ON A QUESTION OF SÁRKOZY AND SÓS FOR BILINEAR FORMS

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1 Submitted exclusively to the London Mathematical Society doi:0/0000/ ON A QUESTION OF SÁRKOZY AND SÓS FOR BILINEAR FORMS JAVIER CILLERUELO and JUANJO RUÉ Abstract We prove that if k k, then there is no infinite sequence A of positive integers such that the representation function r(n #{(a, a ) : n = k a + k a, a, a A} is constant for n large enough This result completes previous work of Dirac and Moser for the special case k = and answers a question posed by Sárkozy and Sós Introduction Given an infinite sequence of positive integers A, the representation functions r(n) and R(n) are defined as the number of solutions of the equations respectively n = a + a, a, a A n = a + a, a, a A, a a, It is obvious that r(n) is odd when n = a, a A, and even otherwise So it is not possible for r(n) to be constant for n large enough This asymmetry disappears in R(n) but Dirac [] gave a beautiful argument that also proves that R(n) cannot be constant for n large enough For k, Moser [3] considered the representation function r(n #{(a, a ), n = a + ka, a, a A} Surprisingly he constructed a sequence A such that r(n for all n 0 Sárkozy and Sós [4] asked for which (k, k ) the representation function r k,k (n, A) #{(a, a ) : n = k a + k a, a, a A} can be constant for n large enough We answer this question by showing that the only cases with affirmative answer are those considered by Moser Theorem Let k, k, k k Then there is no infinite sequence of positive integers A such that is constant for n large enough r k,k (n, A #{(a, a ) : n = k a + k a, a, a A} The question posed in [4] actually concerns general linear forms k x + + k h x h, h The same arguments we use for the case h = can be extended to the general case when the k i s are pairwise coprimes but they are best illustrated in the situation presented in this paper The more general case with arbitrary coefficients requires a different approach and it is considered in a forthcoming paper 000 Mathematics Subject Classification B34 This work was developed during the Doccourse in Additive Combinatorics held in the Centre de Recerca Matemàtica from January to March 008 Both authors are extremely grateful for its hospitality

2 JAVIER CILLERUELO AND JUANJO RUÉ Translation of the problem into generating functions: Dirac s and Moser s arguments As Dirac and Moser did, we use the language of generating functions: to every set A of nonnegative integers, we write the formal power series f A (z) defined as f A (z) := f(z a A z a This formal power series is called the generating function associated to A For every set A, the corresponding generating function defines an analytic function around z = 0 This analytic function is a polynomial if A is finite and has a singularity at z = if A is infinite In fact, if A is infinite, then the Taylor expansion around z = 0 defined by the formal power series has radius of convergence r = We proceed to translate the general problem into the language of generating functions The fundamental equation we use is: f(z k )f(z k z k a+k a = r k,k (n, A)z n () Dirac s argument a,a A We observe that for the functions r(n) and R(n) we have the relation r(n R(n) δ(n), where δ(n if n = a for some a A and 0 otherwise By () we obtain f (z r(n)z n = z a, which can be written in the form f (z) + f(z R(n)z n a A R(n)z n Dirac proved that R(n) cannot be a constant c for n n 0 with an easy but clever argument: suppose that R(n c for n n 0 Then f (z) + f(z Q(z) + c zn 0+ z = P (z) z, where P (z) is a polynomial of finite degree with P () 0 Then we obtain a contradiction by taking the limit for z in both sides of the equation: the left hand side of the equality diverges, but the right hand side has a finite limit Moser s argument Moser [3] studied the case k =, k He wondered if for these cases there exists an infinite sequence of nonnegative integers such that r,k (n, A for all n 0 If this is the case, equation () implies f(z)f(z k z n = z n 0 If we make the change of variables z := z k we get f(z k )f(z k z k

3 ON A QUESTION OF SÁRKOZY AND SÓS FOR BILINEAR FORMS 3 Dividing the initial equation by this one we obtain f(z zk z f(zk ( + z + z + + z k )f(z k ) By iterating we get the relation f(z ( + z (k ) j + z (k ) j + + z (k )(k ) j ) j=0 This product defines an analytic function at the origin, which can be written using its series expansion around z = 0 Moreover, by the unique k adic representation of an integer, the Taylor s coefficients of f(z) are either 0 or So this function f(z) defines a set A which satisfies our assumptions More precisely, the set A is the set of all nonnegative integers such that all its digits in its k adic expansion are smaller than k 3 The general case We want to know if, given k, k, k k, there exists an infinite sequence of non negative integers A and a value (say n 0 ) such that r k,k (n, A) is a positive constant c for n n 0 Since the cases k = have been considered by Dirac (k = with negative answer in both ordered and unordered representations) and Moser (k with affirmative answer) we may assume that k k We may also assume that gcd(k, k, since otherwise we have r k,k (n, A 0 for all n 0 (mod gcd(k, k )) If such a sequence A exists, then by () we have n 0 f(z k )f(z k a n z n + cz n = Q(z) + czn0+ z = P (z) z, n=n 0 where Q(z), P (z) are polynomials in Z[z] with P () 0 This last relation is equivalent to the condition c 0 For convenience, we take the square of the previous equation By writing F (z f (z), we want to show that there is no function F (z), analytic in the disc z <, such that F (z k )F (z k P (z) ( z) Theorem will be a consequence of a more general theorem: Theorem For any integers k, k, k < k with gcd(k, k and any polynomial P (z) Z[z] with P () 0, there is no function F (z), analytic in the disc z <, satisfying F (z k )F (z k P (z) ( z) () In what follows we concentrate in the proof of Theorem 3 Algebraic preliminaries and notation used In our work we will use cyclotomic polynomials Recall that the cyclotomic polynomial of order n is defined by the relation Φ n (z ξ φ n (z ξ) Z[z],

4 4 JAVIER CILLERUELO AND JUANJO RUÉ where φ n denotes the set of primitive roots of order n, φ n = {ξ C : ξ k = if and only if k 0 (mod n)} Many properties of these polynomials are well known The most important one for our present purposes is that they are irreducible over Z[z] As a consequence, it is known that if a polynomial P (z) Z[z] vanishes at a primitive root of order n then there exists a positive integer s such that P (z Φ s n(z)q(z) where Q(z) Z[z] and Q(ξ) 0 for all primitive roots ξ of order n Given k, k, k < k we write ξ j,j to denote a primitive root of order k j kj, 0 j, j, and we let s j,j denote the nonnegative integer such that P (z Φ s j,j (z)p k j k j j,j (z) with P j,j (z) Z[z] and P j,j (ξ) 0 for all ξ φ j k k j It is not true in general that for an analytic function F (z) with integer coefficients there exists r j,j such that F (z Φ r j,j (z)f k j k j j,j (z) with lim z ξ F j,j (z) 0, for any ξ φ j,j However we will prove that there is such a factorization if F (z) satisfies () 4 Proof of theorem The proof of Theorem is a consequence of the two propositions below: Proposition 4 With the notation used in Section 3, r j,0 and r 0,j are well defined for any j 0 and they verify the following recurrence relations r j+,0 = s j+,0 r j,0 r 0,j+ = s 0,j+ r 0,j for any j 0, and initial condition r 0,0 = (horizontal recurrence) (vertical recurrence) Proposition 4 With the notation used in Section 3, r j,j are well defined for any j, j 0 and these numbers verify the recurrence relation for j, j r j,j + r j,j = s j,j (diagonal recurrence) Before proving the above propositions we show how Theorem can be deduced from them Since r 0,0 = is an odd number, from the two propositions above we see that all values are odd numbers As P (z) is a polynomial, it is clear that s j,j = 0 when j + j j for a suitable j Using proposition 4 and for this value of j, we obtain the relations: r j,j r j+,0 = r j,0 r 0,j+ = r 0,j and From proposition 4 we get that r j+,0 = r j, = r j, =, so r j+,0 = ( ) j+ r 0,j+ r j,0 = ( ) j r 0,j and

5 ON A QUESTION OF SÁRKOZY AND SÓS FOR BILINEAR FORMS 5 The above relations are possible if and only if r j,0 = r j+,0 = r 0,j = r 0,j+ = 0, giving a contradiction This proves Theorem and Theorem We next prove the two propositions In what follows all the limits considered are taken along a radius Proof of proposition 4 We deal only with the horizontal recurrence The proof for the vertical recurrence is similar We write r j = r j,0, s j = s j,0 and ξ j = ξ j,0 for simplicity We shall prove by induction that all the r j s are well defined By writing F (z F 0 (z)/(z ) in () we have F 0 (z k )F 0 (z k P (z)( + z + z k )( + z + z k ) For z the left hand side of the above equation clearly goes to F 0 () and the right hand side is neither 0 nor Since Φ (z z we obtain that r 0 = Assume now that r j is well defined Then where F j (ξ) 0, for all ξ φ k j Now we write F (z Φ rj (z)f k j j (z), (4) F (z Φ sj+ rj (z)f k j+ j+ (z) (4) In what follows, we prove that lim z ξj+ F j+ (z) {0, } for any ξ φ k j+ This will show that r j+ exists and that r j+ = s j+ r j We use (4) in F (z k ) and (4) in F (z k ) to write the equation () in the form Pj+ (z)φsj+ (z) F j+ (z k k ( z) Φ sj+ rj (z k j+ k )F j (z k )Φ rj (z k j k ) By making the substitution z = ξ j+ ω we have F j+ (ξ k j+ ωk P j+ (ξ j+ω) ( ξ j+ ω) Φ sj+ rj k j+ F j (ξ k j+ ωk ) Φ s j+ (ξ k j+ j+ ω) (ξ k j+ ωk )Φ r j k j (ξ k j+ ωk ) (43) We let ω and we observe that all the primitive roots of order k j+ can be written in the form ξ k j+ for a suitable ξ j+ since gcd(k, k To conclude, we show that the limit is neither 0 nor In fact, this is the case for each of the three factors on the right hand side of (43) It is clear that P j+ (ξ j+ ) 0 by definition Since ξ k j+ φ k, j we use the induction hypothesis to conclude that the limit of the second factor is neither 0 nor To study the third factor when ω it suffices to analyze the factors in the cyclotomic polynomials which vanish at ω = It should be noticed that ξ k j+ φ k j and ξ k j+ φ k j+ The contribution of these factors is (ξ j+ ω ξ j+ ) sj+ = (ξ k j+ ωk ξ k j+ )sj+ rj (ξ k j+ wk ξ k j+ )rj ξ sj+ j+ ξ k (s j+ r j ) j+ ξ k r j j+ (ω ) sj+ (ω k ) s j+ r j (ω k ) r j

6 6 JAVIER CILLERUELO AND JUANJO RUÉ which tends to ξ sj+( k)+rj(k k) j+ k sj+ rj k rj as ω The relevant fact about this limit is that it is neither zero nor infinity The next proof is quite similar to the previous one: Proof of proposition 4 We will prove, for each diagonal j + j = j, that all r j,j, 0 j j are well defined For each j we will do it by induction on j This is true for r j,0 by Proposition 4 Suppose that r j,j is well defined Thus F (z Φ r j,j j,j (z)f j,j (z) (44) where F j,j (ξ) {0, } for all ξ φ k j k j We prove that r j,j + is also well defined and also that r j,j + = s j,j + r j,j In order to do this we write F (z Φ sj,j + rj,j j,j + (z)f j,j +(z) What we have to prove is that lim z ξ F j,j +(z) {0, } for any ξ φ j,j + We use (44) in F (z k ) and (4) in F (z k ) to write the equation () in the form F j,j +(z k Pj,j + (z)φs j,j + j (z),j + ( z) Φ s j,j + r j,j j,j + (z k )Φ r j,j j,j (z k )F j,j (z k ) Now we make the substitution z = ξ j,j +ω for some arbitrary ξ j,j + φ j k obtain the expression F j,j +(ξ k j,j + ωk P j,j + (ξ j,j +ω) ( ξ j,j +ω) Φ s j,j + r j,j j,j + (ξ k Φ sj,j + j,j + (ξ j,j +ω) j,j + ωk )Φ r j,j F j,j (ξ k j,j + ωk ) j,j (ξ k j,j + ωk ) k j + Now we let ω We observe that all the primitive roots of order k j k j can be written in the form ξ k j for a suitable ξ,j + j,j + φ j k k j + As in the previous proposition, we show that the limit is neither 0 nor, showing that this is the case for every factor in the previous equation For the first factor, it is clear that P j,j +(ξ j,j +) {0, } by definition Since ξ k j,j + φ k j k j +, we use induction hypothesis to conclude that the limit in the second factor does not belong to {0, } Finally, to study the third factor when ω we look at the cyclotomic polynomials which vanish at ω = It should be noticed that ξ k j,j + φ k j k j + and ξ k j,j + φ k j k j The contribution of these factors is (ξ j,j +ω ξ j,j +) sj,j + (ξ k j,j + ωk ξ k j,j + )s j,j + r j,j (ξ k j,j + ωk ξ k j,j + )r j,j which tends to a number which is neither zero nor infinity This concludes the proof We References G A Dirac, Note on a Problem in Additive Number Theory, J London Math Soc 6 (95) pp 3-33

7 ON A QUESTION OF SÁRKOZY AND SÓS FOR BILINEAR FORMS 7 P Erdős, P Turán, On a problem of Sidon in Additive Number Theory, and on some related problems, J London Math Soc 6 (94) pp -5 3 L Moser, An Application of Generating Series, Mathematics Magazine () 35 (96) A Sárkozy, V T Sós, On additive representation functions,the mathematics of Paul Erdős I (eds P Erdős, R L Graham and J Nesetril), Algorithms Combin 3 (Springer, Berlin, 997), pp 9 50 Javier Cilleruelo Departamento de Matemáticas Universidad Autónoma de Madrid 8049 Madrid, Spain franciscojaviercilleruelo@uames Juanjo Rué Departament de Matemàtica Aplicada II Universitat Politècnica de Catalunya Jordi Girona Barcelona, Spain juanjoserue@upcedu