On Arithmetic Structures in Dense Sets of Integers 1. Ben Green 2

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1 On Aithmetic Stuctues in Dense Sets of Integes 1 Ben Geen 2 Abstact We ove that if A {1,..., N} has density at least log log N) c, whee c is an absolute constant, then A contains a tile a, a + d, a + 2d) with d = x 2 + y 2 fo some integes x, y, not both zeo. We combine methods of Gowes and Sákőzy with an alication of Selbeg s Sieve. The esult may be egaded as a ste towads establishing a fully quantitative vesion of the olynomial Szemeédi Theoem of Begelson and Leibman. 1. Intoduction. If one believes that mathematics is the study of attens then it is of no suise that the following esult of Szemeédi [16] is often egaded as one of the highlights of all combinatoics. Theoem 1 Szemeédi) Let α > 0 be a eal numbe and let k be a ositive intege. Then thee is N 0 = N 0 k, α) such that any subset A {1,..., N} of size at least αn contains an aithmetic ogession of length k, ovided that N N 0 k, α). Szemeédi s oof was long and combinatoial but just two yeas late Fustenbug ovided a comletely diffeent oof of Theoem 1 using egodic theoy. Fustenbug s methods have oved extemely amenable to genealisation, and Fustenbug himself oved the following esult. Theoem 2 Fix α > 0 and let A {1,..., N} have size αn. Then ovided N > N 1 α) is sufficiently lage one can find two distinct elements x, x A whose diffeence x x is a efect squae. As with all such alications of egodic theoy Fustenbug s aoach gave no bound on N 1 α). At about the same time Sákőzy [12] oved the same esult in a comletely diffeent manne. Sákőzy s agument took insiation fom a much ealie ae of Roth [10] in which Szemeédi s Theoem was oved fo ogessions of length 3. The method used is analytic in siit and does lead to an effective bound on N 1 α), albeit one which is a geat distance fom the conjectued tuth. The ae [12] was the fist in a seies of thee, and in the final ae [13] of this seies an analytic oof of a genealisation of Theoem 2 was outlined. This genealisation says that the squaes may, in the fomulation of that theoem, be elaced by the set {d) : d N} whee is any olynomial which mas N to itself and has an intege oot. To see that some 1 Mathematics Subject Classification MSC): 11B25 05D10, 11N36). 2 The autho is suoted by an EPSRC eseach gant, and has also enjoyed the hositality of Pinceton Univesity fo some of the eiod duing which this wok was caied out. 1

2 estiction on the olynomial is necessay fo such a esult to hold, we invite the eade to constuct a set with density 1 3 containing no diffeence of the fom x Since the late 1970s thee have been seveal significant advances in ou undestanding of these and elated questions. Fistly we mention the esult of Begelson and Leibman fom 1996 [1], which is an examle of how fa Fustenbug s egodic-theoetic methods have been able to take us. Theoem 3 Begelson-Leibman) Fix α > 0 and let A {1,..., N} have size αn. Let 1,..., be olynomials with N) N and 0) = 0. Then ovided N > N 2 α) is lage enough exactly how lage will deend on the olynomials i as well as on α) we can find a, d N fo which all of the numbes a + i d) lie in A. This extends both Theoem 1 and Theoem 2 and imlies, amongst othe things, that dense subsets of the integes contain abitaily long aithmetic ogessions whose common diffeence is a non-zeo squae. Secondly thee is a esult of Gowes [4], which gives the fist bounds fo Szemeédi s Theoem. Theoem 4 Gowes) Let k > 0 be an intege. Then thee is an effectively comutable constant ck) such that any subset of {1,..., N} of density at least log log N) ck) contains a k-tem aithmetic ogession. Gowes agument takes insiation fom Roth s ae [10]. As we have aleady emaked, Sákőzy s methods also bea some esemblance to those of Roth. It is theefoe natual to ask whethe Gowes techniques can be adated to give bounds fo questions elated to the olynomial Szemeédi Theoem. In 3 we give a new vaiant on Sákőzy s oof of Theoem 2 which we believe to be substantially easie to undestand than the oiginal though it gives a wose bound fo N 1 α)). Pehas moe imotantly we demonstate that this agument can be made to fit almost entiely into the geneal methodology of [4], which we shall outline below. It should be ointed out that in 1985 Sinivasan [15] gave a still diffeent agument which seems to be athe simle than that of Sákőzy, but moe comlex than the one we shall give hee. The est of the ae is devoted to a oof of the following esult. Theoem 5 Thee is a constant c such that any subset of {1,..., N} of density at least log log N) c contains a 3-tem aithmetic ogession whose common diffeence is ositive and of the fom x 2 + y 2. The oof of this esult involves finding a mutual genealisation of the methods of Gowes and Sákőzy, togethe with a slightly suising use of the Selbeg Sieve. The eade will find it had to undestand this section unless she has a woking knowledge of the methods of Gowes, such as can be obtained by eading [3] o, even bette, ats of [4]. 2

3 It would be a sizeable undetaking to summaise those aes in any detail hee, and indeed thee seems little oint in doing so. What we will do is offe a cude outline of the to-level stuctue of Gowes oof of Szemeédi s Theoem. All ou aguments in this ae will have this boad stuctue at thei heat. Suose then that we have a set A {1,..., N} of size δn which contains no aithmetic ogession of length k. Set A 0 = A, δ 0 = δ and N 0 = N. At the ith stage of ou agument we will have a set A i {1,..., N i } with density δ i which contains no aithmetic ogession of length k. The fact that A i contains no ogessions of length k imlies that A i is non-andom in a cetain athe ecise sense involving cetain Fouie coefficients being lage. This means that A i does not satisfy a oety which is ossessed by almost all sets of density δ i. In Gowes oof this oety is known as k 2)-unifomity. If a set is not k 2)-unifom then, by a long and comlicated agument, it is ossible to show that A i has density at least δ i + ηδ i ) on a faily long ogession P, whee η is an inceasing function of δ i. Define A i+1 to be A i P and escale so that P has common diffeence 1. Set δ i+1 = δ i + ηδ i ) and N i+1 = P. Iteating this agument leads to an effective vesion of Szemeédi s Theoem since afte a finite numbe of stes the density δ i will exceed 1, a contadiction. 2. Notation and Basic Concets. We shall make substantial use of Fouie analysis on finite cyclic gous, so we would like to take this ootunity to give the eade a swift intoduction. If nothing else this will seve to claify notation. Let N be a fixed ositive intege, and wite Z N fo the cyclic gou with N elements. Let ω denote the comlex numbe e 2πi/N. Although ω clealy deends on N, we shall not indicate this deendence in the est of the ae, tusting that the value of N is clea fom context. Let f : Z N C be any function. Then fo Z N we define the Fouie tansfom ˆf) = fx)ω x. x Z N We shall eeatedly use two imotant oeties of the Fouie tansfom. The fist is Paseval s identity, which states that if f : Z N C and g : Z N C ae two functions then N fx)gx) = ˆf)ĝ). x Z N 3 Z N

4 The second is the inteaction of convolutions with the Fouie tansfom. If f, g : G C ae two functions on an abelian gou G we define the convolution f g)x) = y G fy)gx y). It is easy to check that f g)ˆ) = ˆf)ĝ). One moe iece of notation: we will often use the same lette to denote both a set and its chaacteistic function. It is now ossible to fomally intoduce the concets of unifomity and quadatic unifomity, the only tyes of unifomity that will featue in this ae. These coesond to what we called 1-unifomity and 2-unifomity in ou ealie outline. Let A Z N be a set of size αn and let f = A α be its balanced function. We say that A is η-unifom if ˆf ηn. To define quadatic unifomity we need some exta notation. If f : G C is any function on an abelian gou we wite f; h)x) = fx)fx h). Let f be the balanced function of A. Then A is said to be quadatically η-unifom if f; h)ˆ 2 ηn 3. 1) h These definitions ae vey simila, though not comletely identical, to those in [4]. Fo a detailed discussion of the basic oeties of unifomity and quadatic unifomity the eade should consult [4]. It tuns out that being quadatically unifom is a stonge equiement than being unifom and so knowing a set fails to be quadatically unifom gives less infomation than knowing it fails to be unifom). We will need the following quantitative statement of this fact late on. Poosition 6 If A is quadatically η-unifom then it is η 1/4 -unifom. Poof By 1) we have f; h)ˆ0) 2 ηn 3. Witing this statement out in full gives fa)fb)fc)fd) ηn 3 h a+b=c+d 4

5 which imlies, by Paseval s theoem, that ˆf) 4 ηn 4. It follows immediately that ˆf 4 ηn 4, which imlies the oosition. 3. A Shot Poof of Sákőzy s Theoem fo Squaes. In this section we ove Theoem 2 using an iteative agument of the tye outlined above. Fo obvious easons we say that A has a squae diffeence if thee ae distinct elements x and x in A with x x a squae. Let us ecall Theoem 2. Theoem 2 Fix α > 0 and let A {1,..., N} have size αn. Then ovided N > N 1 α) is sufficiently lage A has a squae diffeence. As the fist at of ou agument we shall show that a set A {1,..., N} with size αn which contains no squae diffeence fails to be η-unifom fo some easonably lage η. In oving this statement we shall use just one fact about the squaes, an elementay oof of which may be found in [8]. Lemma 7 Let 6 n) denote the numbe of odeed sextules a 1,..., a 6 ) with a a 2 6 = n. Then n 2 6 n) 40n 2. In fact we shall have no use fo the lowe bound in the lemma, but have included it to emhasise that 6 is comaable to n 2. Now let S be the set of non-zeo squaes less than N/2, and let B = A [0, N/2]. Regad S, A and B as subsets of Z N. Assume without loss of geneality that B αn/2. If A A contains no squae then cetainly Bx)Ax + d)sd) = 0, x,d whee the sums ae ove Z N. It follows easily fom Paseval s Theoem and the tiangle inequality that Ŝ) ˆB) Â) Ŝ0) ˆB0) Â0) 1 4 α2 N 5/2. 2) 0 The left-hand side hee is at most su Â) 1/6 0 Ŝ) ˆB) Â) 5/6 5

6 which, by Hölde s inequality, is at most su Â) 1/6 0 Ŝ) 12 ) 1/12 ˆB) 2 ) 1/2 Â) 2 ) 5/12. 3) Now by Paseval again) Ŝ) 12 is equal to N x R 6x) 2, whee R 6 x) is the numbe of solutions to a a 2 6 xmodn) with a 2 i 0, N/2]. An easy execise using Lemma 7 shows that R 6 x) 600N 2 fo all x, which gives the bound Ŝ) N 6. 4) Paseval also gives Â) 2 N 2 and ˆB) 2 N 2. Substituting these and 4) into 3) gives that thee is 0 fo which In othe wods, A is non-unifom. Â) 2 30 α 11/2 A. 5) To comlete ou oof of Theoem 2 by the iteation method we ae going to use 5) to show that A has inceased density on a squae-diffeence AP. If we ass to such a subogession and then escale this subogession to have common diffeence 1, the esulting set A will still not contain a squae diffeence. To find a subogession of the desied tye equies a futhe standad fact about the squaes, which is a quantitative vesion of the fact that the squaes ae a Heilbonn Set see [7]). As noted in [4] it is athe difficult to find a ecise statement of this esult in the liteatue. Fotunately howeve we can use Lemma 5.5 of [4] to simly wite down the next lemma. Lemma 8 Let a Z N, let t N, and suose that t Then thee is t such that 2 a t 1/16 N. Now let be such that Â) is lage. Put t = N 1/4 in Lemma 8 and let N 1/4 be such that 2 N 127/128. Let B be the aithmetic ogession 2, 2 2,..., L 2 whee L = 1 N 1/ One calculates ˆB) ) L su 1 1 ω 2 x x=1,...,l ) L 1 2π 2 L N L/2. Fom this and the fact that Â) 2 30 α 11/2 A we get that N x A B + x) 2 = Â) 2 ˆB) α 11) A 2 L 2. 6) 6

7 Now cetain of the tanslates B +x have the athe unfotunate oety of slitting into two smalle ogessions when we unavel Z N to ecove {1,..., N}. We shall call these values of x bad, and denote the set of good values by G. Now N 1/4 and so B has diamete less than N 2/3. It follows that thee ae at most N 2/3 bad values of x, and so thei total contibution to N x A B + x) 2 does not exceed L 2 N 5/3. Assuming that α 32N 1/39 which it cetainly will be) one sees fom 6) that N x G A B + x) α 11) A 2 L 2. The left hand side hee is at most N su A B + x) A L, x G fom which we deduce that thee is x G fo which A B + x) α α 12) B. We have deduced, fom the assumtion that A A does not contain a squae and that N , that A has density at least α α 12 on a subogession with length at least 1 20 N 1/128 and squae common diffeence. Iteating this agument leads to the following esult. Poosition 9 Thee is a constant C such that, if A is a subset of {1,..., N} with density at least Clog log N) 1/11, then A contains two elements a, a with a a a non-zeo squae. In ode to ove Sákőzy s esult in the simlest ossible manne we have not woied too much about sacificing the quality of the bound obtained. Thee is a vaiation of the above agument in which one shows that A is what I call aithmetically non-unifom, which means that some Â) is lage with aoximately equal to a ational with small denominato. This allows one to efom the iteation moe efficiently, and doing this gives a bound of fom log N) c in Poosition 9. I intend to discuss this and elated mattes in a futue c log log log log N ae. The cuent best known bound of log N) fo this oblem is due to Pintz, Steige and Szemeédi [9] and makes use of some athe involved Fouie aguments. Thee is still a massive ga in ou knowledge, and I cannot esist closing this section with the following oen oblem. Poblem 10 Let ɛ > 0 and let N > N 0 ɛ) be sufficiently lage. Does thee exist a set A {1,..., N} with A N 1 ɛ, such that A does not contain two elements that diffe by a squae? The best that is known is that one can take ɛ = 0.267, a esult due to Ruzsa [11]. 4. APs with Common Diffeence x 2 + y 2. In this section we tun ou attentions to the 7

8 main business of this ae, a oof of Theoem 5. Theoem 5 Thee is a constant c such that any subset of {1,..., N} of density at least log log N) c contains a 3-tem aithmetic ogession whose common diffeence is non-zeo and of the fom x 2 + y 2. Ou oof of Theoem 5 will be by the iteation method, and as such will fall into two ats. In the fist at, containing most of ou oiginal wok on the oblem, we show that a quadatically unifom set contains oughly the exected numbe of ogessions a, a + d, a + 2d) with d = x 2 + y 2. The second at of the oof follows [4] extemely closely. Ou objective thee is to show that a set which fails to be quadatically unifom has inceased density on a long subogession with squae common diffeence. We shall stat ou teatment in quite a geneal setting. Let D be a subset of N, and fo N N egad D N = D {1,..., N} as a subset of Z N in the natual way. Suose that fo some k N we have ˆD N ) 2k C D N 2k and D 2N C D N, whee C is indeendent of N. Then we shall say that D is unifomly k-dense. The obvious non-tivial examle in view of ou ealie discussions is the set S of squaes, which is unifomly 6-dense by 4). Ou nomenclatue is non-standad, but quite convenient. Much of ou late wok deends on anothe instance of this henomenon. Poosition 11 The set E of imes of fom 4k + 1 is unifomly 2-dense. We will ove this oosition in a numbe of stages. We begin with a bief esumé of the esults fom Sieve Theoy we will need both hee and late on. The standad efeence fo this subject is [6] but in ou unbiased oinion a good way to undestand the necessay backgound is to ead [5], which has been secially udated fo this uose. We will only be concened with sieving olynomial sequences, which is the simlest situation coveed by the Selbeg Sieve. Let h be a olynomial with intege coefficients, and let A denote the sequence {h1),..., hn)}. Let P denote the set of imes. Then the Selbeg Sieve gives ue bounds fo SA, P, z), which is defined to be the numbe of x A which ae not divisible by any ime z. This ue bound is given in tems of a function ω defined at imes to be the numbe of elements in {h1),..., h)} which ae divisible by. To ut it anothe way, the ootion of elements of A which ae divisible by is oughly ω)/. The key esult we shall equie is the following, which is Theoem 11 in [5] we should also note that it can be ead out of [6]). 8

9 Theoem 12 Suose that ω) C fo all imes. Then SA, P, N 1/16C ) N 1 ω) ), whee the imlied constant deends only on C. N 1/16C Poosition 13 Let n N. Then the numbe of eesentations of n as the sum of two elements of E, 2 E, n) satisfies 2 E, n) n ). log n) 2 Poof Clealy 2 E, n) 2 P, n). Conside the olynomial hx) = xn x) and let A = {h1),..., hn)}. We wish to count the numbe of x n fo which x and n x ae both ime. Fo such x we eithe have x n 1/2, x n n 1/2 o else hx) has no ime facto less than n 1/2. It follows that 2 P, n) is bounded above by 2n 1/2 + SA, P, n 1/2 ). It is easy to see that, in the notation of ou otted intoduction to sieve theoy, one has ω) = 2 fo all excet when n, in which case ω) = 1. Thus by Theoem 12 one has 2 E, n) n 1/2 + n 1 2 ) 1 2 ) ). 7) n 1/32 n n 1/32 n Now ecall that ) 1 m 1 1 log m and that Amed with these two facts one sees fom 7) that 2 E, n) n log n) 2 n 1 λ 2 ) ) conveges fo any eal λ. as claimed. This is of couse a vey standad deduction fom the Selbeg Sieve, but we have included it to ensue that the eade is hay with ou notation. It tuns out to be extemely convenient to wite ξn) fo the quantity ) n aeaing hee. In a shot while we will use Poosition 13 to show that E is unifomly 2-dense. Befoe doing this howeve it is necessay to give a cude estimate fo the moments u N ξu)s of ξ. Lemma 14 Let s 1 be eal. Then u N ξu) s 2 2s22s N. 9

10 Poof Fist obseve that fo any x 1 one has 1 + x) s s x. Secondly, if 1,..., ae distinct imes then fo any C one has i=1 C = C i i i C 2 C C 1 C2 C C2 i i i >C 2 i >C 2 i=1 1 i. Thus ξu) s = ) s u N u N u N u u ) 1 + 2s 2 s22s u N d u 1 d, which is at most 2 s22s N d N d 3/2. This imlies the esult. Poof of Poosition 11 Regad E N = E {1,..., N} as a subset of Z N, as we must do to even make sense of what it means to be unifomly 2-dense. It is easy to see that ÊN) 4 = N Ea)Eb)Ec)Ed), a+b=c+d whee the equation a + b = c + d is taken in Z N. This is clealy at most N n N 2 E, n) + 2 E, N + n)) 2, which by Poosition 13 is at most a constant times N 3 log N) 4 n N ξn) + ξn + n)) 2. The sum, by the Cauchy-Schwaz inequality, is at most 4 n 2N ξn)2, a quantity which we know to be ON) by Lemma 14. Thus ÊN) 4 ) 4 N. log N Since E N N/2 log N it follows that E is indeed unifomly 2-dense. If A {1,..., N} is a set of density α then we wite f = A α fo its balanced function. Wite B = A {1,..., N/3}, let β be the density of B and let g = B β be its balanced function. Finally if D is any set then we say that an aithmetic ogession x, x + d, x + 2d) with d D is a D-ogession. 10

11 Poosition 15 Suose that A does not contain a D-ogession, whee D is unifomly k-dense. Then eithe su Â) α2k N 8) 0 o else we have gx)fx + d)fx + 2d) α 3 N U. 9) x d U whee the sum is ove Z N and U = D {1,..., N/3}. Poof To begin with we show that the conclusion holds with oom to sae if β is much smalle than exected. Suose that β α/12, and let I be the chaacteistic function of { N/12,..., N/12} Z N. Then 6 N Ax)I I)x + N/6) βn αn/12. x Taking Fouie tansfoms gives ω N/6 Â) Î) 2 αn 3 /72, which imlies by the tiangle inequality that Â) Î) 2 αn 3 /72. 0 Howeve Paseval s identity gives Î) 2 = N 2 /6, fom which it follows immediately that su Â) αn/12. 0 Assume now that β α/12. It is easy to see that thee is no modula ogession of fom x, x + d, x + 2d), d U, in B A A, and so we have Bx)Ax + d)ax + 2d) = 0. x Z N d U Witing A = f + α and B = g + β we may exand this as a sum of eight tems. Of these we have a tem T = x d U gx)fx + d)fx + 2d), a tem α2 βn U and thee tems which ae identically zeo. If T α 2 βn U /4 then we ae done. Failing this the tiangle inequality imlies that one of the othe thee tems must be at least α 2 βn U /4, which in tun is not less than α 3 N U /48. These othe thee tems ae vey simila to one anothe, and bushing a little wok unde the caet we suose without loss of geneality that α gx)fx + d) α 3 N U /48. 10) x d U 11

12 The LHS may be witten in tems of Fouie coefficients as αn 1 ĝ) ˆf )Û). This sum may be estimated using Hölde s Inequality exactly as in 3. It is at most αn 1 su ˆf) 1/k ) 1/2k ) 1/2 Û) 2k ĝ) 2 ) ˆf) k 1)/2k 2 11) To deal with this obseve that since D is unifomly k-dense we have Û) 2k = N Ua 1 )... Ua k )Ub 1 )... Ub k ) a 1 + +a k =b 1 + +b k N D N a 1 )... D N a k )D N b 1 )... D N b k ) a 1 + +a k =b 1 + +b k = ˆD N ) 2k D N 2k U 2k. Using Paseval s identity on the othe two factos in 11) we can bound that exession above by a constant multile of su ˆf) 1/k N 1 1/k U. It follows fom 10) that su ˆf) α 2k N, and we ae done. Suose now that we have a set A {1,..., N} with density α which contains no D- ogessions, whee D is the set of sums of two squaes. A classical numbe-theoetic fact allows us to make the key obsevation of this ae, namely that A does not contain any E ogessions whee E, as befoe, is the set of imes of the fom 4k + 1). Using Poositions 11 and 15, we have the following. Poosition 16 Let A {1,..., N} have density α, and suose that A does not contain a tile a, a + d, a + 2d) with d = x 2 + y 2. Then eithe o whee V = E {1,..., N/3}. su Â) α4 N 12) 0 gx)fx + d)fx + 2d) α3 N 2 x d V log N. 13) 12

13 If 12) holds then futhe ogess is comaitively easy, so we assume fo the moment that 13) is satisfied. The Cauchy-Schwaz inequality togethe with the fact that g 1 gives that 2 fx + d)fx + 2d)V d) α6 N 3 log N). 2 x d Multilying out, eaanging and changing the summation vaiables, this imlies that f; h)x) f; 2h)x + d) V ; h)d) α6 N 3 14) log N) 2 h x d We ae going to use this to show that, fo many h, f; h) has a lage Fouie coefficient. We will do this by alying the Selbeg Sieve again to show that V ; h) is unifomly 2-dense on aveage, a concet we shall not define ecisely. Poosition 17 Let h n) denote the numbe of ways of exessing n as a diffeence of 2 elements of V ; h). Then h n) N ξh) 2 ξn) 2 ξn + h)ξn h). log N) 4 ξn, h)) 3 Poof We bing the Selbeg Sieve to bea on this by obseving that h n) is at most the numbe of x N fo which x, x n, x h and x n h ae all ime. With the excetion of at most 8N 1/2 of these x the olynomial hx) = xx n)x h)x n h) has no ime facto less than N 1/2. Witing A = {h1),..., hn)} this imlies that h n) 8N 1/2 + SA, P, N 1/2 ). We would clealy like to aly Theoem 12, but fist we must think about ω). Suose that 3. Then it is easonably easy to see that ω) = 4 unless one of the following fou ossibilities occus: i) n; ii) h; iii) n + h) and iv) n h). Futhemoe these ossibilities ae mutually exclusive unless divides both n and h, in which case they all occu. If they do all occu then ω) = 1. If i) o ii) occus then ω) = 2, and if iii) o iv) occus then ω) = 3. If = 2 the behaviou is moe subtle, but we will not concen ouselves with this as each individual ime only contibutes a bounded multilicative facto to the sieve estimate of Theoem 12. Theoem 12 with C = 4) cetainly alies to this situation, then, and with a modicum of effot one can veify that the key quantity N 1 ω) ) N 1/64 13

14 is equal, u to a oduct of tems of the fom 1 λ 2 ), to the delightful exession 1 4 ) 1 2 ) ) 1 N 1/64 n n+h) h 1 1 ) 1 n h) 1 1 ) ) 3 n,h) whee the final five oducts ae also constained to be ove N 1/64. Since an intege less than N cannot have moe than 64 ime factos N 1/64, this estiction can be emoved at the exense of intoducing anothe bounded multilicative constant. The esulting exession is easily seen to be bounded above by a constant multile of the one aeaing in the statement of the oosition. Poosition 18 Fo any h we have n N whee the imlied constant is indeendent of h. Poof By Poosition 17 we have n N h n) 2 N 3 log N) 8 ξh)4, h n) 2 N 2 log N) 8 ξh)4 ξn) 4 ξn + h) 2 ξn h) 2. n N By Hölde s Inequality the sum ove n is at most n ξn)8. We ae theefoe done by Lemma 14. The next esult claifies the sense in which V ; h) is, on aveage, unifomly 2-dense. In the following oosition V ; h) is egaded as a subset of Z N. Poosition 19 V ; h)ˆ) 4 N 4 log N) 8 ξh)4. 15) Poof Immediate fom Poosition 18 and the fact that V, h) is suoted in an inteval of size N/3 so that thee is no oblem with modula addition not being the same as odinay addition). Now we come to use 14). We shall use it to teat each h seaately, which we do by noting that fom 14) follows f; h)x) f; 2h)x + d) V ; h)d) γh) α6 N 2 log N), 2 x d 14

15 whee h γh) = N. Taking Fouie Coefficients, this imlies that Alying Hölde gives su f; h)ˆ) 1/2 f; h)ˆ) f; 2h)ˆ ) V ; h)ˆ) γh) α6 N 3 log N). 16) 2 γh) α6 N 3 log N) 2. f; h)ˆ) 2 ) 1/4 f; 2h)ˆ) 2 ) 1/2 V ; h)ˆ) 4 ) 1/4 The fist two backeted exessions may be bounded above using Paseval, and the thid is subject to the ue bound 15). One gets Thus thee is a function φ : Z N Z N such that su f; h)ˆ) γh)2 α 12 N. 17) ξh) 2 f; h)ˆφh)) 2 h α 24 N 2 h γh) 4 ξh) 4. This would imly that A fails to be quadatically unifom if we could show that the sum hee is N. To do this, we ecall that h γh) = N and use Hölde, getting ) 3 γh) 4 N 4 ξh) 4/3. ξh) 4 This is indeed N by Lemma 14. h We have now shown that if 14) holds then A is not quadatically Cα 24 -unifom fo some C. We also know that if A does not contain any E-ogessions then eithe 12) o 14) must hold. Howeve 12) is just the statement that A is not Cα 4 -unifom fo some C. Thus by Poosition 6 we may incooate eveything we have done so fa into the following. Poosition 20 Suose that A {1,..., N} has density α yet does not contain a ogession x, x + d, x + 2d) with d a ositive sum of two squaes. Then A is not quadatically Cα 24 -unifom fo some C. To sell it out, the conclusion of this oosition imlies that f; u)ˆφu)) 2 α 24 N 3 u 15 h

16 fo some function φ : Z N Z N. 5. Inceasing the Density on a Secial Subogession. Sets that fail to be quadatically unifom have an inteesting stuctue, as was established by Gowes [4] in the couse of oving Szemeédi s Theoem fo ogessions of length 4 a slightly weake esult was established in [3]). In that ae a esult along the following lines is oved. Theoem 21 Gowes Invese Theoem) Suose that A {1,..., N} has density α and that A fails to be quadatically C 1 α m 1 -unifom. Then thee is an aithmetic ogession L of length L N C 2α m 2 on which A has density at least α + C 3 α m 3, whee C 2, C 3, m 2 and m 3 deend only on m 1 and C 1. The main esult of this section is a vesion of Theoem 21 in which L has squae common diffeence. Unfotunately we have found it necessay to go a fai way into the detailed wokings of Gowes agument in ode to obtain this modification. Theefoe in ode to fully undestand this section the eade will need to be convesant with Chates 6, 7 and 8 of [4]. We shall equie one additional ingedient, which is a vesion of the following simultaneous aoximation esult of Schmidt [14]. Theoem 22 Schmidt) Let 1,..., h Z N and let t N. Suose that t N 0 h) is sufficiently lage. Then thee is t such that 2 i t 1/3h2 N fo all i, 1 i h. Unfotunately thee does not seem to be any lace in the liteatue whee an exlicit value of N 0 h) is deived. It would be ossible to wok though the oof in [14] and deive such an exlicit value which would obably not be too lage) but we will ove ou own, substantially weake, vesion of Theoem 22 with exlicit constants. Poosition 23 Let 1,..., h Z N and let t N. Suose that t 2 27h+129. Then thee is t such that 2 i t 2 7h+6) N fo all i, 1 i h. Poof We make extensive use of Lemma 8. Choose 1, 2,... inductively so that i t 2 7i+1) 18) and 2 i 2 i i t 2 7i+5) N 19) fo each i = 1,..., h. Let = h. It is easy to check that t. Futhemoe 2 i h 2 2 h i+1 2 i i t 2 7i+6) t 2 7i+5) N t 2 7h+6) N 16

17 as equied. The eeated alications of Lemma 8 equie cetain things to be sufficiently lage. The most difficult condition to satisfy is that aising fom the last ste in ou inductive constuction, whee we equie t 2 7h+1) This is whee the estiction in the oosition comes fom. Let us assume then that f : Z N C has f 1 and that f; k)ˆφk)) 2 2βN 3 20) k fo some function φ : Z N Z N. We stat with a tivial deduction fom this, which is oved by a simle aveaging agument: thee is a set B Z N with B βn and f; )ˆφ)) β 1/2 N 21) fo all k B. We have then f; k)ˆφk)) 2 β 2 N 3. 22) k B If K Z N and η 0, 1) wite BK, η) fo the set of all n Z N such that nk ηn fo all k K. We call such a set a Boh Neighbouhood. The following may be deduced fom Poosition 6.1, Coollay 7.6, Lemma 7.8 and Coollay 7.9 of [4]. This deduction is logically extemely staightfowad, but just a touch messy when it comes to actually woking out values. Poosition 24 Let δ = β Thee is a set B B with B δn, and a set K Z N with K 16δ 2, such that the following is tue. If m is any ositive intege and d BK, δ/100m) then thee is c such that φx) φy) = cx y) wheneve x, y B and x y belongs to the set {jd : m j m}. Wite P 0 fo the aithmetic ogession {d, 2d,..., md}. Choose a tanslate P = P 0 + z fo which P B δm, and let H = P B. If x and y lie in H then x y is in {jd : m j m} and so φ H is the estiction of a linea function fom Z N to itself, by Poosition 24. We will soon find ouselves dealing with some easonably lage numbes, and it is convenient to have a shothand notation fo them. If n 1 is a aamete then we wite C 0 n) fo any olynomial in n. C 1 n) will mean a function of tye 2 n), whee is a olynomial, and finally C 2 n) will be a function of tye 2 2n). We will feel at libety to use these symbols seveal times, sometimes in the same fomula, to denote diffeent functions. Poosition 23 allows us, ovided K and δ/100m satisfy a cetain inequality, to conclude 17

18 that d may be taken to be a small squae numbe. Take t = N 1/4 in Poosition 23 and ecall that K 16δ 2. Then thee is N 1/4 such that 2 BK, δ/100m) ovided that δ 100m N 1/C 1δ 1 ) and that N C 2 δ 1 ). Fo N geate than some C 2 δ 1 ) we can ick m = N 1/C 1δ 1) so that this is satisfied by a colossal magin. Putting eveything togethe, and ecalling that δ is β 5821, we have the following. Poosition 25 Suose that 22) holds and that N C 2 β 1 ). Then thee is a ogession P Z N with common diffeence 2, whee N 1/4, and length at least N 1/C 1β 1), with the following oety. Thee is H P B with H β 5821 P and λ, µ Z N such that φs) = λs + µ fo all s H. Let us now make a deduction fom Poosition 8.1 of [4]. This is done using the evious oosition and 21). Poosition 26 Suose that 20) holds and that N C 2 β 1 ). Then thee is an aithmetic ogession P with common diffeence 2, whee N 1/4, and length at least N 1/C 1β 1) and quadatic olynomials ψ 0, ψ 1,..., ψ N 1 such that fz)ω ψsz) β 5822 N P. s z P +s Befoe stating and oving the next Lemma we need a vesion of Lemma 8 fo fouth owes. Once again we can simly ead it fom [4]. Lemma 27 Let a Z N, let t N and suose that t Then thee is t such that 4 a t 1/128 N. Poosition 28 Let ψx) = ax 2 + bx + c be a quadatic olynomial with coefficients in Z N. Let L N, and let P Z N be an aithmetic ogession of length L with squae common diffeence. Let W L Then thee is a atition of P into subogessions R 1,..., R m, with R i having length between W and 2W and squae common diffeence, such that DiamψR i )) L 2 20 N fo all i. Poof We may escale and assume that P = {1,..., L} with no loss of geneality. Fo any x 0, λ, d we have ψx 0 + λd 2 ) ψx 0 ) = aλ 2 d 4 + 2ax 0 + b)λd 2. 18

19 By Poosition 27 we may choose d L 1/8 such that ad 4 L 2 10 N. Patition {1,..., L} into ogessions P 1,..., P l with common diffeence d 2 and lengths lying between L 2 13 and L The diamete of ψ on P i then satisfies DiamψP i )) L 2 11 N + Diamθ i P i )), 23) whee θ i is a linea olynomial deending on i. Fix i, and fo ease of notation escale P i to {1,..., K} by the squae scaling facto d 2 ) whee K L Suose that θ i x) = x + s unde this escaling. Clealy θ i x 1 + µe 2 ) θ i x 1 ) = µe 2. By Lemma 8 we may choose e K 1/4 so that e 2 K 1/64 N. Divide P i into futhe subogessions E j with common diffeence e 2 and lengths lying between K 1/256 and K 1/128. On these subogessions we will have Diamθ i E j )) K 1/128 N. Do this fo each i, escale the esultant ogessions by the facto d 2, and then efom a futhe subdivision to satisfy the technical condition on the lengths of the R i in the statement. Recalling 23) we get the esult. Call an aithmetic ogession Q Z N nice if it does not wa in Z N, by which we mean that the length LQ) and the common diffeence dq) satisfy LQ)dQ) N. Obseve that the ogession P found in Poosition 25 is nice, because we constucted it to have small common diffeence. We obseve that any tanslate of a nice ogession is nice, as is any subogession. Let us now combine Poositions 26 and 28 in the obvious way. Poosition 29 Suose that 20) holds and that N C 2 β 1 ). Then thee is W = N 1/C 1β 1), nice aithmetic ogessions R i,s, 1 i m, 1 s N, each having length between W and 2W and squae common diffeence, and quadatic olynomials ψ 0,..., ψ N 1 such that i s β 5822 R i,s. 24) i,s z R i,s fz)ω ψsz) Futhemoe evey oint of Z N lies in the same numbe of R i,s and Diamψ s R i,s )) W 2 N. 25) Let us use this oosition immediately. Fo each i, s choose z 0 R i,s. Then using 25) and 19

20 the fact that f 1 we have fz) ω ψsz) ω ) ψsz0) z R i,s 2πW 2 fz) z R i,s 4πW β 5822 R i,s ovided that W C 0 β 1 ). This will be the case if N is at least some suitably lage C 2 β 1 ) though N might need to be lage than it had to be befoe). It now follows fom 24) that fz) i s z R i,s β 5822 R i,s. i,s We ae now almost home. The fact that evey oint lies in the same numbe of R i,s imlies that fz) = 0, i s z R i,s and so fz) i s z R i,s + fz) β 5822 R i,s. z R i,s i,s Fom this it follows immediately that fz) β 5822 R fo some R = R i,s. z R The one emaining obstacle is the fact that whilst R is nice, it might still staddle 0 in Z N. Thus when Z N is unwaed, R might become two aithmetic ogessions R 1 and R 2. Both of these will still have squae common diffeence, so we can be hoeful of exticating ouselves. The following lemma, oved by a simle aveaging agument, coves the situation. Lemma 30 Suose that f 1 and that z R fz) η R, whee R is a nice aithmetic ogession in Z N. Suose that R = R 1 R 2, whee neithe R 1 no R 2 staddles 0. Then, fo some i {1, 2}, R i has length at least η R /3 and z R i fz) η R i /3. Finally, we can deduce the following vaiant of Gowes Invese Theoem. 20

21 Theoem 31 Let f : Z N C have f 1, and suose that thee is a function φ : Z N Z N such that f; )ˆφ)) 2 βn 3. Suose that N C 2 β 1 ). Then thee is an aithmetic ogession S Z with squae common diffeence and length at least N 1/C1β) such that fx) β 5822 S. x S The sudden change in the owes of two hee comes fom the fact that we elaced 2β in 20) with β. By the main esult of 4 we have shown that if A contains no tiles a, a + d, a + 2d) with d = x 2 + y 2 then A has density α + Ω α ) on a ogession of size N 1/C 1α 1) with squae common diffeence. The new set A must have the same oety - it cannot contain an aithmetic ogession with sum-of-two-squaes common diffeence. How often can this agument be iteated? Afte O α ) iteations we will have eached density 1, by which time the length of ou subogession will still be of the fom N = N 1/C 1α 1). In ode fo the iteation ste to wok we equie that N C 2 α 1 ) at all times. We theefoe have a contadiction ovided that N 1/C 1α 1) C 2 α 1 ), and it is easy to see that this is imlied by some bound α log log N) c. This concludes the oof of Theoem 5. Accoding to my calculations, c = 10 6 is admissible. Refeences [1] Begelson, V. and Leibman, A. Polynomial Extensions of Van de Waeden s and Szemeédi s Theoems, J. Ame. Math. Soc ), no 3, [2] Fustenbug, H. Egodic behaviou of diagonal measues and a theoem of Szemeedi on aithmetic ogessions, J. Analyse. Math ) [3] Gowes, W.T. A New Poof of Szemeédi s Theoem fo Pogessions of Length 4, Geom. Funct. Anal ), no. 3, [4] Gowes, W.T. A New Poof of Szemeédi s Theoem, to aea in Geom. Funct. Analysis. Available at htt:// wtg10. [5] Geen, B. J. Notes on Sieve Theoy: The Selbeg Sieve, available at htt:// bjg23/exos.html. 21

22 [6] Halbestam, H. and Richet, H. -E. Sieve Methods, Academic Pess [7] Montgomey, H.L. Ten Lectues on the Inteface between Analytic Numbe Theoy and Hamonic Analysis, CBMS Regional Confeence Seies in Mathematics 84, AMS [8] Nathanson, M.B. Elementay Methods in Numbe Theoy, Singe [9] Pintz, J., Steige, W.L. and Szemeédi, E. On Sets of Natual Numbes whose Diffeence Set Contains No Squaes, J. London Math. Soc. 2) ) [10] Roth, K.F. On Cetain Sets of Integes, J. London Math Soc ) [11] Ruzsa, I.Z. Diffeence Sets Without Squaes, Peiod. Math. Hunga ), no. 3, [12] Sákőzy, A. On Diffeence Sets of Sequences of Integes I, Acta. Math. Acad. Sci. Hunga ), nos. 1 2, [13] Sákőzy, A. On Diffeence Sets of Sequences of Integes III, Acta Math. Acad. Sci. Hunga ), nos [14] Schmidt, W.M. Small Factional Pats of Polynomials, Regional Confeence Seies in Mathematics 32, AMS [15] Sinivasan, S. On a Result of Sakozy and Fustenbug, Nieuw. Ach. Wisk 4) ), no. 3, [16] Szemeédi, E. On Sets of Integes Containing No k Elements In Aithmetic Pogession, Acta. Aith ),