 (Theorem B) . Let q ∈ Q > 0 , a, b ∈ N and n, m ∈ Z \ { 0 } . Ass ume partition regularit y of the pattern x, y , q ( ax + n )( by + m ) . Then there is t ∈ Z suc h that (a) q am = t 2 and q bn = ( t + 1) 2 , or (b) q am = t ( t − 1) and q bn = t ( t + 1) . Remark 3.6. Note that, b y replacing t with − t − 1 or − t resp ectiv ely , one sees that b oth configurations in the conclusion are symmetrical in x, y . In the pro of of Theorem 3.5, we will use the following standard arithmetic fact, of which w e provide a pro of for the reader’s conv enience. F act 3.7. An in teger ∆ ∈ Z is a square in Z if and only if, for cofinitely many p , its residue is a square in F p . 10 Pr o of. If ∆ = 0 this is true, so assume this is not the case. By Gauss’ Lemma, ∆ is a square in Z if and only if it is a square in Q . By the Grun wald– W ang Theorem, this holds if and only ∆ is a square in Q p for cofinitely man y p . Equiv alently , if and only if it is a square in Z p for cofinitely man y p (since v ( x 2 ) = 2 v ( x ) and Z p = { x ∈ Q p | v ( x ) ≥ 0 } ). By assumption, for cofinitely many p , the residue of ∆ in F p is a square, say z 2 . The deriv ative of the p olynomial x 2 − ∆ is 2 x . As p is large enough, p ∤ ∆ , so z  = 0 , hence 2 z  = 0 . By Hensel’s Lemma, ∆ is a square in Z p . Pr o of of The or em 3.5. W rite q = k /h , for suitable p ositive in tegers k , h . By replacing a and n by k a and k n resp ectiv ely , we may assume that k = 1 . Assume that α, β ∈ ∗ N are infinite suc h that hα ∼ hβ ∼ ( aα + n )( bβ + m ) . Fix a sufficiently large p . W e ha ve v ( α ) = v ( hα ) ∼ v (( aα + n )( bβ + m )) whic h, if v ( α ) > 0 (hence also v ( β ) > 0 ), results in the contradiction v ( α ) ∼ v ( n ) + v ( m ) = 0 . Therefore, w e m ust hav e v ( α ) = 0 = v ( β ) . Let r b e the residue class of α in F p . W e hav e hr = abr 2 + ( am + bn ) r + nm. (4) View this as a degree 2 equation in r , and let ∆ : = ( am + bn − h ) 2 − 4 abnm b e its discriminan t. Because (4) has a solution in F p (namely , r ), its discriminant ∆ must b e a square in F p . As this happ ens for every sufficien tly large p , by F act 3.7, it follows that ∆ is a square in Z . Set A : = am and B : = bn . Observe that, since n, m  = 0 b y assumption, we also hav e A, B  = 0 . A routine calculation shows ∆ = ( A − B − h ) 2 − 4 B h . Let L ∈ Z be suc h that ∆ = (( A − B − h ) + L ) 2 . F rom ( A − B − h ) 2 − 4 B h = (( A − B − h ) + L ) 2 it follows that L ( L − 2 h ) = B (2 L − 4 h ) − 2 LA , so L is ev en. Set ℓ : = L/ 2 and observe that the previous equalit y yields ℓ ( ℓ − h ) = B ( ℓ − h ) − Aℓ. (5) In particular w e hav e ℓ  = 0 and ℓ  = h . View (5) as an affine equation in A, B , and parameterise its solutions as  A B  =  0 ℓ  + t  ℓ − h ℓ  =  t ( ℓ − h ) ( t + 1) ℓ  . By substituting this in ∆ = (( A − B − h ) + L ) 2 w e obtain ∆ = ( ℓ − ( t + 1) h ) 2 . It follows that the solutions of (4) are r 1 = − ℓt AB and r 2 = (1 + t )( h − ℓ ) AB . Note that ℓ, t do not dep end on the c hoice of p . Therefore, w e may assume that p is large also with resp ect to ℓ and t . W e first consider the solution r 1 . W e hav e AB α + ℓt ≡ 0 (mo d p ) . Set γ : = AB α + ℓt, δ : = AB β + ℓt. By substituting in the original pattern, m ultiplying by AB and adding hℓt , hγ ∼ hδ ∼ γ δ − htγ + ℓδ. By Lemma 3.3 we hav e that v ( γ ) , v ( δ ) > 0 . W e apply Lemma 2.8. 11 • If v ( δ ) > v ( γ ) then either ht = 0 or h = − ht . Both are con tradictory as they imply A = 0 and B = 0 resp ectiv ely . • If v ( γ ) > v ( δ ) then ℓ = 0 or ℓ = h , and again this implies the contradiction AB = 0 . • Supp ose now v ( γ ) = v ( δ ) . If ℓ = ht then am = A = t ( ℓ − h ) = ht ( t − 1) and bn = B = ( t + 1) ℓ = ht ( t + 1) , so then (b) holds. Otherwise, ℓ = h ( t + 1) , so am = A = t ( ℓ − h ) = ht 2 and bn = B = ℓ ( t + 1) = h ( t + 1) 2 , as in (a). W e now consider the solution r 2 . In this case AB α ≡ (1 + t )( h − ℓ ) (mo d p ) . Set γ : = AB α + ( ℓ − h )(1 + t ) , δ : = AB β + ( ℓ − h )(1 + t ) . Similarly as ab o ve, we obtain hγ ∼ hδ ∼ γ δ + ( h − ℓ ) γ + h (1 + t ) δ. By Lemma 3.3 v ( γ ) , v ( δ ) > 0 . As ab o ve, we apply Lemma 2.8. • If v ( δ ) > v ( γ ) , then either h = ℓ or h = h − ℓ , so ℓ = 0 . Both imply AB = 0 , a con tradiction. • If v ( γ ) > v ( δ ) then either h (1 + t ) = 0 or h = h (1 + t ) , so ht = 0 . Again, b oth imply the contradiction AB = 0 . • Assume no w that v ( γ ) = v ( δ ) . If ( h − ℓ ) + h (1 + t ) = 0 then h (2 + t ) = ℓ , hence am = A = t ( ℓ − h ) = ht ( t + 1) and bn = B = ( t + 1) ℓ = h ( t + 1)( t + 2) , so by shifting t we fall in case (b) and we are done. If instead h = h − ℓ + h (1 + t ) then h (1 + t ) = ℓ , that is ∆ = 0 , so r 2 = r 1 and we fall back to the previous case. Remark 3.8. One can chec k that, in case (a), either t or − t equals ( q bn − qam − 1) / 2 . Similarly , in case (b), either t or − 1 − t equals ( q bn − q am ) / 2 . F rom this, one may derive certain p olynomial relations b etw een q am and q bn . As we will never use this, we leav e details to the reader. Example 3.9. There are no h, n ∈ N such that the equation hz = ( x + n )( y + n ) is PR. Consider now the equations z = ( x + 1)( y + 4) and z = (4 x + 1)( y + 1) . They b oth fall in case (a) of Theorem 3.5 with t = 1 , hence the partition regularity of neither of them is excluded by this result. Nevertheless, while we leav e it op en whether z = ( x + 1)( y + 4) is PR (Question 5.1), easy parity considerations sho w that z = (4 x + 1)( y + 1) is not. Our Theorem C identifies, amongst the configurations resp ecting the conclusion of The- orem B, which ones admit obstructions of this sort. Sp ecifically , if suc h a configuration is PR, then a certain com bination of the parameters needs to b e an integer and, if this is the case, then the pattern may b e equiv alently rewritten in a certain canonical form. The pro of of Theorem C is split in the tw o lemmas b elo w, which are stated in a technical w ay that will find further use in Section 4. Here w e need to work in ∗ Q , and Remark 2.5 b ecomes relev ant. Lemma 3.10. Let t ∈ Z b e suc h that q am = t 2 and q bn = ( t + 1) 2 . Let α 0 , β 0 ∈ ∗ Q . Define • d : = − ( q am + q bn − 1) / 2 q ab ∈ Q . • α 1 : = q ab ( α 0 − d ) and β 1 : = q ab ( β 0 − d ) . • α 2 : = α 1 − t ( t + 1) and β 2 : = β 1 − t ( t + 1) . The following hold. 12 (A) W e hav e α 2 = q abα 0 . (B) The following are equiv alen t. (i) α 0 ∼ β 0 ∼ q ( aα 0 + n )( bβ 0 + m ) . (ii) α 1 ∼ β 1 ∼ α 1 β 1 + ( t + 1) β 1 − tα 1 . (iii) α 2 ∼ β 2 ∼ ( α 2 + ( t + 1) 2 )( β 2 + t 2 ) . (C) If the equiv alent conditions in p oin t (B) hold, the following are equiv alen t. (1) α 0 ∈ ∗ N . (2) d ∈ Z and α 1 ∈ ∗ N . (3) d ∈ Z and α 2 ∈ ∗ N . (D) If the conditions in p oints (B) and (C) hold, then α 1 is divis ible by ev ery natural n umber. Pr o of. W e leav e it to the reader to chec k part (A) and to p erform the changes of v ariables sho wing that (i), (ii) and (iii) are equiv alen t. The equiv alence (2) ⇔ (3) is immediate b y definition of α 2 and β 2 . F or (1) ⇒ (2) , observe that d = − t ( t + 1) /q ab . Let p b e an arbitrary prime and let ρ b e the common class of α 0 , β 0 in Z p . W e hav e ρ = q ( aρ + n )( bρ + m ) = q abρ 2 + ( q am + k bn ) ρ + k nm whence q abρ 2 + 2 t ( t + 1) ρ + q nm = 0 . Solving with the usual quadratic formula in Q p , we obtain that ∆ = 0 and that ρ = − t ( t + 1) /qab , whic h equals d . If d / ∈ Z , then there is a prime p such that v p ( d ) < 0 , contradicting d = ρ ∈ Z p . By construction, α 0 − d is divisible b y every p o wer of every prime, hence by every natural num b er. This implies that α 1 ∈ ∗ N , completing the pro of of (2), and that α 1 is itself divisible by every natural num b er, that is, p oin t (D). F or (2) ⇒ (1) , it suffices to observe that, for every p , the class of α 1 in Z p satisfies ρ = ρ 2 + ρ , that is, ρ = 0 . Therefore, α 0 = α 1 /q ab + d ∈ ∗ N . Lemma 3.11. Let t ∈ Z \ { 0 } b e such that q am = t ( t − 1) and qbn = t ( t + 1) . Let α 0 , β 0 ∈ ∗ Q . Define • d : = − ( am + bn ) / 2 ab ∈ Q . • α 1 : = q ab ( α 0 − d ) and β 1 : = q ab ( β 0 − d ) . • α 2 : = α 1 − t 2 and β 2 : = β 1 − t 2 . The following hold. (A) W e hav e α 2 = q abα 0 . (B) The following are equiv alen t. (i) α 0 ∼ β 0 ∼ q ( aα 0 + n )( bβ 0 + m ) . (ii) α 1 ∼ β 1 ∼ α 1 β 1 + tβ 1 − tα 1 . (iii) α 2 ∼ β 2 ∼ ( α 2 + t ( t + 1))( β 2 + t ( t − 1)) . (C) If the equiv alent conditions in p oin t (B) hold, the following are equiv alen t. (1) α 0 ∈ ∗ N . (2) d ∈ Z and α 1 ∈ ∗ N . 13 (3) d ∈ Z and α 2 ∈ ∗ N . (D) If the conditions in p oints (B) and (C) hold, then α 1 is divisi ble by ev ery natural n umber. Pr o of. As in the pro of of Lemma 3.10, we prov e (1) ⇔ (2) , as well as p oin t (D), and leav e it to the reader to chec k the remaining parts. W rite q = k /h , for suitable coprime k , h ∈ N . F or the implication (1) ⇒ (2) , if p is an y prime and ρ is the common class of α 0 , β 0 in Z p , b y solving the degree 2 equation given by (i) in Q p , we see that ρ must satisfy k abρ + ht 2 = 0 or k abρ + ( t 2 − 1) h = 0 . Claim 3.11.1. F or every prime p we are in the case k abρ + ht 2 = 0 . Pr o of of Claim. In fact, if p is such that the other case holds, set γ : = kabα 0 + ( t 2 − 1) h and δ : = k abβ 0 + ( t 2 − 1) h and observe that v p ( γ ) , v p ( δ ) are b oth infinite by Lemma 3.3. A routine calculation shows that (i) implies hγ ∼ hδ ∼ γ δ + h (1 − t ) γ + h (1 + t ) δ. (6) Let s : = smo d( γ ) . W e hav e three cases, each split into v arious sub cases. Assume first that t = 1 . Then (6) giv es hδ ∼ γ δ + 2 hδ . If p = 2 , the 2 -adic v aluation of the left hand side is v ( h ) + v ( δ ) , while that of the right hand side is 1 + v ( h ) + v ( δ ) , a con tradiction since there are no equiv alent p oin ts at finite nonzero distance. If p  = 2 , then smo d( γ δ + 2 hδ ) = 2 smo d( h ) s , therefore smo d( h ) s = 2 smod( h ) s , again a contradiction. The case t = − 1 is analogous, so w e no w assume that t is neither 1 nor − 1 (nor 0 , by assumption). Let smo d p ω : ∗ N → Z p \ { 0 } b e the function sending x to the class in Z p of x/p v p ( x ) . In tuitively , smo d p ω ( x ) is calculated by writing x in base p , discarding the rightmost 0 digits and taking the remaining rightmost ω digits. If v p ( γ ) < v p ( δ ) , then v p ( δ ) − v p ( γ ) > N . This together with (6) implies v ( h ) + v ( γ ) ∼ v ( h ) + v (1 − t ) + v ( γ ) , hence v (1 − t ) = 0 , so p ∤ 1 − t . By applying smo d p ω w e then find smo d p ω ( γ ) = smo d p ω (1 − t ) smod p ω ( γ ) , hence smo d p ω (1 − t ) = 1 , whic h implies t = 0 , a con tradiction. The argument in the case v p ( γ ) > v p ( δ ) is completely analogous, and left to the reader. If v p ( γ ) = v p ( δ ) , let us rewrite (6) as hγ ∼ hδ ∼ γ δ + ht ( δ − γ ) + h ( γ + δ ) . Observ e that, b ecause smo d p ω ( γ ) = smo d p ω ( δ ) and v ( γ ) = v ( δ ) , w e ha v e v ( δ − γ ) − v ( γ + δ ) > N . If p = 2 , b y c hecking digits in base 2 we see that v ( γ ) ∼ v ( γ + δ ) = v ( γ ) + 1 , a con tradiction. If p  = 2 , w e obtain smo d( γ ) = 2 smod( γ ) , again a con tradiction. claim T o show that d ∈ Z , observ e that b y Claim 3.11.1 and definition of d we hav e d = ρ . Hence, for every p , the p -adic v aluation of d is nonnegative, and we hav e the conclusion. It remains to verify that α 1 ∈ ∗ N . But, by construction and the fact that d = ρ , we see that α 0 − d and β 0 − d are divisible by every element of N , and we hav e the conclusion, as w ell as p oin t (D). F or (2) ⇒ (1) , since α 0 = α 1 /q ab + d , it suffices to show that α 1 is divisible b y every natural num b er. F or every prime p , the class ρ in Z p of α 1 satisfies ρ = ρ 2 , hence it must equal 0 or 1 . W e show that the second case never happ ens. In fact, if p is a counterexample and w e set γ : = α 1 − 1 and δ : = β 1 − 1 , we obtain γ ∼ δ ∼ γ δ + (1 − t ) γ + ( t + 1) δ . W e conclude by essentially the same pro of as that of Claim 3.11.1. Remark 3.12. In Lemmas 3.10 and 3.11, if we replace t b y its symmetric parameter giv en b y Remark 3.6, the changes of v ariables relating α 0 to α 1 , α 2 remain unchanged. 14 Pr o of of The or em C. By Lemmas 3.10 and 3.11 we are only left to deal with the case where q am = q bn = 0 . But this is Remark 3.1. Example 3.13. By Theorem C and partition regularit y of z = x ( y + 1) (Corollary 1.5), the equation z = x (18 y + 1) is PR. More generally , for every coprime b, m ∈ N , we ha ve partition regularity of the pattern mx, my , x ( by + m ) . 4 Sev eral pro ducts of t w o linear p olynomials In this section we study configurations of the form x, y , q 1 ( a 1 x + n 1 )( b 1 y + m 1 ) , q 2 ( a 2 x + n 2 )( b 2 y + m 2 ) (7) for q i ∈ Q > 0 , a i , b i ∈ N and m i , n i ∈ Z , with the goal of proving Theorems D and F and Corollary E. F or patterns where all constan t terms are nonzero, Theorem D ma y b e prov en b y a quic k argumen t using Theorem B. Even if w e do not know whether the conclusion of the latter holds when some constant term is null, we will still b e able to prov e Theorem D b y slightly differen t arguments. W e b egin b y an easy observ ation that will b e used without mention throughout the rest of the section. Lemma 4.1. If the pattern (7) is PR, then q 1 a 1 b 1 = q 2 a 2 b 2 . Pr o of. Let α, β b e witnesses of partition regularity . Observe that q i ( a i α + n i )( b i β + m i ) is asymptotic to q i a i b i αβ , that is, their ratio has standard part 1 . By F act 2.3 st  q 2 ( a 2 α + n 2 )( b 2 β + m 2 ) q 1 ( a 1 α + n 1 )( b 1 β + m 1 )  = 1 hence 1 = st( q 2 a 2 b 2 αβ /q 1 a 1 b 1 αβ ) and the conclusion follows. Prop osition 4.2. In the case m 1 n 1 m 2 n 2  = 0 , the conclusion of Theorem D holds. Pr o of. Assume that (7) is PR and let α 0 , β 0 b e witnesses. Apply Theorem B to th e tw o 3- piece patterns x, y , q i ( a i x + n i )( b i y + m i ) for i = 1 and i = 2 , obtaining t 1 , t 2 ∈ Z witnessing its conclusion. By Lemma 4.1 we hav e q 1 a 1 b 1 = q 2 a 2 b 2 . It follows that the c hanges of v ariables in Lemmas 3.10 and 3.11 bringing α 0 to α 2 = q 1 a 1 b 1 α 0 coincide. Fix a prime p sufficiently large with resp ect to t 1 , t 2 . By part (D) of Lemmas 3.10 and 3.11 the class of α 2 in F p is either − t i ( t i + 1) (if w e are in case (a) of Theorem B) or − t 2 i (if we are in case (b)). • If we are in case (a) for b oth i = 1 and i = 2 then we get t 1 ( t 1 + 1) = t 2 ( t 2 + 1) . It follo ws that either t 1 = t 2 , hence we are in case 1 of the conclusion, or t 1 = − t 2 − 1 , hence we are in case 2. • If we are in case (b) for b oth i = 1 and i = 2 then we get t 2 1 = t 2 2 . It follows that either t 1 = t 2 , hence we are in case 1 of the conclusion, or t 1 = − t 2 , hence we are in case 2. • In the remaining cases w e get t 1 ( t 1 + 1) = t 2 2 or t 2 1 = t 2 ( t 2 + 1) , hence t 1 = 0 = t 2 . As all q i , a i , b i are nonzero, this implies that some n i or some m i is zero, against the assumptions 5 . 5 Note that, at any rate, this would bring us in case 3 of the conclusion. 15 W e no w deal with the case where m 1 n 1 m 2 n 2 = 0 . The argument will b e split in to v arious sub cases. Prop osition 4.3. If the pattern (7) is PR and n 1 m 1 = 0 , then n 2 m 2 = 0 . Pr o of. W e ma y assume without loss of generality that n 1 = 0 . T ow ards a contradiction, assume that n 2 m 2  = 0 . Let t ∈ Z b e given by applying Theorem B to the configuration obtained by ignoring the third piece. Fix a sufficien tly large prime p and let r b e the common class of α, β in F p . It follows from (7) that r = q 1 a 1 r ( b 1 r + m 1 ) = q 2 ( a 2 r + n 2 )( b 2 r + m 2 ) . (8) If r = 0 , then q 2 n 2 m 2 = 0 , a contradiction. Therefore r  = 0 and in particular b 2 r + m 2  = 0 . (9) By dividing the first equalit y in (8) by r , we then find that 1 = q 1 a 1 b 1 r + q 1 a 1 m 1 , hence that q 1 a 1 m 1  = 1 . (10) Let γ : = q 1 a 1 b 1 α − 1 + q 1 a 1 m 1 , δ : = q 1 a 1 b 1 β − 1 + q 1 a 1 m 1 . As v ( γ ) is p ositiv e by Lemma 3.3, so is v ( δ ) ∼ v ( γ ) . W e obtain from (7) that 6 γ ∼ δ ∼ γ δ + γ + (1 − q 1 a 1 m 1 ) δ ∼ γ δ + (1 − q 1 a 1 m 1 + q 2 a 2 m 2 ) γ + (1 − q 1 a 1 m 1 + q 2 b 2 n 2 ) δ. (11) W e hav e three cases, and in each we apply Lemma 2.8. • Assume v ( δ ) < v ( γ ) . Applying Lemma 2.8 to the second and third piece in (11), we obtain that either 1 = q 1 a 1 m 1 , contradicting (10), or 1 = 1 − q 1 a 1 m 1 , which implies m 1 = 0 . By the same lemma applied to the second and fourth piece of (11), our assumptions give that either 1 + q 2 b 2 n 2 = 0 , or 1 = 1 + q 2 b 2 n 2 . In the latter case, we obtain the contradiction q 2 b 2 n 2 = 0 . In the former, q 2 b 2 n 2 = − 1 . By writing q 2 b 2 n 2 in terms of t , we get a con tradiction since − 1 is neither a square nor a pro duct of tw o consecutiv e integers. • Assume v ( γ ) < v ( δ ) . Comparing the first and fourth piece in (11) yields that either 1 − q 1 a 1 m 1 + q 2 a 2 m 2 = 0 , or q 1 a 1 m 1 = q 2 a 2 m 2 . In the second case, plugging the equalit y in to (8) gives us 0 = q 2 b 2 n 2 r + q 2 n 2 m 2 , hence since q 2 n 2  = 0 we get b 2 r + m 2 = 0 , against (9). The same contradiction can b e obtained in the first case, by expanding the first equality in (8) and then substituting q 2 a 2 b 2 for q 1 a 1 b 1 and 1 + q 2 a 2 m 2 for q 1 a 1 m 1 . • Finally , assume v ( γ ) = v ( δ ) . Comparing the first and third piece gives that q 1 a 1 m 1 equals either 2 or 1 , the latter b eing excluded by (10). Similarly , comparing smo d in the second and fourth piece and using that q 1 a 1 m 1 = 2 gives us that q 2 a 2 m 2 + q 2 b 2 n 2 equals either 2 or 3 . Now observe that q 2 a 2 m 2 + q 2 b 2 n 2 equals either t 2 + ( t + 1) 2 , or 2 t 2 , dep ending on whether we are in case (a) or in case (b) of Theorem B. As neither 2 nor 3 is a sum of tw o consecutiv e squares, and as 2 t 2 = 3 has no solutions in Z , it only remains to consider the case where q 2 a 2 m 2 = t ( t − 1) , q 2 b 2 n 2 = t ( t + 1) and t 2 = 1 . This gives either m 2 = 0 or n 2 = 0 , and we are done. In the case where n 1 = 0 = m 1 w e can fully characterise the PR patterns: they are either the 3 -piece pattern from Remark 3.1 or rescalings of the 4 -piece pattern in Corollary 1.5. 6 The reader may wan t to recall Remark 3.4, esp ecially in calculating the fourth piece of (11). 16 Prop osition 4.4. F or i ∈ { 1 , 2 } , let q i ∈ Q > 0 and a i , b i ∈ N . Let n 2 , m 2 ∈ Z . The pattern x, y , q 1 a 1 b 1 xy , q 2 ( a 2 x + n 2 )( b 2 y + m 2 ) is PR if and only if q 1 a 1 b 1 = q 2 a 2 b 2 and either 1. n 2 = m 2 = 0 , or 2. n 2 = 0 and q 2 a 2 m 2 = 1 , or 3. m 2 = 0 and q 2 b 2 n 2 = 1 . Pr o of. By F act 2.3, if the pattern is PR then q 1 a 1 b 1 = q 2 a 2 b 2 , therefore we assume this in the rest of the pro of. Let q : = q 1 a 1 b 1 = q 2 a 2 b 2 . By Proposition 4.3, if the pattern is PR then at least one of n 2 , m 2 has to b e zero. If n 2 = m 2 = 0 , then the pattern is PR b y Remark 3.1. Let us deal with the case n 2 = 0 , m 2  = 0 , the case m 2 = 0 , n 2  = 0 , b eing symmetrical. Note that, if α ∼ β ∼ q αβ ∼ q αβ + q 2 a 2 m 2 α and α, β b elong to ∗ N , then so do the other items ab o ve. Therefore, α ′ : = α and β ′ : = q αβ belong to ∗ N and satisfy α ′ ∼ β ′ / ( q α ′ ) ∼ β ′ ∼ β ′ + q 2 a 2 m 2 α ′ . (12) This is PR if and only if q 2 a 2 m 2 = 1 by Theorem 1.7. Conv ersely , if q 2 a 2 m 2 = 1 and α ′ , β ′ ∈ ∗ N satisfy (12) then we conclude by setting α : = α ′ and β : = β ′ /q α ′ , and observing that the latter lies in ∗ N since it is equiv alent to α ′ . Example 4.5. By Prop osition 4.4, we hav e that the configuration 2 x, 2 y , 3 xy , x (3 y + 2) is PR, whilst 2 x, 2 y , 3 xy , x (3 y − 2) is not. When m 1  = 0 w e are left with the cases n 2 = 0  = m 2 and m 2 = 0  = n 2 . It turns out that in the first case there are no PR 4-piece patterns, and that PR in the second case implies symmetry in x, y . Prop osition 4.6. Let m 1 m 2  = 0 . If the pattern x, y , q 1 a 1 x ( b 1 y + m 1 ) , q 2 a 2 x ( b 2 y + m 2 ) is PR then q 1 a 1 x ( b 1 y + m 1 ) = q 2 a 2 x ( b 2 y + m 2 ) . Pr o of. As usual let α, β b e witnesses of partition regularity and observe that by Lemma 4.1 q 1 a 1 b 1 = q 2 a 2 b 2 . The common class r of α, β mo dulo a sufficiently large prime p satisfies r = q 1 a 1 b 1 r 2 + q 1 a 1 m 1 r = q 1 a 1 b 1 r 2 + q 2 a 2 m 2 r , whence q 1 a 1 m 1 r = q 2 a 2 m 2 r . If r  = 0 , as p is large, we obtain q 1 a 1 m 1 = q 2 a 2 m 2 and, as q 1 a 1 b 1 = q 2 a 2 b 2 , we obtain the conclusion. If instead r = 0 , then v ( α ) , v ( β ) are p ositiv e. If s : = smo d( α ) = smo d( β ) we obtain q 1 a 1 m 1 s = q 2 a 2 m 2 s and, again b ecause p is large, this means q 1 a 1 m 1 = q 2 a 2 m 2 , so we conclude as ab o ve. Example 4.7. The configuration x, y , x ( y + 1) , x ( y + 2) is not PR. Prop osition 4.8. If m 1 n 2  = 0 and the pattern x, y , q 1 a 1 x ( b 1 y + m 1 ) , q 2 ( a 2 x + n 2 ) b 2 y is PR, then q 1 a 1 m 1 = q 2 b 2 n 2 . Pr o of. Again, b y F act 2.3 partition regularit y implies q 1 a 1 b 1 = q 2 a 2 b 2 . Let α, β b e nonstand- ard witnesses of partition regularity and r their common class mo dulo a sufficiently large prime p . W e hav e r = q 1 a 1 b 1 r 2 + q 1 a 1 m 1 r = q 1 a 1 b 1 r 2 + q 2 b 2 n 2 r , whence q 1 a 1 m 1 r = q 2 b 2 n 2 r . If r  = 0 we immediately get q 1 a 1 m 1 = q 2 b 2 n 2 . If r = 0 , then v ( α ) , v ( β ) are p ositiv e. As p is large if s : = smo d( α ) , we obtain q 1 a 1 m 1 s = q 2 n 2 b 2 s , hence the conclusion. 17 Example 4.9. The configuration x, y , x ( y + 1) , ( x + 2) y is not PR. This was the last case to consider in the proof of Theorem D, and Corollary E follo ws easily . Pr o of of The or em D. By combining the previous results in this section. Pr o of of Cor ol lary E. Any PR configuration with 5 or more pieces induces v arious PR 4- piece configurations. By insp ecting the conclusion of Theorem D, w e see that the only compatible 4-piece configurations are those in case 3, with the same q , from which the first part of the conclusion follows. F or the “moreo ver part”, if p is an arbitrary prime and α, β witness partition regularity of the pattern, then their class ρ in Z p satisfies q ρ 2 + ρ = q ρ 2 . Therefore, α and β are divisible by ev ery natural num b er, and b y setting α ′ : = q α and β ′ : = q β we obtain witnesses of partition regularity of x, y, xy , xy + x, xy + y . The conv erse is analogous. Finally , we prov e Theorem F. Pr o of of The or em F. By Lemmas 3.10 and 3.11 and Remark 3.12, it suffices to sho w that w e are in the assumptions of said lemmas. If nm  = 0 , this is Theorem B. If nm = 0 , since b y assumption the third and fourth piece of the pattern differ, precisely one of n, m is null, without loss of generality n = 0  = m . T o conclude, w e prov e that q am equals either 1 or 2 , so w e are in the assumptions of Lemma 3.10 with t = − 1 or of Lemma 3.11 with t = 1 resp ectiv ely . Let α , β ∈ ∗ N witness partition regularit y . W e mov e to in teger parameters, b y writing q · a · ( b, m ) = A/h for suitable coprime integers A, h and setting B : = b/ ( b, m ) and M : = m/ ( b, m ) . By m ultiplying by h w e obtain hα ∼ hβ ∼ Aα ( B β + M ) ∼ A ( B α + M ) β where ( h, A ) = 1 = ( B , M ) . After this renaming, the conclusion amoun ts to showing that h = AM or 2 h = AM . Fix a sufficien tly large prime p . If v ( α ) , v ( β ) > 0 , by Lemma 2.8 we obtain immediately h = AM . If instead v ( α ) = 0 = v ( β ) , set γ : = AB α + AM − h and δ : = AB β + AM − h. Similarly as previously done in this pap er, we obtain hγ ∼ hδ ∼ γ δ + hγ + ( h − AM ) δ ∼ γ δ + ( h − AM ) γ + hδ and v ( γ ) , v ( δ ) > 0 . If v ( γ ) > v ( δ ) , we apply Lemma 2.8 to the second and third piece of the pattern. If h = h − AM then AM = 0 , a con tradiction, therefore h − AM = 0 . If v ( γ ) < v ( δ ) the argument is analogous, but we apply Lemma 2.8 to the first and fourth piece, so we are left with the case v ( γ ) = v ( δ ) > 0 . Again b y Lemma 2.8, applied to the first and third piece, this implies either 2 h = AM or h = 2 h − AM , that is, h = AM . Example 4.10. F or all n ∈ Z \ { 0 , 1 , 2 } the configuration x, y , x ( y + n ) , ( x + n ) y is not PR. 5 Op en problems It remains op en to determine which patterns ob eying the restrictions given by our main theorems are PR. F or 3-piece patterns, this amounts to answ ering the follo wing. Question 5.1. 18 1. F or which t ∈ Z is the equation z = ( x + t 2 )( y + ( t + 1) 2 ) PR? 2. F or which t ∈ Z is the equation z = ( x + t ( t − 1))( y + t ( t + 1)) PR? 3. F or which q ∈ Q > 0 , b ∈ N and m ∈ Z is the equation z = q x ( by + m ) PR? The abundance of parameters in Question 5.1.3 is due to the lack of a canonical form when n = 0 . Problem 5.2. Find a canonical form, preserving (non-)partition regularity , for equations of the form z = q x ( by + m ) . As for patterns with more than 3 pieces, our w ork reduces the general problem to the follo wing. Question 5.3. 1. F or which t ∈ Z is x, y , ( x + t 2 )( y + ( t + 1) 2 ) , ( x + ( t + 1) 2 )( y + t 2 ) PR? 2. F or which t ∈ Z is x, y , ( x + t ( t − 1))( y + t ( t + 1)) , ( x + t ( t + 1))( y + t ( t − 1)) PR? 3. Is x, y , xy , xy + x, xy + y PR? As the pattern x, y , xy, x ( y + 1) is PR, one wonders if this is the case when w e replace x + y for xy . Question 5.4. Is the pattern x, y , x + y , x ( y + 1) PR? Let us observe that 1 ab ov e cannot b e replaced by other nonzero integers. Remark 5.5. If x, y , x + y, x ( y + m ) is PR then m = 1 . This can b e easily shown by fixing a large prime p , observing that the class r of x, y in F p m ust satisfy r = r + r , hence r = 0 , colouring with v alues of smo d and observing that 1 = smo d( y + m ) = smo d( m ) . All of the questions ab o ve also ha ve Ramsey versions, again in the sense of [7]. Problem 5.6. Study the v arian ts of Questions 5.1, 5.3 and 5.4 given by replacing “PR” b y “Ramsey PR”. 6 App endix: some standard pro ofs In this section w e give standard versions of some previously mentioned statements and of the pro of Theorem 1.7. W e b egin with a counterpart to F act 2.4. F act 6.1. The configuration f 1 ( x 1 , . . . , x k ) , . . . , f n ( x 1 , . . . , x k ) is partition regular if and only if there is w ∈ β N k \ N k suc h that f 1 ( w ) = . . . = f n ( w ) ∈ β N \ N . The follo wing lemma is the translation in standard terms of the fact that equiv alen t nonstandard points must b e at infinite distance. Let π 1 , π 2 : N 2 → N be the usual co ordinate pro jections. Lemma 6.2. Let w ∈ β N 2 b e suc h that π 1 ( w ) = π 2 ( w ) , and let f : N → N . If { ( x, y ) | | f ( x ) − f ( y ) | ≤ ℓ } ∈ w for some ℓ ∈ N then { ( x, y ) | f ( x ) = f ( y ) } ∈ w . Pr o of. Let X k : = { ( x, y ) | | f ( x ) − f ( y ) | = k } . Since S ℓ k =0 X k ∈ w , there exists k such that X k ∈ w . Assume for the sake of contradiction that k  = 0 , and hence k + 1 ≥ 2 . F or s = 0 , . . . , k let C s : = { x | f ( x ) ≡ s mo d ( k + 1) } . Since N = S k s =0 C s there exists s such that C s ∈ π 1 ( w ) = π 2 ( w ) . But then ( C s × C s ) ∈ w while ( C s × C s ) ∩ X k = ∅ , a con tradiction. 19 The standard reformulation of F act 2.3 is the following. Lemma 6.3. Let w ∈ β N 2 b e such that π 1 ( w ) = π 2 ( w ) . If there is n such that { ( x, y ) | 1 n x < y < nx } ∈ w , then for all ε > 0 we hav e { ( x, y ) | | y /x − 1 | < ε } ∈ w . Pr o of. F or a ∈ N , let L ( a ) b e the length of the binary representation of a , that is, the natural num b er such that 2 L ( a ) − 1 ≤ a < 2 L ( a ) . A straigh tforward computation shows that if 1 n x < y < nx then | L ( y ) − L ( x ) | ≤ L ( n ) . By this and assumption it follows that { ( x, y ) | | L ( y ) − L ( x ) | ≤ L ( n ) } ∈ w and so, b y Lemma 6.2, Λ : = { ( x, y ) | L ( x ) = L ( y ) } ∈ w . F or s ≥ 1 , let λ s : N → { 0 , 1 } b e the function where λ s ( a ) is the s -th digit in the binary expansion of a (w e agree that λ s ( a ) = 0 if s > L ( a ) ). Since { ( x, y ) | | λ s ( x ) − λ s ( y ) | ≤ 1 } = N × N ∈ w , again by Lemma 6.2 we obtain that Λ s : = { ( x, y ) | λ s ( x ) = λ s ( y ) } ∈ w . Pic k k ∈ N such that 2 − k < ε . W e will reach the conclusion by showing that [2 k − 1 , + ∞ ) 2 ∩ Λ ∩ k \ s =1 Λ s ⊆  ( x, y )         x y − 1     < ε  . (13) Indeed, since π 1 ( w ) = π 2 ( w ) is non-principal, [2 k − 1 , + ∞ ) 2 ∈ w ; b esides, we already noticed that Λ , Λ 1 , . . . , Λ k ∈ w . Let us finally show (13). Pic k any ( x, y ) in the intersection on the left hand side. Since ( x, y ) ∈ Λ , the num b ers x = P ℓ s =1 x s 2 ℓ − s and y = P ℓ s =1 y s 2 ℓ − s ha ve binary represen tations of the same length ℓ . Note that x s = y s for s = 1 , . . . , k , since ( x, y ) ∈ T k s =1 Λ s . If w e let z : = P k s =1 x s 2 ℓ − s , then x ′ : = x − z = P ℓ s = k +1 x s 2 ℓ − s < 2 ℓ − k and y ′ : = y − z = P ℓ s = k +1 y s 2 ℓ − s < 2 ℓ − k , and we obtain the desired inequalit y:     x y − 1     =     z + x ′ z + y ′ − 1     = | x ′ − y ′ | z + y ′ ≤ max { x ′ , y ′ } y ≤ 2 ℓ − k 2 ℓ = 1 2 k < ε. Standar d pr o of of The or em 1.7. The pro of of right to left is the same as in the one originally pro vided at page 9 (and do es not use nonstandard metho ds). Left to righ t, let f ( x, y ) = cx + dy and g ( x, y ) = q · y/x if q · y /x is an integer, and 1 otherwise. By F act 6.1 there exists w ∈ β N 2 \ N 2 suc h that π 1 ( w ) = π 2 ( w ) = f ( w ) = g ( w ) . As g ( w ) is nonprincipal, it follows that, for every n ∈ N , we hav e { ( x, y ) | y > nx } ∈ w . Therefore, for n = d + 1 , w e hav e that { ( x, y ) | 1 n y < cx + dy < ny } ∈ w and, b y Lemma 6.3 (applied to the pushforward of w along ( x, y ) 7→ ( y , cx + dy ) ), it follo ws that for all ε > 0 w e ha ve { ( x, y ) | | ( cx + dy ) /y − 1 | < ε } ∈ w ; since for every n we also hav e { ( x, y ) | y > nx } ∈ w , it follo ws that d = 1 . Fix a sufficien tly large prime p , let v b e the p -adic v aluation and smod the func- tion sending n to ( n/p v ( n ) mo d p ) ∈ F × p . Observ e that, as π 1 ( w ) = π 2 ( w ) , w e hav e (smo d ◦ π 1 )( w ) = (smod ◦ π 2 )( w ) . Moreov er, as π 1 ( w ) = f ( w ) and smo d has finite image, we ha ve { ( x, y ) | smod( x ) = smo d( cx + y ) } ∈ w . (14) Since { ( x, y ) | q y /x ∈ N ) } ∈ w and p is large, so v ( q ) = 0 , w e hav e { ( x, y ) | v ( y ) ≥ v ( x ) } ∈ w . W rite { ( x, y ) | v ( y ) ≥ v ( x ) } = { ( x, y ) | v ( y ) > v ( x ) } ∪ { ( x, y ) | v ( y ) = v ( x ) } . W e consider tw o cases dep ending on what is in w . If { ( x, y ) | v ( y ) > v ( x ) } ∈ w , then b ecause p ∤ c we hav e { ( x, y ) | smo d( cx + y ) = smo d( cx ) } ∈ w . Recall that smod( cx ) = smo d( c ) smod( x ) , hence { ( x, y ) | smod( x ) = smo d( c ) smo d( x ) } ∈ w 20 and it follows that smo d( c ) = 1 . Since p is sufficiently large, this entails c = 1 . Assume now that { ( x, y ) | v ( y ) = v ( x ) } ∈ w . Note that, as v ( q ) = 0 , this set equals { ( x, y ) | v ( q y/x ) = 0 } . Since f ( w ) = g ( w ) , we hav e { ( x, y ) | v ( y + cx ) = 0 } ∈ w . Because { ( x, y ) | smod( y ) = smo d( x ) } ∈ w , this implies c  = − 1 , as otherwise { ( x, y ) | v ( y + cx ) > 0 } ∈ w . 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